1
Function
1
FUNCTION
Exercise-1 : Single Choice Problems
1. f( x) = log 2 (2 - 2 log 2 (16 sin 2 x + 1))
0 £ log 2 (16 sin 2 x + 1) £ log 2 17
Þ
Þ 2 - 2 log 2 17 £ 2 - 2 log 2 (16 sin 2 x + 1) £ 2
0 < 2 - 2 log 2 (16 sin 2 x + 1) £ 2 Þ f( x) £ 1
2. For any b Î R e| x -b| is
e|x–b|
|e|x–b|–3|
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|
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1
b
min
–2
|e|x -b| - a| has four distinct solutions a > 3 so a Î (3 , ¥)
3. Domain = [-1, 1] and both are increasing functions.
\ x = -1, we get minimum value & x = 1, we get maximum value.
é p p p pù é p pù
êë - 4 - 4 , 4 + 4 úû = êë - 2 , 2 úû
2
2
2
2
2
4. æç 2 2 x + 2 y - 2 2 x + 2 y ö÷ = 1 - 2 2 x + 2 y + 2 x + 2 y +1 ³ 0
è
ø
2
Þ
2
1ö
1ö
æ
æ
çx + ÷ +çy + ÷ £0
2ø
2ø
è
è
2
Solution of Advanced Problems in Mathematics for JEE
y
1
5.
O
x
(0,–a)
æ 5
ö
2
÷
7. sec -1 ç - +
ç 2 2( x 2 + 2) ÷
è
ø
x2 + 2 ³ 2
æ 5
ö
1
÷
= sec -1 ç - +
ç 2 ( x 2 + 2) ÷
è
ø
1ö
æ 1
£ ÷
ç 2
2
è x +2
ø
£ sec -1 ( -2) = p - sec -1 (2)
1
5 1
æ 5
ö
£ - + = -2 ÷
ç- + 2
2 2
è 2 x +2
ø
=
2p
3
8. f ¢( x) = x 2 + ax + b is injective if D £ 0
a 2 - 4b £ 0
If
Number of pair = 5
a = 1, b = 1, 2 , 3 , 4 , 5
Number of pair = 5
a = 2 , b = 1, 2 , 3 , 4 , 5
Number of pair = 3
a = 3, b = 3, 4, 5
Number of pair = 2
a = 4, b = 4, 5
a =5
b has no value
9. f( x) = log x [ x] Þ f( x) Î [0 , 1]
g( x) =|sin x| + |cos x|
Þ
g( x) Î [1, 2 ]
10. f( x) = 2 x 3 - 3 x 2 + 6
f ¢( x) = 6 x 2 - 6 x ³ 0
Þ x Î [1, ¥)
and f( x) Î [5, ¥)
11. 0 £ { x} < 1
{ x}({ x} - 1)({ x} + 2) ³ 0
Þ { x} = 0 Þ x Î z
14. 1 + sin 2 x Î [1, 2]
1
é1 ù
Î ê , 1ú
1 + sin x ë 2 û
1
æ
ö ép pù
sin -1 ç
÷Îê , ú
2
è 1 + sin x ø ë 6 2 û
2
3
Function
Kp é p p ù
Îê , ú
6
ë6 2 û
K Î[1, 3]
15. f( x - y) = f( x) f( y) - f( a - x) f( a + y)
Put x = y = 0
f(0) = [ f(0)]2 - f( a) f( a)
Þ
f( a) = 0
Put x = a and y = x
[Q f(0) = 1]
f( a - x) = f( a) f( x) - f(0) f( a + x)
- f ( a - x) = f ( a + x) Þ f (2 a - x) = - f ( x)
Þ
18. f( x) = 4 x - x 2 = y
x2 - 4x + y = 0
f -1 ( x) = 2 - 4 - x
19. [5 sin x] + [cos x] = -6
Þ -1 £ cos x < 0 and -5 £ 5 sin x < -4
4
-1 £ sin x < 5
20. f( x) = ax + cos x
f ¢( x) = a - sin x
if f( x) is invertible, then
f ¢( x) ³ 0 or f ¢( x) £ 0
Þ
a ³ 1 or a £ -1
xù é
xù
xù
é
é
21. f( x) = [1 + sin x] + ê2 + sin ú + ê3 + sin ú +¼ + ên + sin ú
2û ë
3û
nû
ë
ë
xù é
xù
xù
é
é
= (1 + 2 + 3 +¼ + n) + [sin x] + êsin ú + êsin ú +¼ + êsin ú
2
3
n
ë
û ë
û
ë
û
22. y =
x 2 + ax + 1
x2 + x + 1
( y - 1) x 2 + ( y - a) x + ( y - 1) = 0
D³0
( y - a) 2 - 4( y - 1) 2 ³ 0
-3 y 2 + y(8 - 2 a) + a 2 - 4 ³ 0 " y Î R
Not possible
23. f( x) = [ x] + [- x]
4
Solution of Advanced Problems in Mathematics for JEE
f( x) = ìí 0 x Î I
î-1 x Ï I
g( x) = { x}
h( x) = f[ g( x)] = f({ x})
f ( 0)
h( x) = ìí
î f({ x})
xÎI
xÏI
Hence, the option (b).
é x ù é 15 ù
24. f( x) = ê ú ê - ú
ë 15 û ë x û
Þ
{ x} = 0
xÎI
{ x} = { x} x Ï I
h( x) = ìí 0 x Î I
î-1 x Ï I
x Î(0 , 90)
0 £ x < 15
f( x) = 0
15 £ x < 30
f( x) = -1
30 £ x < 45
f( x) = -2
45 £ x < 60
f( x) = -3
60 £ x < 75
f( x) = -4
75 £ x < 90
f( x) = -5
Total integers in range f( x) = {0 , - 1, - 2 , - 3 , - 4 , - 5}
1
25. g( x) =
f(| x|)
g( x) Þ even functions Þ symmetric about y-axis
Þ
x ® ¥ f( x) ® 0
at x = x 1
f( x) = 0 Þ g( x 1 ) ® ¥
26. Homogeneous function Þ f(tx , ty) = t n f( x , y)
2x + 3 x £ 1
27. f( x) = é 2
êë a x + 1 x > 1
For x £ 1
f( x) £ 5
So for range of f( x) to be R.
Þ
a 2 + 1 £ 5 and a ¹ 0
Þ a Î [-2 , 2]
Hence, a = {-2 , - 1, 1, 2}
28. log 1/ 3 (log 4 ( x - 5)) > 0
0 < log 4 ( x - 5) < 1
1< x - 5 < 4
6< x<9
æ
ö
4
29. f( x) = log 2 ç
÷;-2 £ x £ 2
è 2+x + 2-xø
5
Function
2+x + 2-x =y
4 + 2 4 - x2 = y2
y Î[2 , 2 2 ]
é
4
4ù
Range f( x) = êlog 2
, log 2 ú
2û
2 2
ë
é1 ù
f( x) lies between ê , 1ú
ë2 û
30. | x 2 + 5 x| + | x - x 2| =|6 x| Þ | x 2 + 5 x| + | x - x 2| =|( x 2 + 5 x) + ( x - x 2 )|
|a| + |b| =|a + b| Þ ab ³ 0
( x 2 + 5 x)( x - x 2 ) ³ 0
x( x + 5) × x( x - 1) £ 0 Þ -5 £ x £ 1
æ 1ö
æ 1ö
31. f( x) + f ç ÷ = f( x) f ç ÷
x
è ø
è xø
Þ
f( x) = 1 ± x n
f(2) = 33 Þ n = 5
Hence,
f( x) = 1 + x 5
Here,
f ( x ) + f ( - x ) ¹ 0.
Hence not an odd function.
sin x + sin 7 x
2 sin 4 x cos 3 x
32. g( x) =
+ |sin x| =
+ |sin x|
cos x + cos 7 x
2 cos 4 x cos 3 x
= tan 4 x + |sin x|
g( x) period = p
éx -1
x = odd
ê
33. f( x) = ê 2
x
x = even
ê ë 2
f( x) : N ® Z
Let x = odd = (2n + 1); n > 0
2n + 1 - 1
f( x) =
= n Þ +ve integer
2
Let x = even = 2m ; m > 0
2m
f( x) = = -m Þ –ve integer
2
Þ Range = codomains Þ onto and clearly f( x) is one-one function.
Hence, bijective.
34. y =
2 x + 1 - 2 1- x
2 x + 2 -x
=
2 2 x +1 - 2
2 2x + 1
6
Solution of Advanced Problems in Mathematics for JEE
y(2 2 x + 1) = (2 2 x - 1) 2
2 2x × y + y = 2 2x × 2 - 2
2 2 x ( y - 2) = - ( 2 + y )
2 2x =
( y + 2)
(2 - y)
2 x = log 2
x=
(2 + y)
1
log 2
2
(2 - y)
f -1 ( x) =
35.
(2 + y)
(2 - y)
1
(2 + x)
log 2
2
(2 - x)
(Q x > 0)
| y| = x
Þ | y| = x
Þ
2
y = x2
y = -x
y³0
2
y<0
36. f( x) = log [ x] (9 - x 2 )
Domains =
ü
9 - x2 > 0
Þ x Î[2 , 3) Þ f( x) = log 2 (9 - x 2 )
[ x] > 0 and [ x] ¹ 1ýþ
Range = ( -¥, log 2 5]
37. Gives
e x + e f( x) = e
e f( x) = e - e x
Þ
f( x) = log e ( e - e x )
Domain e - e x > 0
x < 1 Þ x Î ( -¥, 1)
Range ( -¥, 1)
38. Gives
y + | y| = x + | x|
If
x > 0, y > 0 Þ 2 y = 2 x Þ y = x
x < 0, y > 0 Þ 2 y = 0 Þ y = 0
x > 0, y < 0 Þ 0 = 2 x Þ x = 0
x < 0, y < 0 Þ 0 = 0
Þ whole region of III quadrant.
For person to be safe there should not be point common to the given curves and the voltage
field graph. Only y = m +| x| does not have any point of intersection with the curve.
39. Gives
Þ
| f( x) + 6 - x 2| =| f( x)| + |4 - x 2| + 2
| f( x) + 2 + ( 4 - x 2 )| =| f( x)| + |4 - x 2| + 2
7
Function
Since
|a + b + c| =|a| + |b| + |c|
If a ³ 0 , b ³ 0 , c ³ 0 or a £ 0 , b £ 0 , c £ 0
Þ
f( x) ³ 0 and 4 - x 2 ³ 0 Þ -2 £ x £ 2 and f( x) ³ 0
40. f( x) = cos px + sin x
æ 2p 2p ö
Period : L.C.M. of çç
÷÷
,
è p 1 ø
For period to exist p should be a rational number.
41. y = f( e x ) + f(ln| x|)
Domain f( x) = (0 , 1)
Þ
and
0 < ex < 1
Þ
x<0
0 < ln| x|< 1
Þ 1 < | x| < e Þ x Î ( -e, - 1) È (1, e)
Taking intersection x Î ( -e, - 1)
42. Givens
f(1) = 2 , f(2) = 3 , f(3) = 4 , f( 4) = 1, g(1) = 3 and f [ g( x)] = g [ f( x)]
at x = 1 f [ g(1)] = g [ f(1)]
Þ f(3) = g(2) Þ g(2) = 4
at x = 2 f [ g(2)] = g [ f(2)]
Þ f( 4) = g(3) Þ g(3) = 1
at x = 3 f [ g(3)] = g [ f(3)]
Þ f(1) = g( 4) Þ g( 4) = 2
43. Gives
[ y + [ y]] = 2 cos x Þ [ y] + [ y] = 2 cos x Þ 2 [ y] = 2 cos x; [ y] = cos x
1
where
y = [sin x + [sin x + [sin x]]]
3
1
y = [sin x + [sin x] + [sin x]]
3
1
y = (3 [sin x])
3
y = [sin x]
From eqn. (1) & (2),
[sin x] = cos x
Þ
cos x = 0 , 1, - 1
Hence, no solution.
1
é 1
- ù
êex - e x ú
x 2n
44. f( x) =
x ¹ 0 and f(0) = 1
ê
1 ú
( x 2n sgn x) 2n +1 ê 1
- ú
ëe x + e x û
1
é 1
- ù
x
ê
(x )
e -e x ú
f( x) =
; x>0
2n 2n +1 ê 1
1 ú
(x )
ê x
ú
ëe + e x û
2n
when
…(1)
…(2)
…(1)
…(2)
8
Solution of Advanced Problems in Mathematics for JEE
1
é 1
- ù
x
êe - e x ú
x
f( x) =
; x<0
ê
1 ú
- ( x 2n ) 2n +1 ê 1
- ú
ëe x + e x û
2n
Clearly, f( x) = f( - x). Hence, f( x) is even function.
45. f(n) = 2( f(1) + f(2) ¼¼ + f(n - 1))
f(2) = 2 f(1)
é f ( 2)
ù
f(3) = 2 [ f(1) + f(2)] = 2 ê
+ f ( 2) ú = 3 f ( 2)
2
ë
û
é f ( 3)
ù
f( 4) = 2 [ f(1) + f(2) + f(3)] = 2 ê
+ f ( 3) ú = 3 f ( 3) = 3 2 f ( 2)
ë 2
û
M
M
m
å f( r) = f(1) + f(2)¼¼ + f(m) = f(1) + f(2) + 3 f(2)¼¼ +
r =1
= f(1) + f(2)[1 + 3 + 3 2 ¼¼ 3 m -2 ]
= f(1) + 2 ×
46. Gives
f( x) =
f( f( x)) =
f( f( f( x))) =
(3 m -1 - 1)
= 3 m -1
(3 - 1)
x
1+ x2
x
1 + 2x2
x
1 + 3x2
M
M
fofo¼¼ fof( x) =
1442443
n times
x
1 + nx 2
=
x
æ n ö
1 + ç 1÷ x 2
ç
÷
è r =1 ø
å
47. f( x) = 2 x + |cos x|
Range f( x) = R = codomain Þ onto.
Clearly, f( x) is increasing function Þ one-one function.
48. Gives f( x) = x 3 + x 2 + 3 x + sin x
Since, f( x) is continuous function.
and f( x) = ¥ as x ® ¥
f( x) = -¥ as x ® -¥
9
Function
Range f( x) = R = codomains Þ onto function
2
1ö
8
æ
and f ¢( x) = 3 x 2 + 2 x + 3 + cos x = 3 ç x + ÷ + + cos x Þ f ¢( x) > 0
3ø
3
è
Hence, f( x) is one-one.
49. f( x) = { x} + { x + 1} +¼ { x + 99}
Since
where I = integer
{ x} = { x + I }
f( x) = { x} + { x} ¼ { x}
1442443
100 times
f( x) = 100{ x} Þ f( 2 ) = 100{ 2 } = 100 ´ 0.414 = 41.4
[ f( 2 )] = 41
50. |cot x + cosec x| =|cot x| + |cosec x|; x Î [0 , 2p]
Þ
cot x ³ 0 and cosec x ³ 0 Þ 1st quadrant
or
cot x £ 0 and cosec x £ 0 Þ 4 th quadrant
Hence,
æ p ù é3p
ö
x Î ç 0, ú È ê
, 2p ÷
è 2û ë 2
ø
51. If f( 4 + x) = f( 4 - x)
Þ f( x) is symmetric about x = 4.
Roots of f( x) = 0 are of the form
4 - a , 4 + a , 4 - b, 4 + b, 4 - g , 4 + g , 4 - d, 4 + d
52. f( x) + x - 6 = ( x - 1)( x - 2)( x - 3)( x - 4)( x - 5)
Þ
f(6) = 120
53. f( x) = x - 2 + 4 - x
f(x)
2
2
x
1
2
3
4
éxù é x ù
54. ê ú = ê ú
ë 9 û ë 11û
x Î [1, 9) È [11, 18) È [22 , 27) È [33 , 36) È [44 , 45)
55. log é
1 ù (2 x
êx+ 2ú
ë
û
2
+ x - 1)
1ù
1ù
é
é
2
êë x + 2 úû > 0 , êë x + 2 úû ¹ 1 & 2 x + x - 1 > 0
10
Solution of Advanced Problems in Mathematics for JEE
1
³ 2 & (2 x - 1)( x + 1) > 0
2
3
æ1 ö
x ³ & x( -¥, - 1) È ç , ¥ ÷
2
è2
ø
Þ
x+
Þ
é3
ö
x Î ê , ¥÷
ë2
ø
56. [ x 2 ] + [ x] - 2 = 0
Let [ x] = t
Þ
t2 + t -2 =0
Þ (t + 2)(t - 1) = 0
Þ
t = -2 or t = 1
Þ [ x] = -2 or [ x] = 1
Þ
x Î [-2 , - 1) È [1, 2)
58. f( x) is many one function.
59. f( f( x)) = 2 + f( x)
f( x) ³ 0
= 2 - f( x)
f( x) < 0
f( f( x)) = 4 + x
x³0
= 4 - ( x)
x<0
2
60. f ¢( x) =
7(3 x - 2 x + 3)
(3 + 3 x - 4 x 2 ) 2
> 0 Þ f( x) ­
y
61.
O
x
y
62.
–2 a –1
[a ] = -2
[b] = 3
3b4
x
11
Function
63. f( x) = sin(log 7 (cos(sin x)))
cos(sin x) £ 1 Þ cos(sin x) = 1 Þ f( x) = 0
64. - 3 £|[ x]| £ 2 Þ - 2 £ [ x] £ 2 Þ -2 £ x < 3
p
65. f( x) = + cot -1 {- x}
2
p
p
0 £ {- x} < 1 Þ < cot -1 {- x} £
4
2
66. f( f( x)) = x
3x + 5 2x2 + 5
=
2x - 3 2x - 3
1
1 ö
1
1
æ
öæ
2
67. f( x) = ç x + + 1÷ ç x 2 +
÷ ; x + 2 ³ 2 ; x + + 1 ³ 3 Þ f( x) ³ 6
2
x
x
è
øè
x ø
x
f2008 ( x) + f2009 ( x) = x + f( x) = x +
3
2
68. f( x) = e x -3 x -9 x + 2
3
2
f ¢( x) = e( x -3 x -9 x + 2) 3 ( x - 3)( x + 1)
Þ
f( x) is many one.
at x = -1, f( x) = e7
at x ® -¥, f( x) ® 0
Range of f( x) is (0 , e7 ].
69. D f : ( -2 , 1)
æ 4 - x2 ö
÷<¥
-¥ < log ç
ç 1- x ÷
è
ø
æ æ 4 - x 2 öö
÷÷ £ 1
-1 £ sin ç log ç
ç ç 1 - x ÷÷
øø
è è
70. f ¢( x) ³ 0 " x Î R Þ 3 x 2 + 2 ( a + 2) x + 3 a ³ 0 " x Î R
Þ D£0
Þ 4 ( a + 2) 2 - 4 × 9 a £ 0
Þ a 2 - 5a + 4 £ 0 Þ ( a - 1)( a - 4) £ 0
Þ a Î[1, 4]
2
71. Min. value of 3 x + bx + c = 0
Þ
D =0
72. f( x) = sin -1 x - cos -1 x = 2 sin -1 x -
p
2
12
Solution of Advanced Problems in Mathematics for JEE
3p/2
p
p/2
x
–1
1
73. is one-one when
2 3 = ln 1 + b 2 - 3 b + 10
Þ
b2 - 3b + 2 = 0
Þ
b = 1, 2
8
x=1
–1
80. We have, [ x]2 - 7 [ x] + 10 < 0
Þ
Þ
Þ
Þ
([ x] - 5)([ x] - 2) < 0
2 < [ x] < 5
[ x] = 3 or 4
and
4[ y]2 - 16 [ y] + 7 < 0
Þ
(2 [ y] - 7)(2 [ y] - 1) < 0
1
7
< [ y] <
2
2
[ y] = 1 or 2 or 3
x Î[3 , 5)
Þ
Þ
y Î[1, 4)
Therefore,
x + y Î[4 , 9)
[ x + y ] Î { 4 , 5, 6 , 7 , 8}
Hence, [ x + y] cannot be 9.
81. f : R ® R
ì e x - e-x
ï x
-x
ï
f( x) = í e- x + e - x
ïe - e
ïî e x + e - x
f( x) =
e| x| - e - x
e x + e-x
y
1
if x ³ 0
if x < 0
Many one into function.
O
x
13
Function
82. f( x) such f(1 - x) + 2 f( x) = 3 x " x Î R
æ1
ö
x ® ç + x÷
2
è
ø
æ1
ö
æ1
ö
æ1
ö
f ç - x÷ + 2 f ç + x÷ =3ç + x÷
è2
ø
è2
ø
è2
ø
æ1
ö
x ® ç - x÷
2
è
ø
1
æ
ö
æ1
ö
æ1
ö
f ç + x÷ + 2 f ç - x÷ =3ç - x÷
è2
ø
è2
ø
è2
ø
(1) + (2)
æ æ1
ö
æ1
öö
3 ç f ç + x÷ + f ç - x÷÷ = 3 ;
2
2
ø
è
øø
è è
æ1
ö
æ1
ö
f ç - x÷ =1- f ç + x÷
2
2
è
ø
è
ø
æ1
ö 3
æ1
ö 1
1 + f ç + x÷ = + 3x ; f ç + x÷ = + 3x
2
2
2
è
ø
è
ø 2
1
1 3
x = - Þ f(0) = - = -1
2
2 2
83. f : [0 , 5] ® [0 , 5]
f( x) = ax 2 + bx + c
0
5
a , b, c Î R , abc ¹ 0
or
0
5
c¹0
ax 2 + bx + c = 0 (a )
æ 1ö
cx 2 + bx + a = 0 ç ÷
èaø
æ 1ö
So, roots are ç a, ÷ .
è 5ø
25a + 5b + c = 0
f(5) = 0
c
= 5 ´b
a
1
b=
a
84. f( x) = x 2 + lx + m cos x
f( x) = x
85. f( k) = odd
f( k + 1) = even
k = 1, 2 , 3
…(1)
…(2)
14
Solution of Advanced Problems in Mathematics for JEE
f(1) Þ odd
f(2) Þ even
f(3) Þ odd
f( 4) Þ even
ü
ï
ý Hence , 4 functions.
ï
þ
f(1) Þ even
f(2) Þ odd
f(3) Þ even
f( 4) Þ odd
ü
ï
ý Hence , 4 functions.
ï
þ
f(1) Þ odd
f(2) Þ even
f(3) Þ even
f( 4) Þ odd
ü
ï
ý Hence , 4 functions.
ï
þ
86. y = tan(sin x).
Here function is continuous and differentiable and y max = tan(1); y min = - tan 1.
87. f( x) =
y =2 +
2x
x -1
y
(0,2)
(1,2)
2
( x - 1)
(1,0)
x
( y - 2)( x - 1) = 2
88. R f = [-2 , 4]
R g = [-1, 2]
89. f( x) = ( x 4 + 1) +
1
2
x + x +1
y
90. 0 £ f( x) £ 1
0 £ 7 f( x) £ 7
1
-1 £ sin(7 f( x)) £ 1
91. ln|ln| x|| ³ 0
O
Ç
| x|2 - 7| x| + 10 £ 0
|ln| x|| ³ 1
(| x| - 2)(| x| - 5) £ 0
ln| x|Î ( -¥, - 1) È [1, ¥)
2 £ | x| £ 5
æ 1ù
| x|Î ç 0 , ú È [e, ¥)
è eû
x Î ( -5, - 2] È [2 , 5]
é 1 ö æ 1ù
x Î ( -¥, - e] È ê - , 0 ÷ È ç 0 , ú È [e, ¥)
ë e ø è eû
x
1
2
3
15
Function
2
ææ
5ö
3ö
92. log [ x] + 3{ x} ç ç [ x] - ÷ + ÷ ³ 0 Þ [ x] + 3 { x} > 1
çè
2ø
4÷
è
ø
93. x - 3 = X
| X | + |Y | = 5
y -1=Y
x + y -4=X +Y
|X + Y|= 5
number of pairs of ( x , y) = 12
(0,
5)
(5,
(–5,
0)
0)
|x+y|
|x|
(0,
+
|y|
=
=
5
5
–5)
f(1) = 1
95.
f(2) = 3
f(4) = 2
f(3) = 4
f(2) = 4
f(3) = 2
f(4) = 3
f(3) = 3
f(4) = 2
96. x 2 - x ¹ 0 Þ x ¹ 0 , 1
97. Total one-one function – (at least one get right place) + (at least two get right place)
– (at least three get right place) + (at least four get right place)
= 6 C 4 ´ 4 ! - 4 C 1 ´ 5 C 3 ´ 3 ! + 4 C 2 ´ 4 C 2 ´ 2 ! - 4 C 3 ´ 3 C 1 + 4 C 4 = 181
98. f( x) = x 2 - 2 x - 3
g( x) = f -1 ( x) = 1 +
x+4
x ³ -4
f( x) = g( x) = f -1 ( x) Þ f( x) = x
Þ
x2 - 3x - 3 = 0
Þx=
3 + 21
2
16
Solution of Advanced Problems in Mathematics for JEE
Exercise-2 : One or More than One Answer is/are Correct
1. f( -4) = f( 4) = 40
f( -13) = f(13) = f(3) = 19
f( -11) = f(11) = f(1) = 2
y
2
2.
1
0
x
1
2
3
4
æ 1 - tan 2 ( x 2) ö
÷ is defined when
3. f( x) = cos -1 ç
ç 1 + tan 2 ( x 2) ÷
è
ø
x
p
¹ (2n - 1)
2
2
Þ
x ¹ (2n - 1)p
Domain = R - {(2n - 1)p : n Î I }
\
Range = [0 , p)
f( x) = cos -1 (cos x)
f( x) is even function.
when x Î( p , 2 p), then f( x) = 2p - x is differentiable.
2
4.
–e2
–1
1
e2
0 < |k - 1| - 3 < 2
Þ
k Î ( -4 , - 2) È ( 4 , 6)
5. (a) D f Î R
(b) D f Î R
17
Function
(c) f( x) = 2 cos 2 x + cos x +
1
8
Df ÎR
(d) ln(1 + | x|) ³ 0
ì(2n + 1) p ü
Df Îí
ý
2
î
þ
3
9
æ ö
6. f ç ÷ =
è2ø 4
æ æ 3 öö 3
f ç f ç ÷÷ =
è è 2 øø 2
æ æ æ 3 ö öù 9
f çç f ç f ç ÷ ÷ ú =
è è è 2 ø øû 4
æ 5ö
f ç ÷ =2
è2ø
æ æ 5 öö
f ç f ç ÷÷ = 1
è è 2 øø
æ æ æ 5 ööö
f çç f ç f ç ÷ ÷ ÷÷ = 1
è è è 2 øøø
8. f -1 ( f( x)) = f( f -1 ( x)) = x
if f( f -1 ( x)) = f -1 ( x) Þ x = f -1 ( x)
if f( f -1 ( x)) = f -1 ( x) Þ f( f -1 ( f( x))) = f -1 ( f( x)) Þ f( x) = f -1 ( f( x)) = x
æx
3 × 1 - x 2 ö÷
9. f( x) = cos -1 x + cos -1 ç +
ç2
÷
2
è
ø
Let x = cos q
æ1
ö
3
f( x) = cos -1 (cos q) + cos -1 çç cos q +
sin q ÷÷
2
è2
ø
p öö
æ
æ
= cos -1 (cos q) + cos -1 ç cos ç q - ÷ ÷
3 øø
è
è
=
p
3
= 2q -
0£q£
p
3
p
3
p
<q£ p
3
18
Solution of Advanced Problems in Mathematics for JEE
10. f( x) = cos -1 ( - {- x})
ép ö
- {- x} Î ( -1, 0] Þ cos -1 ( - {- x}) Î ê , p ÷
ë2 ø
12. h( x) = [ln x - 1] + [1 - ln x]
-1, ln x - 1 Ï I
Þ h( x) = é
êë 0 , ln x - 1 Î I
14. f( x) =
1
, f( x) is periodic & constant function.
2
16. f( x) =
e-x
1+ x
f ¢( x) =
- e - x ( x + 2)
1
(1 + x) 2
–2
–e–2
17. [ x] =
2 x{ x} 2{ x}([ x] + { x})
=
x + { x}
[ x] + 2 { x}
Þ
[ x ] 2 = 2{ x } 2
Þ
x =1+
1
(Q x Î R + )
2
18. || x - 1| + a| = 4
if a > 0
if a < 0
y
y
–a
a
x
x
1
1
(a) if eq. has three distinct real root then a < 0 and a = -4
(b) 4 distinct roots for a Î ( -¥, - 4)
(c) if -4 < a < 4 , there are two distinct real roots
(d) if a > 4, no real root.
19. (a) f2 ( x) = (sin x) 1/ 2 + (cos x) 1/ 2
sin x > sin 2 x;
(b) f2 ( x) = (sin x)
1/ 2
cos x > cos 2 x Þ
+ (cos x)
1/ 2
sin x + cos x > 1
Þ f2 ( x) = 1 at x = 2 kp
(c) f2 ( x) = (sin x) 1/ 2 + (cos x) 1/ 2 ; f3 ( x) = (sin x) 1/ 3 + (cos x) 1/ 3
if x Î (2 kp , 2 kp + p / 2)
0 < sin x < 1 and 0 < cos x < 1
As power increases, value of function decreases.
19
Function
Þ
f2 ( x) < f3 ( x)
(d) f3 ( x) = (sin x) 1/ 3 + (cos x) 1/ 3
f5 ( x) = (sin x) 1/ 5 + (cos x) 1/ 5
f3 ( x) < f5 ( x)
2
æ x2 ö
÷£1 Þ 1£ x £3
20. -1 £ log 3 ç
ç 3 ÷
3
3
è
ø
Þ
Range is [0 , 1].
3x - 1
21.
=n
2
é 4n + 5 ù é 4n + 5 1 ù
êë 9 úû + êë 9 + 2 úû = n
n =2 ü
M ï
ý
M ï
n = 10 þ
22. sin 6 q + cos 6 q = 1 - 3 sin 2 q cos 2 q
3 æ 1 - cos 4q ö 5 3
=1- ç
÷ = + cos( 4q)
4è
2
ø 8 8
5 3
= + cos( x)
8 8
23. (a) g( f( x)) = ln(sin x)
(b) x 2 + ( a - 1) x + 9 > 0 " x Î R
( a - 1) 2 - 36 < 0 Þ -5 < a < 7
(c) f( f( x)) = (2011 - (2011 - x 2012 )) 1 2012 = x
é 1 150 ù é 1 151 ù
é 1 199 ù
24. ê +
+ê +
+¼ + ê +
= 50
ú
ú
ë 4 200 û ë 4 200 û
ë 4 200 úû
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 4 to 6
Sol. f( x) = qx 2 - 2 (q 2 - 3) x - 12q
g( x) = ln( x 2 - 49)
if domain of f + g is same as domain of g. Then
qx 2 - 2 (q 2 - 3) x - 12q ³ 0 " x Î ( -¥, - 7) È (7 , ¥)
20
Solution of Advanced Problems in Mathematics for JEE
Þ
é6 7 ù
qÎ ê , ú
ë7 2 û
éq
ù
h(q) = ln ê 4 cos 2 t dt - q 2 ú = ln[2q + sin 2q - q 2 ]
êë 0
úû
ò
Paragraph for Question Nos. 7 to 8
7. For x Î[5 4 , 5 5 ]
é
ù
x
f( x) = a 4 ê2 -3 ú
4
5
ë
û
a =2
f( x) max = 32
8. a = 5
é
ù
x
f( x) = 5 4 ê2 -3 ú
4
5
ë
û
2007
é
ù
f(2007) = 5 4 ê2 + 3 ú = 1118
625
ë
û
Paragraph for Question Nos. 9 to 10
x
0£x£2
–x
–2 £ x < 0
9. f(x)
f ( x ) = f ( x + 4)
{ f(5.12)} = { f(1.12)} = 0.12
{ f(7.88)} = { f(3.88)} = { f( -0.12)} = 0.12
10.
2
–2p –6
–4 –p
–2
O
p/2 2
p
4
6 2p
Paragraph for Question Nos. 13 to 14
13. f( x) = 3
3 + ln b1 , 3 + ln b2 , 3 + ln b3 are in A.P.
14. y = 3 x 2
Let slope of tangent be m.
21
Function
y = m( x - 2)
Þ
m( x 1 - 2) = 3 x 12
Þ
Also,
Þ
m = 6 x1
6 x 1 ( x 1 - 2) = 3 x 12
x1 = 4
m = 24
Paragraph for Question Nos. 15 to 16
4
2
15. y = 2 x -4 x Þ x 4 - 4 x 2 = log 2 y
x2 =
4 + 16 - 4 log 2 y
Þ x = 2 + 4 - log 2 y
2
6
16. g( x) = 1 +
Þ Range [-5, - 2]
sin x - 2
Exercise-4 : Matching Type Problems
1.
…(i)
…(ii)
…(iii)
[ x] + { x} + [ y] + {z } = 12.7
[ x] + { y} + [z ] + {z } = 4.1
{ x } + [ y ] + { y } + [z ] = 2
Adding (i), (ii) & (iii),
Þ
[ x] + { x} + [ y] + { y} + [z ] + {z } = 9.4
Þ
Þ
{ y} + [z ] = -3.3, { x} + [ y] = 5.3, [ x] + {z } = 7.4
{ y} = 0.7, [z ] = -4 , { x} = 0.3, [ y] = 5
[ x] = 7 , {z } = 0.4
4. (A) f( x) = sin 2 2 x - 2 sin 2 x = 2 sin 2 x cos 2 x
Function is even, hence many one, function is also periodic.
f( x) = (1 - cos 2 x) cos 2 x =
1 æ
1ö
- ç cos 2 x - ÷
4 è
2ø
1ù
é
Range of function is ê -2 , ú .
4
ë
û
(B) f( x) = 4 x
(C) f( x) = ln(cos(sin x))
ln(cos(sin x)) ³ 0
Þ cos(sin x) = 1
Þ
f( x) = 0
2
22
Solution of Advanced Problems in Mathematics for JEE
æ x2 + 1 ö
÷
(D) f( x) = tan -1 ç
ç x2 + 3 ÷
è
ø
f( x) is even & hence many one.
ép p ö
Range is ê , ÷.
ë6 4 ø
7. (A) Domain of g( x) is [0, 3].
(B) Range of g( x) is [0, 3].
(C) f( f( f(2))) = 1
f( f( f(3))) = 2
(D) m = 3
Exercise-5 : Subjective Type Problems
1. f( x) - 2 x + 1 = ( x - 1)( x - 2)( x - 3)( x - 4)( x - 5)(2009 x - a )
2. f( x) = x 3 - 3 x + 1
f( f( x)) = 0
Let f( x) = t
Þ f (t ) = 0
Þ t = a ,b, g
Þ f( x) = a , a Î ( -2 , - 1)
No. of solution = 1
f( x) = b , b Î (0 , 1)
No. of solution = 3
f( x) = g , g Î (1, 2)
No. of solution = 3
3. Put x = y = 0
f(1) = 4
Put x = 0 , y = 1
f(2) = 9
2x
-3
3
4. -1 £
£1
Þ
£ x£
3
2
2
27
12 - 3 x ³ 0 Þ (3 x - 3)(3 x - 9) £ 0 Þ 1 £ x £ 2
x
3
4
-1
5. sin (0) + cos -1 ( -1) = p
0 £ x2 <
9
4
13
-1
-1
2
sin (1) + cos (0) = p
£x <
9
9
(0,3)
(0,1)
b
–2
a
–1
0
–1
1
g 2
23
Function
P( x) = ax 4 + bx 3 + cx 2 + dx + 2
8. Let
Þ
Þ
P(1) = a + b + c + d + 2 = 5
P( -1) = a - b + c - d + 2 = 5
b + d = 0 and a + c = 3
P(2) = 16 a + 8 b + 4 c + 2 d + 2 = 2
P( -2) = 16 a - 8 b + 4 c - 2 d + 2 = 2
4 a + c = 0 and 4 b + d = 0
and a = -1, c = 4
b= d =0
Þ
P( x ) = - x 4 + 4 x 2 + 2
Þ
9. ( x + 1) 2 + y 2 = 1
…(1)
…(2)
…(3)
…(4)
(Q y > 0)
x + y =k
k+1
<1
2
y
(–2,0) (–1,0)
0
x
- 2 - 1< k < 2 - 1
Þ
10.
(Q k > 0)
0 < k< 2 -1
éxù éxù
[ x] + ê ú + ê ú = 3
ë2 û ë3 û
\
éxù
éxù
when ê ú is an integer then definitely [ x] + ê ú is also an integer.
3
ë û
ë2 û
So,
éxù
éxù
[ x] + ê ú = 2 and ê ú = 1 (and check like this)
ë2 û
ë3 û
éxù
éxù
[ x] + ê ú = 4 , ê ú = 1 Þ x Î[3 , 6)
2
ë û
ë3 û
when x Î[3 , 4)
éxù
[ x] = 3 , ê ú = 1
ë2 û
So, x Î[3 , 4) satisfies.
éxù
éxù
when x Î [4 , 5) [ x] = 4 ê ú = 2 Þ [ x] + ê ú = 6 ¹ 4 not satisfies, similarly on checking all
ë2 û
ë2 û
possibilities we have only x Î[3 , 4).
\ a = 3, b = 4
2011
1
11. f( f( x)) =
2011 1 -
1
1 - x 2011
=
1 - x 2011
-x
24
Solution of Advanced Problems in Mathematics for JEE
-1
æ
2011 ç 1 -
2011
è
1- x
-1
2011
1 - x 2011
f( f( f( x))) =
ö
÷
ø
-x
2011
1 - x 2011
=x
-1
2011
1 - x 2011
=
\ f2013 ( x) = x = {- x}
12. f( x) = 0
0< x<6
= -1
6 £ x < 12
12 £ x < 18
18 £ x < 24
24 £ x < 30
x = 30
1
13. ( f( x , y)) 2 - ( g( x , y)) 2 =
2
= -2
= -3
= -4
= -5
f( x , y) × g( x , y) =
Þ
3
4
f( x , y) = x 2 - y 2 = ±
g( x , y) = 2 xy = ±
14. f( x) =
3
2
1
2
x +5
x2 + 1
y
26
5
1
–5
–1
1/5
1ù
æ
15. f( x) is injective for x Î ç -¥, ú
5û
è
é 1ù
[l] = ê ú = 0
ë5û
16. f : R ® R f( x) =
x3
+ (m - 1) x 2 + (m + 5) x + n
3
f ¢( x) = x 2 + 2(m - 1) x + (m + 5) ³ 0
D£0
x
25
Function
4(m - 1) 2 - 4(m + 5) £ 0
m 2 - 3m - 4 £ 0
(m - 4)(m + 1) £ 0
-1 £ m £ 4
( x - 1)( x - 3)
17. f( x) =
- ex
( x - 2)( x - 4)
f( x) = 0 has three solutions.
y
1
0
f( - x) =
2
1
3
x
4
( x + 1)( x + 3)
- e - x = 0 has three solutions.
( x + 2)( x + 4)
x 3 = cos x
–p/2
there are total 7 solutions.
æ
ö
2
18. cos -1 ç
- 1÷ = p (1 - { x})
ç (1 + x) 2
÷
è
ø
there are total 76 solutions.
0
p/2
one solution
y
p
0
19. f( x) = x 2 - bx + c = 0
p1 + p 2 = b (odd no.)
Þ p1 = 2
p1
p2
x
1
2
3
4
26
Solution of Advanced Problems in Mathematics for JEE
p1 p 2 = c
b + c = ( p 2 + 2) + 2 p 2 = 35
Þ p 2 = 11
Þ f( x) = x 2 - 13 x + 22
81
4
æ xö
f( x) - f ç ÷
è7ø 1
20. f ¢( x) = lim
=
x
x ®0
6
x7
x
Þ f( x) = + 1 Þ f( 42) = 8
6
1
21. g( x) = f( x)
0£ x<
2
1
1
=
£ x£1
4
2
l = f( x) min = -
=3 - x
1< x £ 2
1
1/4
0
100
22. x =
y
2
10
1 ö 10 æ
1 1 1
1
1
1
1 ö
æ 1
ç
÷=
ç1 + + + ÷
4 r =3 è r - 2 r + 2 ø 4 è
2 3 4 102 101 100 99 ø
å
1
1
1 ö
æ 1
= 5 ´ 49 ç
+
+
+
÷
è 99 200 303 408 ø
[ x] = 5
23. f( x) = x has two real roots.
7
cx 2 + ( d - a) x - b = 0
11
a-d
-b
= 18 and
= 77
c
c
if f( f( x)) = x " x Î R Þ ( ac + cd) x 2 + ( d 2 - a 2 ) x - ( a + d)b = 0
Þ a + d = 0 Þ a = -d
a
f( x) will not attain the value = 9.
c
24. A = (1, 3)
p £ -2 1-1 , p £ -2 1-3
1 - 2( p + 7) + 5 £ 0 and 9 - 6 ( p + 7) + 5 £ 0 Þ p Î [-4 , - 1]
x
1/2
1
27
Function
1
x
1
x25. y =
x3 y=
=
x3
t
Let t = x +2
x3 -
2
t(t + 3) + 2
1
x
3
1
> 0 for x > 1
x
= t(t 2 + 3)
t
3
t + 3t + 2
2
1 1
æ
2
2
ç t + =t + + ³ 3
t
t t
ç
2
2
çç\t + + 3 ³ 6 ( AM ³ GM)
è
t
1
=
2
2
t + +3
t
y max =
1
æ 2 2
ö
çt + + 3÷
t
è
ø min
=
1
6
p = 1, q = 6
28. a + ar + ar 2 = 1
a 2 r + a 2 r 2 + a 2 r 3 = b = ar ( a + ar + ar 2 ) = ar
a 3 r 3 = -g
29. m = 6 C 4 ´ 1 = 15
n=
6!
6!
´ 4 !+
´ 4 ! = 1560
3 !1!1!1!3 !
(2 !) 4
n
30.
+ 3 +¼ + 3) +¼
å[log 2 r] = 0 + 1 + 1 + (2 + 2 + 2 + 2) + (13 4
42443
r =1
8 times
= 2 × 1 + 4 × 2 + 8 × 3 +¼ +
32. 3
|( x - 2 y)( y + x)( x + 3 y)| = f( x , y)
No rain, then f( x , y) = 0 hence 3 lines.
33. Cubic = ( x 2 - 5 x + 6)( x + a ) + 2 ( Bx + 100 - 4a )
( x 2 - 5 x + 4)( x + a ) + Bx + 100 - 4a
Both identical B = -2
a = 50
2
Cubic = ( x - 5 x + 6)( x + 50) - 4 x - 200
28
Solution of Advanced Problems in Mathematics for JEE
34. f(q) = 0 Þ q = -5 ± 5
Þ
f( f( f( x))) = -5 ± 5
Since f( x) = ( x + 5) 2 - 5
f( f( f( x))) = -5 ± 5
(( f( f( f( x)))) + 5) 2 = -5 ± 5
( f( f ) + 5) 2 = 5
f ( f ) + 5 = ± 51/ 4
f( f ) = -5 ± 51/ 4
( f + 5) 2 - 5 = -5 ± 51/ 4
( f + 5) 2 = 51/ 4
f + 5 = ± 51/ 8
35. Let ln x = t
y=
2t 2 + 3t + 3
2
t + 2t + 2
Þ ( y - 2) t 2 + ( 2 y - 3) t + ( 2 y - 3) > 0
D ³ 0 Þ (2 y - 3)(2 y - 5) £ 0 Þ
3
5
£ y<
2
2
36. P( x) = ( x - 3)Q 1 ( x) + 6 Þ P(3) = 6
…(1)
P( x) = ( x 2 - 9) Q ( x) + ( ax + b)
P ( 3) = 3 a + b = 6
If equation of odd degree polynomial, then b = 0 , a = 2.
37. f( x) = 2 x 3 - 3 x 2 + P
1
f ¢( x) = 6 x 2 - 6 x = 6 x( x - 1)
0
f(0) ³ 0 Ç f(1) £ 0
Þ P ³ 0 Ç P - 1£ 0
1
38. f( x) =
ln(cos -1 x)
ln(cos -1 x) > 0 Þ cos -1 x > 1
❑❑❑
x
29
Limit
2
LIMIT
Exercise-1 : Single Choice Problems
æ x - tan x ö
æ tan x + x ö
2 sin ç
÷ sin ç
÷
2
2
è
ø
è
ø ´ æ x - tan x ö æ x + tan x ö ´ 1
1. lim
ç
֍
÷
x ®0
x
æ x - tan x ö æ tan x + x ö
ø 4
x3
è
øè
ç
֍
÷
2
2
è
øè
ø
=
1 æ 1ö
1
´ ç - ÷ ×2 = 2 è 3ø
3
2
æ ln(1 + cos 2 x - 1) ö (cos 2 x - 1)
3. a = lim ç
=÷
2
x ®0 è
cos 2 x - 1
3
ø
3x
æ sin 2 2 x ö
4x2
÷
b = lim ç
= -4
x
x ®0 ç 4 x 2 ÷
è
ø x 2 æç 1 - e ö÷
ç x ÷
è
ø
c = lim
x (1 - x)
x ®1 æ ln(1 + x - 1) ö
ç
÷ ( x - 1)( x + 1)
x -1
è
ø
p
-1
4. f( x) = - 3 tan x
2
g( x) = 2 tan -1 x
f( x) - f( a) f ¢( a)
3
=
=x ® 0 g( x ) - g( a)
g¢( a)
2
lim
=
-1
2
(use expansions)
30
5.
Solution of Advanced Problems in Mathematics for JEE
æ æ ln(1 + x ) ö ö
4
4 ç 2 çè x -1 ÷ø ÷
lim
-1 ÷
çe
æ 2 ln (1+ x ) - 2 ö sin x
x ® 0 sin x ç
÷
ç
÷
è
ø
lim e x
=e
÷
x ®0 ç
è
ø
=e
æ 2 é ln (1 + x ) -1 ù ö
ú
ç ê
÷
û -1
4 çe ë x
÷ ´ 2 é ln(1+ x ) -1ù
lim
úû
ç
x ® 0 sin x
x
æ ln(1+ x ) ö ÷ êë
-1 ÷ ÷
ç 2ç
x
ø ø
è è
2
æ
ö
ç x- x +¼ ÷
1
2
lim 8 ç
-1 ÷´
÷ sin x
x ®0 ç
x
ç
÷
è
ø
=e
-
8
= e 2 = e -4
3 æ x ì x üö 3
3
ç - í ý ÷÷ = - 0 =
4
4
4
î þø
6. lim
x ®¥ x ç
è4
p + q=7
n
æ
æ p ö ö÷
x ç1 + ç
÷
ç
è 3 x ø ÷ø
p
7. f( x) = lim è
= x; x >
n -1
n ®¥
3
æ p ö
1+ ç
÷
è 3x ø
Þ
\
\
8. lim
n
ö
p æç æ 3 x ö
ç
÷ + 1÷
÷
3 çè p ø
ø = p;x< p
= lim è
n
1
n ®¥ æ
3
ö 3
ç æç 3 x ö÷
+ 1÷
çè p ø
÷
è
ø
p
p
=
x=
3
3
p
f( x) = x ; x ³
3
p
p
= ; x<
3
3
Option (d) is wrong.
sin( p - p cos 2 (tan(sin x))
x2
x ®0
9.
x ®3
x ®3
+
p/3
sin[p sin 2 (tan(sin x))]
= lim
p sin 2 (tan(sin x))
x ®0
lim f( x) = lim f( x) Þ lim
-
x
p/3
x ®3
(27)
-
( x + 3) x
27
-9
3
x
- 27
= lim l
x ®3
+
2
æ sin(tan(sin x)) ö
´ pç
÷ =p
x
è
ø
1 - cos( x - 3)
( x - 3) 2
31
Limit
æ
2ç
x 2+ 3 x
ö
-2
÷
3 9
-1
3 ç
÷
ç
÷ l
è
ø=
lim
3
x -3
2
x ® 33 (3
- 1)
Þ
1 x 2 + 3 x - 18 l
=
x ®3 3
9( x - 3)
2
lim
Þ
Þ
1
l
×9 =
27
2
Þ l=
2
3
æp
ö
æp
ö
æp
ö
æp
ö
ç - x÷
ç - x÷
ç - x÷
ç - x÷
3
3
3
3
÷ cos ç
÷
÷ cos ç
÷
2 sin ç
sin ç
ç 2 ÷
ç 2 ÷
ç 2 ÷
ç 2 ÷
ç
÷
ç
÷
ç
÷
ç
÷
è
ø
è
ø = lim
è
ø
è
ø
10. lim
p
p
p
p
p
æ
ö
æ
ö
æ
ö
x®
x®
2 ç cos x - cos ÷
ç - x÷
ç + x÷
3
3
3
3
3
è
ø
÷ sin ç
÷
2 sin ç
ç 2 ÷
ç 2 ÷
ç
÷
ç
÷
è
ø
è
ø
1 1
1
=
=
2 3
3
2
p
sin
sin x
2 = 1 =2
11. lim
Þ
-1
p cos -1 [sin 3 x ]
cos (0) p 2 p
x®
2
13.
lim { x} = lim x - [ x] = 1 ;
x ®I -
x ®I -
lim
e{ x} - { x} - 1
x ®I -
{ x} 2
= e -2
c
é
ù
7
2 ö
5c -1 æ
16. lim x ê x
ç1 + +
÷ - 1ú = l
x ®¥ ê
x x5 ø
úû
è
ë
Case-I: 5c - 1 > 0 , then l ® ¥
Case-II: 5c - 1 < 0 , then l ® -¥
Since limit is finite and non-zero so 5c - 1 = 0 Þ c =
\
1
5
1/ 5
éæ
ù
7
2 ö
l = lim x êç 1 + +
- 1ú
÷
x ®¥ êè
x x5 ø
úû
ë
é æ 1öæ 7
ù
2 ö
= lim x ê1 + ç ÷ ç +
÷ +¼¼ - 1ú
5
x ®¥ ë
è 5øè x x ø
û
=
7
5
(by binomial approximation)
32
Solution of Advanced Problems in Mathematics for JEE
cos x - 1 æç cos x - 1 ( e x - 1) ö÷
= 0 Þ n = 1, 2 , 3
x ®0
x 2 çè x n -2
x n -2 ÷ø
17. lim
lim
¥
18. 1 (form) = e
1
æ sin x - x ö
ç
÷
x
ø = e 2 ´ -1/ 6 = e -1/ 3
x ® 0 1- cos x è
19. lim [ x 2 - x + 1 - ( ax + b)] = 0
x ®¥
So a > 0, on rationalizing
é( x 2 - x + 1) - [a 2 x 2 + b 2 + (2 ab) x] ù
lim ê
ú =0
x ®¥ ê
úû
x 2 - x + 1 + ax + b
ë
So, 1 - a 2 = 0
-1 - 2 ab = 0
a =1
2
lim sec [k ! p( - 1 2)] = 1 = a
n ®¥
20. f( x + T ) = f( x + 2T ) = ¼¼ = f( x + nT ) = f( x)
æ n(n + 1) ö
nç
÷
nf( x)(1 + 2 + 3 +¼ + n)
2
è
ø =3
lim
= lim
n ®¥ f ( x )(1 + 2 2 + 3 2 +¼ + n 2 ) n ®¥ n(n + 1)(2n + 1)
2
6
é
ù
ê
ú
h2 + 3
3
53 ´ 3
ú = -265 ´
21. 265 ê lim
=h
®
0
f ¢(1) × 5
f ¢(1)
ê
æ f(1 - h) - f(1) ö æ sin 5h ö ú
ç
֍
÷ú
ê
h
h
øû
è
øè
ë
=-
53 ´ 3
-53
=3
2
22. lim
cos x - 1
x ® 0 cos x × x 2 × ( x + 1)
æ sin 2 x ö
-1
÷
lim - ç
= -1
2
ç
÷
x ®0
è x
ø cos x( x + 1)
23. f( x + y) = f( x) × f( y)
f( x + h) - f( x)
f( x) f( h) - f( x)
f( h) - 1 ö
æ
f ¢( x) = lim
= lim
= f( x) ç lim
÷
h ®0
h ®0
h
h
h
è h ®0
ø
hP( h) + h 2 Q ( h)
= P ( 0) f ( x )
h ®0
h
If f( h) = 1 + hP( h) + h 2 Q ( h) Þ f ¢( x) = f( x) lim
[Q f ¢(1) = -53]
33
Limit
xö
æ
ç 1 - tan ÷ (1 - sin x)
2
ø
24. lim è
pæ
x
3
x ® ç 1 + tan ö
÷ ( p - 2 x)
2
2ø
è
= lim
x®
æ p x öæ
æp
öö
tan ç - ÷ çç 1 - cos ç - x ÷ ÷÷
4
2
2
è
øè
è
øø
( p - 2 x) 3
p
2
p
+h
2
æ hö
tan ç - ÷ (1 - cos h)
è 2ø
Let x =
lim
x ®0
( -2 h)
3
=
1
32
æ -5 ö
x
lim x ç
÷
æ x -3 ö
x ®¥ è x + 2 ø
25. lim ç
= e -5
÷ =e
x ®¥ è x + 2 ø
27. ln c = I ,
(I Î integer)
I
Þ c=e
c is rational when I = 0
a sin bx ö
æ
28. lim ç 1 +
÷
x ®0 è
cos x ø
1/ x
lim
1 é a sin bx ù
ê1+ cos x -1úû
= e x ®0 x ë
= e ab
1 ö
x2 -1
x + 1 x -1
æ x
30. a = lim ç
= lim
×
=2
÷ = lim
x ®1 è ln x
x ln x ø x ®1 x ln x x ®1 x
ln x
b = -4 , c = 1, d = -2
32. f( x) = x 2
-1 < x < 0
=1
1
=
x2
x =0
0< x<1
lim { f( x)} + lim { f( x)} + lim { f( x)} = 0
x ® 0-
x ®1-
x ®-1-
33. Let sin -1 x = q
Þ
cos -1 sin 2q
lim
= lim
p
p+
p+
sin q - sin
q®
q®
4
4
4
2q -
p
2
pö
pö
æ
æ
çq - ÷
çq + ÷
4
4÷
÷ cos ç
2 sin ç
ç 2 ÷
ç 2 ÷
ç
÷
ç
÷
è
ø
è
ø
=2 2
34
Solution of Advanced Problems in Mathematics for JEE
cos -1 sin 2q
= -2 2
p
psin q - sin
q®
4
4
lim
n
n
p
p
p
p ö
æ
ö
æ
- sin
- cos
ç sin
÷ + lim
ç cos
÷
n ®¥
2k
2( k + 2) ø n ®¥ k =1 è
2( k + 2)
2k ø
k =1 è
34. lim
å
å
p
p
p
p
p
p
p
p
æ
ö
= lim ç sin - sin + sin - sin + sin - sin
+¼ + sin
- sin
÷
n ®¥ è
2
6
4
8
6
10
2n
2(n + 2) ø
p
p
p
p
p
p
p
p ö
æ
+ lim ç cos - cos + cos - cos + cos
- cos +¼ + cos
- cos
÷
n ®¥ è
6
2
8
4
10
6
2(n + 2)
2n ø
1
1
=1+
+2=3
2
2
1
36. lim
1 1
(cos x) m n - 1
x2
x ®0
æ
2 xöm
ç 1 - 2 sin
÷
2ø
è
= lim
x ®0
x2
æ 1 1ö
= lim - 2 ç - ÷
x ®0
èm nø
37. lim
-
1
n
-1
x
2 = m -n
2
2mn
x
sin 2
x + xa cos x - b sin x
=1
x3
Using expansion,
æ
æ
x 2 ö÷
x 3 ö÷
ax 3
bx 3
x + xa ç 1 - bç x x + ax - bx +
ç
÷
ç
÷
2
!
3
!
è
ø
è
ø Þ lim
2!
3!
Þ lim
3
3
x ®0
x
®
0
x
x
Clearly, 1 + a - b = 0 for limit to be finite
x ®0
…(1)
3
b a
æ b aöx
lim ç - ÷
= 1 Þ - = 1 Þ b - 3a = 6
2 !ø x 3
6 2
5
3
From (1) and (2), a = - , b = 2
2
Þ
…(2)
x ®0 è 3 !
Þ
a cos ax 38. lim
x ®0
e x (cos x - sin x)
e x × cos x
sin bx + bx × cos bx
=
1
2
cos 2 x - cos x + sin x
1
=
x ® 0 cos x (sin bx + bx cos bx ) 2
Þ lim
(Q a = 1)
2
n (2n + 1)(n + 1)
æ n(n + 1) ö
ç
÷ ( 13 + 2 3 + 3 3 ¼ + n 3 ) - ( 12 + 2 2 ¼ + n 2 )
2
6
ø
39. a = lim
= lim è
4
4
n ®¥
n ®¥
n
n
35
Limit
2
é1æ
(2n + 1)(n + 1) ù 1
1ö
lim ê ç 1 + ÷ ú=
3
n ®¥ ê 4 è
nø
6
n
úû 4
ë
æ sin x + x ö
æ x - sin x ö
2 sin ç
÷ sin ç
÷
2
2
cos(sin x) - cos x
è
ø
è
ø
40. lim
Þ lim
4
4
x® 0
x
®
0
x
x
2 (sin x + x) ( x - sin x)
x - sin x
ì sin x + x
ü
Þ lim
×
® 0;
® 0ý
íQ
4
x® 0 x
2
2
2
2
î
þ
Þ
1æ
sin x ö æ x - sin x ö
1
÷Þ
ç1 +
֍
3
x® 0 2 è
x øè x
6
ø
lim
Þ
1 2
3
n
+
+
¼+
2 22 23
2n
…(1)
1
1
2
n -1
n
u n =¼ +
+
¼+
+
2
3
n
n
2
2
2
2
2 +1
…(2)
42.
un =
Substracting equation (1) and (2),
1æ
1 ö
ç1 ÷
un 1
un 2 è
1
1
1
n
n
2n ø
= +
+
+¼ +
Þ
=
2
3
n
n
+
1
n
2 2 2
2
1ö
æ
2
2
2
2 +1
ç1 - ÷
2ø
è
1 ö
n
æ
un = 2 ç 1 ; lim u = 2
÷n
n +1 n ® 0 n
2 ø 2
è
Þ
lim
(cos x -1)
2
43. e x ®0 sin x
sin 2 x
tan -1 3 x
×2x + 6x ×
+ 3x2
1
2
x
3
x
+ lim
=
+2
x ® 0 ln(1 + 3 x + sin 2 x )
e
2
x
× (3 x + sin x) + xe
3 x + sin 2 x
x
(1 + sec x) = tan x
2
x
fn ( x) = tan (1 + sec x)(1 + sec 2 x) ¼(1 + sec 2 n x) = tan 2 n x
2
44. tan
45. lim (1 + [ x])
x®
p
4
46. lim
x ®0
Þ
1
ln (tan x )
= lim (1)
x®
1
ln (tan x )
p
4
=1
{( a - n) nx - tan x} sin nx
xn = 0
nx
x2
{( a - n) n - 1}n = 0 Þ a = n +
1
n
36
Solution of Advanced Problems in Mathematics for JEE
3n 3 + 4
æ n ! ö 4n 4 -1
47. y = lim ç
÷
n ®¥ è n n ø
3n 3 + 4 n
r
3
1
-3
æ ö
ln ç ÷ = ò ln x dx =
Þ y = e -3/ 4
å
4
n ®¥ 4n - 1
4
ènø 4
ln y = lim
r =1
0
2
ax + b + ( c x)
ax + bx + c
bö
æa
= lim
= lim ç x + ÷
x
®¥
x
®¥
dx + e
d + ( e x)
dø
èd
48. lim
x ®¥
æ aö
= + ¥ if ç ÷ is positive.
è dø
æ aö
= -¥ if ç ÷ is negative.
è dø
Alternate solution :
a + ( b x) + ( c x 2 )
ax 2 + bx + c
= lim
x ®¥
x ®¥ ( d x ) + ( e x 2 )
dx + e
lim
Here
e
x
2
<<
d
. Therefore,
x
ax 2 + bx + c
a
= lim
x ®¥
x
®¥
dx + e
d x
lim
ì a
ïï + if d > 0
= í0
a
ï
if d < 0
ïî 0 -
ì+ ¥ if a > 0 and d > 0
í -¥ if a < 0 and d > 0
î
ì -¥ if a > 0 and d < 0
í+ ¥ if a < 0 and d < 0
î
2
æ
ö
x ö
æ
ç
÷
ç sin
÷
2
x ö
æ
-1 ç
2ç
2n ÷ × x ÷ = tan -1 (2 x 2 )
49. f( x) = lim tan -1 ç 4n 2 × 2 sin 2
=
lim
tan
8
n
÷
ç
n ®¥
2n ø n ®¥
ç x ÷ 4n 2 ÷
è
çç
÷÷
ç
÷
è 2n ø
è
ø
æ æ
2x
öö
ç ln ç 1 + cos 2
- 1÷ ÷
n
n2 ç è
ø ÷ æ cos 2 x - 1ö = x 2
g( x) = lim
÷
ç
÷ çè
n ®¥ 2
n
2 2x
ø
cos
1
ç
÷
n
è
ø
sin 2 x 1
= Þ f( x) = x 2 ( ax + 3) ; a ¹ 0
x ® 0 f( x)
3
50. lim
51. lim
x ®0
(2 e 2 sin x - e sin x - 1)
2
( x + 2 x) e
sin x
= lim
x ®0
(2 e sin x + 1)( e sin x - 1)
x ( x + 2) e
sin x
=
3
2
37
Limit
52. x n + ax + b = ( x - x 1 )( x - x 2 )( x - x 3 ) ¼( x - x n )
x n + ax + b
= ( x 1 - x 2 )( x 1 - x 3 ) ¼( x 1 - x n )
x ® x1
x - x1
lim
1
1
æ
ö æ
ö
2
ç 1 + sin x +¼ ÷ - ç 1 - (2 tan x) +¼ ÷
3
4
è
ø
è
ø=1
53. lim
2
x ®0
2
sin x + tan x
2 sin x
cos x
tan x
1 2 0
x
f( x)
54. lim
= lim
1
1
1 = 1
1 1 = -1
x ®0 x 2
x ®0
1 2 1
1
2
1
Exercise-2 : One or More than One Answer is/are Correct
lim
1. e
1
x ®0 3 x 2
( p tan qx 2 - 3 cos2 x + 3)
pq 3 (1- cos2 x )
+
x ®0 3
3x2
lim
e
Þ
pq
5
+ 1= ;
3
3
pq = 2
3. a ³ e > 2
x
x
æ
æ2ö
æ eö ö
(a) L = a lim ç 1 + ç ÷ + ç ÷ ÷
x ®¥ ç
è aø
è a ø ÷ø
è
\
1/ x
1
æ2ö
æ eö
x ® a, ç ÷ ® 0 , ç ÷ ® 0 , ® 0
x
è aø
è aø
So, L = a
(b) If a = 2 e > 2
L = lim (2
x ®¥
x
x
x 1/ x
+ (2 e) + e )
x
éæ 1 ö x
æ 1ö ù
= 2 e lim êç ÷ + 1 + ç ÷ ú
x ®¥ êè e ø
è 2 ø úû
ë
(c) If 0 < a £ e
1/ x ö
æ
ææ 2 ö x æ a ö x
ö
ç
÷
ç
÷
L = e ç lim ç ÷ + ç ÷ + 1
÷=e
÷
è eø
ç x ®¥ çè è e ø
÷
ø
è
ø
1/ x
= 2 e(1) = 2 e
38
Solution of Advanced Problems in Mathematics for JEE
(d) a >
e
>1
2
x
é
ù
æ 2a ö
x
L = lim ê2 x + ç
÷ +e ú
x ®¥ ê
è 2 ø
úû
ë
1/ x
ææ 1ö x æ 1ö x æ e ö x ö
= 2 a lim ç ç ÷ + ç ÷ + ç
÷ ÷
x ®¥ ç è a ø
2ø
2a ø ÷
è
è
è
ø
1/ x
=0
5. f( x) = cos(sin x)
Range is [cos 1, 1].
æ3 3
ö
8. f( x) = x ç + [cos x]÷
è2 2
ø
1
9. If x ¹
2
2n
2
1
Also, if x =
2
11.
2n
then 2 x ¹
x ® 0+
x ®0
22
n
cos ( - x) sin
-
-1
( - x)
2( x + 1) ( - x)
2n
then lim f( x) = lim ( -1) n , hence does not exist.
x ®0
=
n ®¥
Þ f (2 x) = 0
= lim
æ sin -1 2 x - x 2 ö
ç
÷ 2 x - x 2 × sin -1 (1 - x)
ç
÷
2
2x - x
è
ø
x ® 0+
2 x (1 - x)
-1
lim
1
cos -1 (1 - x) sin -1 (1 - x)
lim
1
then f( x) = 0 but if x =
2 x (1 - x)
p
2 2
æ sin x - x ö
æ sin x + x ö
2 sin ç
÷ × cos ç
÷
2
2
è
ø
è
ø = -1
12. lim
3
5
x ®0
12
ax + bx + c
æ
æ sin x - x ö ö
ç sin ç
÷÷
2
è
ø ÷ æ sin x - x ö × cos æ sin x + x ö
ç
2
÷
ç
÷
ç sin x - x ÷ çè
2
2
ø
è
ø
ç
÷
2
-1
ø
lim è
=
3
5
x ®0
12
ax + bx + c
pö
æ
14. cos 2 ç np + ÷
3ø
è
15. sin a + sin b = 16.
sin b
1
Þ sin a = sin b = sin a
2
lim [5 - 2 x] = 0
x ® 2+
lim [| x - 2| + a 2 - 6 a + 9] = 0 Þ ( a - 3) 2 < 1
x ® 2-
=
p
2
39
Limit
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1. S 1 = 1, S 2 = 7 , S 3 = 19
Þ S n = 1 + 3n(n - 1)
S
lim n = 3
n ®¥ n 2
1
1
2. r1 = 1, r2 = , r3 =
3
5
1
or rn =
2n - 1
1
1
lim n ´
=
n ®¥
2n - 1 2
Paragraph for Question Nos. 3 to 4
3. x > 0, x < tan x
x < 0, x > tan x Þ x - tan x > 0
\ [ x - tan x] = 0
\
lim f([ x - tan x]) = f(0) = 4
x ® 0-
4. x > 0
x < tan x
x
<1
tan x
x
ì x ü
lim í
® 1ý=
+ tan x
x ®0 î
þ tan x
\
æ ì x üö
æ x ö
lim çç f í
÷ = f (1 ) = 2 + 5 = 7
ý ÷÷ = lim+ f ç
+
x ® 0 è îtan x þ ø x ® 0
è tan x ø
Paragraph for Question Nos. 5 to 6
5. f( x) = 1 -| x - 2|
x ® 2 + , f( x) ® 1- and x ® 2 - , f( x) ® 1-
R.H.L. =
1
f ( x )-1
lim
+
æ px ö
æ px ö
x
®
2
sin ç
sin ç
÷
÷
è 2 ø
lim ( f( x)) è 2 ø = e
x ® 2+
40
Solution of Advanced Problems in Mathematics for JEE
lim
1-( x - 2)-1
x ® 2+ sin p æ 1- x ö
ç
è
=e
÷
2ø
lim
+
= e x ®2
( x - 2)
p
´ (2 - x)
p
æ
ö 2
ç sin (2 - x) ÷
2
ç
÷
ç p
÷
(2 - x) ÷
ç
è 2
ø
= e 2/p
L.H.L. =
1
f ( x ) -1
lim
æ px ö
x ® 2- sin æ px ö
sin xç
÷
ç
÷
è 2 ø =e
è 2 ø
lim ( f( x))
x ® 2lim
-
= e x ®2
lim
1 + ( x - 2) - 1
x -2
= e x ®2
p
p
sin (2 - x)
(2 - x)
2
2
= e -2/p
Limit does not exist.
\
6. [1, 3]
as f(3 x) = af( x)
;
x Î[1, 3]
3 x Î[3 , 9] ;
f( x) Î [0 , 1]
f(3 x) = af( x) Î [0 , a ]
9 x Î[9 , 27] ;
f(9 x) = af(3 x) Î [0 , a 2 ]
1
1
´2 ´1=1
2
1
area between [3 , 9] is D 2 = ´ 6 ´ a = 3a
2
1
area between [9 , 27] is D 3 = ´ 18 ´ a 2 = 9a 2
2
area between [1, 3] is D 1 =
\
æ 1 1ö
1, 3a , 9a 2 , ¼¼ is converges when (g.p.) |3a| < 1 a Î ç - , ÷
è 3 3ø
Paragraph for Question Nos. 7 to 9
æ
x x 2 x 3 ö÷
(1 + bx) - (1 + ax) ç 1 + +
ç
2
8
16 ÷ø
[(1 + bx) - (1 + ax) 1 + x ]
è
7. lim
= lim
x ®0
x ®0
x3
x3
bx = lim
x ®0
x x2 x3
ax 2 ax 3
+
- ax +
2
8
16
2
8
x3
1
3
9
41
Limit
Þ
coefficient of x and x 2 = 0 Þ b - a =
Þ
a=
1
a 1
and =
2
2 8
1
3
,b=
4
4
8. a + b = 1
1
3
9. l = ; b=
32
4
Paragraph for Question Nos. 10 to 11
Sol. sin x + sin y = 1
- cos x
y¢ =
y ¢¢ =
Þ
2 sin x - sin 2 x
sin 2 x - sin x + 1
(2 sin x - sin 2 x) 3/ 2
Exercise-5 : Subjective Type Problems
éæ 1 - tan bx
ù
ö
æp
ö
ln êçç
- 1÷÷ + 1ú
ln tan ç - bx ÷
ø
è4
ø = - lim ëè 1 + tan bx
û
1. lim x ®0
x ®0
tan ax
tan ax
bö
æ
= -1ç -2 ÷ = 1
aø
è
a
=2
b
Þ
3. a( x 3 - 1) + ( x - 1) = 0
( x - 1)( ax 2 + ax + a + 1) = 0
a , b ¹ 1 so, a , b are roots of ax 2 + ax + a + 1 = 0
a + b = -1, ab =
lim
x®
a+1
a
(1 + a) x 3 - x 2 - a
1 ( e 1- ax - 1)( x - 1)
a
= lim
x®
1
a
= lim
x®
1
a
( x 3 - x 2 ) + a( x 3 - 1)
( e1-ax - 1)( x - 1)
[ x 2 + a( x 2 + x + 1)]
( e1-ax - 1)
= lim
x®
(1 + a) x 2 + ax + a
1 æ 1- ax
e
- 1 ö÷
(1 - ax)
a ç
ç 1 - ax ÷
è
ø
42
Solution of Advanced Problems in Mathematics for JEE
éæ 1 + a ö 2
ù
êç a ÷ x + (1) x + 1ú
2
è
ø
û = lim a (abx - (a + b) x + 1)
= lim a ë
1
1
(1 - ax)
(1 - ax)
x®
x®
a
a
(1 - (a ) x)(1 - (b) x) a(a - b)
= lim a
=
1
(1 - ax)
a
x®
a
4. lim
x ®0
5. lim
x ®0
(4
x
x
- 1)(5 - 1)(7
x
2
x sin x
ax cos x + b sin x
2
x sin x
=
- 1)
= 2 ln 2 ln 5 ln 7
1
3
æ
æ
x 2 x 4 ö÷
x 3 x 5 ö÷
ax ç 1 +
¼ + bç x +
¼
ç
ç
2!
4 ! ÷ø
3!
5 ! ÷ø 1
è
è
lim
=
x ®0
3
x 2 sin x
a b 1
a + b = 0 and - - =
2 6 3
0
y=
x–
1
y=
x
Þ b = 1, a = -1
é sin x ù
7. lim ê
=0
+ x - 1ú
x ®a ë
û
1
❑❑❑
a2
3 p
Continuity, Differentiability and Differentiation
3
43
CONTINUITY, DIFFERENTIABILITY
AND DIFFERENTIATION
Exercise-1 : Single Choice Problems
f( x + h) - f( x)
f( x) + f( h) + 3 hx( h + x) - f( x)
= lim
h ®0
h ®0
h
h
1. f ¢( x) = lim
f ¢( x) = 3 x 2 + f ¢(0) Þ f ¢¢( x) = 6 x
1
2.
–2 –1
0 1
–1
2
f( x) is non-differentiable at five points.
x
3.
is integer at 21 points in [0 , 100]
5
x
is integer at 51 points in [0 , 100]
2
\ But when x is a multiple of 10 then f( x) is continuous.
So that respective points should be subtract from both i . e. , multiple of 10 are 11 points in
[0 , 100].
21 + 51 - 11 - 11 = 72 - 22 = 50
4. f( x) has isolated point of discontinuity but| f( x)|is continuous at
x =a
So, lim f( x) and f( a) has opposite sign, with same magnitude.
x ®a
So, lim f( x) = - f( a)
x ®a
lim f( x) + f( a) = 0
x ®a
5. lim
x® 0
f ( 4 x) - 3 f (3 x) + 3 f (2 x) - f ( x)
x3
= 12
44
Solution of Advanced Problems in Mathematics for JEE
lim
4 f ¢( 4 x) - 9 f ¢(3 x) + 6 f ¢(2 x) - f ¢( x)
3x2
x® 0
= 12
4 2 f ¢¢( 4 x) - 27 f ¢¢(3 x) + 12 f ¢¢(2 x) - f ¢¢( x)
= 12
x® 0
6x
lim
4 3 f ¢¢¢( 4 x) - 81 f ¢¢¢(3 x) + 24 f ¢¢¢(2 x) - f ¢¢¢( x)
= 12
x® 0
6
lim
\
( 4 3 - 81 + 24 - 1) f ¢¢¢(0) = 12 ´ 6
6 f ¢¢¢(0) = 12 ´ 6
f ¢¢¢(0) = 12
6. y =
1
1 + (tan q)
sin q - cos q
+ (tan q)
cot q - cos q
1
+
1 + (tan q)
cos q - sin q
+
y=
(tan q) cos q
(tan q) cos q + (tan q) sin q + (tan q) cot q
+
(tan q) cos q + (tan q) sin q + (tan q) cot q
y =1
dy
=0
dx 0 = p/ 3
7. f ¢( x) = sin( x 2 )
y = f( x 2 + 1)
dy
= f ¢( x 2 + 1)2 x
dx
dy
= 2 × f ¢(2) = 2 sin 4
dx x =1
7p
6
So, f( x) = - (sin x + cos x)
f ¢( x) = (sin x - cos x)
9. 2 sin x cos y = 1
cos x cos y - sin x sin y × y ¢ = 0 Þ y (¢ p/ 4, p/ 4) = 1
y ¢ = cot x cot y
y ¢¢ = - cot x cosec 2 y ´ y ¢ - cot y cosec 2 x
y (¢¢p/ 4, p/ 4) = - (1 ´ 2 ´ 1) - (1 ´ 2) = 0
1 + (tan q) cos q - cot q + (tan q) sin q - cot q
(tan q) sin q
+
8. Clearly sin x , cos x are negative at x =
+ (tan q) cot q - sin q
1
(tan q) cot q
(tan q) cos q + (tan q) sin q + (tan q) cot q
45
Continuity, Differentiability and Differentiation
10.
dy
dx
= 2t f ¢(t 2 ),
= 3t 2 f ¢(t 3 )
dt
dt
dy 3 tf ¢(t 3 )
=
dx 2 f ¢(t 2 )
d2 y
dx 2
=
3 æç f ¢(t 2 )( f ¢(t 3 ) + 3t 3 f ¢¢(t 3 )) - 2t 2 f ¢(t 3 ) × f ¢¢(t 2 ) ö÷ dt
÷ dx
2 çè
( f ¢(t 2 )) 2
ø
d2 y
dx
=
2
t =1
3 æç f ¢(1)( f ¢(1) + 3 f ¢¢(1)) - 2 f ¢(1) × f ¢¢(1) ö÷ 1
3 æ f ¢¢(1) + f ¢(1) ö÷
= ç
2
2
ç
÷
ç
÷
2è
( f ¢(1))
ø 2 f ¢(1) 4 è ( f ¢(1))
ø
11. L.H.L. = a + 1
R.H.L. = b + 1
\ they are continuous L.H.L. = R.H.L.
g
b
x
x2
12. y =
+
+
1
1
1
1
1
1
- a æç - a ö÷ æç - b ö÷ æç - a ö÷ æç - b ö÷ æç - g ö÷
x
èx
øè x
ø èx
øè x
øè x
ø
1
x
1
g
2
2
1
x
x
x3
+
=
æ1
öæ 1
ö æ1
öæ 1
öæ 1
ö æ1
öæ 1
öæ 1
ö
ç - a ÷ç -b÷ ç - a ÷ç -b÷ç - g÷ ç - a ÷ç -b÷ç - g÷
x
x
x
x
x
x
x
x
è
øè
ø è
øè
øè
ø è
øè
øè
ø
æ1
ö
æ1
ö
æ1
ö
log y = -3 ln x - ln ç - a ÷ - ln ç - b ÷ - ln ç - g ÷
èx
ø
èx
ø
èx
ø
=
1
1
1
2
2
2
1
-3
x
y¢ =
+
+ x
+ x
y
x
æ1
ö æ1
ö æ1
ö
ç - a ÷ ç -b÷ ç - g÷
x
x
x
è
ø è
ø è
ø
æ
1
1
1 ö÷
ç
y
÷
x
x
x
y ¢ = ç -3 +
+
+
xç
æ1
ö æ1
ö æ1
ö÷
ç - a ÷ ç -b÷ ç - g÷ ÷
ç
èx
ø èx
ø èx
øø
è
y¢ =
yæ a
b
g ö
ç
÷
+
+
x çè 1 x - a 1 x - b 1 x - g ÷ø
13. f( x) =
1 + sin -1 x
1 - tan -1 x
1
ln f( x) = [ln(1 + sin -1 x) - ln(1 - tan -1 x)]
2
46
Solution of Advanced Problems in Mathematics for JEE
ù
f ¢( x) 1 é
1
1
ú
= ê
+
f( x) 2 ê(1 + sin -1 x) 1 - x 2 (1 - tan -1 x)(1 + x 2 ) ú
ë
û
f ¢(0) = 1
\
2
14. sin x = - sin 2 x Þ 2 sin 2 x = 0 Þ x = np
tan x
tan x < cot x
15. f(x)
cot x
tan x ³ cot x
p 3 p 5p 7 p
Points of non-derivability = ,
,
,
4 4 4 4
16. g( x) =||| x - 1| - 1| - 1|
= x -3
x>3
= - ( x - 3)
2< x<3
18.
d2 x
dy 2
d2 y
1
=-
æ dy ö
ç ÷
è dx ø
3
dx 2
dy
d2 y
=1+ ex ,
= ex
dx
dx 2
at x = ln 2 ,
d2 x
dy
2
=
dy
d2 y
= 3,
=2
dx
dx 2
-2
27
19. g¢( f( x)) =
1
f ¢( x)
f( x) = -4 at x = -2
1
1
Þ g¢( -4) =
=
f ¢( -2) 2
20. f( x) = 2 - x
=x
= -x
=x +2
x³1
0£ x<1
-1 £ x < 0
x < -1
21. f( x) = cos x 2
f ¢( x) = -2 x sin x 2
22. f( g( x)) = x Þ f ¢( g( x)) g¢( x) = 1 Þ g¢( x) =
g¢¢( x) = 5( g( x)) 4 g¢( x)
1
= 1 + ( g( x)) 5
f ¢( g( x))
47
Continuity, Differentiability and Differentiation
23. f( x) = x 2
x³1
=x
= 2x
0£ x£1
-1 £ x £ 0
= x -1
x £ -1
Clearly it is non-differentiable at x = 0 , - 1 and 1.
x
x
x
x ö
æ
24. f( x) = lim ç cos × cos
× cos
¼¼ cos
÷ = lim
2
3
n ®¥ è
2
2
2
2 n ø n ®¥
æ p- ö
æ +ö
÷ = f æç p ö÷ = f ç p ÷
25. f ç
ç 4 ÷
ç 4 ÷
è4ø
è
ø
è
ø
æp
ö
tan ç - x ÷ × (1 + tan x)
4
1
è
ø
=2
pö
æ
4ç x - ÷
4
4ø
è
æ 1 - tan x ö
lim ç
÷ = lim
p
p
x® è 4 x - p ø x®
4
-
f(0 + h) - f(0)
= lim
h ®0
h ®0
h
26. f ¢(0) = lim
27.
e
1
h 2 sin 1
h
h
dy
= 2 y + 10
dx
dy
= 2 dx
y +5
ò
ò
ln( y + 5) = 2 x + c
y = 5 ( e 2 x - 1)
(Q c = ln 5)
f( x) + 5 sec 2 x = 0 Þ e 2 x + tan 2 x = 0
æ p- ö
÷ = lim sin{cos x} = lim sin(cos x) = -1
28. f ç
ç 2 ÷
p
p
pè
ø x® p
xxx®
2
2
2
2
æ p+ ö
÷ = lim sin{cos x} = lim sin(cos x + 1)
fç
ç 2 ÷
+
p
p
p+
è
ø x® p
xxx®
2
2
2
2
29. Let g( x) = f( e x )
g¢( x) = f ¢( e x ) × e x
g¢¢( x) = f ¢¢( e x ) × e 2 x + f ¢( e x )e x
30. e f ( x ) = ln x Þ f( x) = ln(ln x) Þ g( x) = f -1 ( x) = e e
x
x
g¢( x) = e e × e x = e e + x
x
sin x
sin x
=
x
x
æ
ö
2 n sin ç
÷
n
è2 ø
48
Solution of Advanced Problems in Mathematics for JEE
32. ln f( x) = 4 ln( x - 1) + 3 ln( x - 2) + 2 ln( x - 3)
f ¢( x)
4
3
2
=
+
+
f( x) x - 1 x - 2 x - 3
3
2 ö
æ 4
f ¢( x) = f( x) ç
+
+
÷
è x -1 x -2 x -3ø
34. f(2 + ) = 0 Þ c = 0
f (2 - ) =
b sin{- x}
= f (2 + ) = 0 Þ b = 0
{- x }
e tan x - e x + ln(sec x + tan x) - x
x® 0
tan x - x
35. f(0) = lim
= lim e x
x® 0
ln(sec x + tan x) - x
( e tan x - x - 1)
sec x - 1
1 3
+ lim
= 1 + lim
=1+ =
2
x
®
0
x
®
0
tan x - x
tan x - x
2 2
sec x - 1
36. f(0 - ) = e a
f ( 0) = b
c =1
f(0 + ) =
37.
2
2
Þ b = ea =
3
3
x+ y+
y - x =5
x + y =5 -
y-x
Sq. both sides,
Þ
x + y = 25 + y - x - 10 y - x
Þ
25 - 2 x = 10 y - x
Þ
-2 =
Þ
-2 y - x = 5 ( y ¢ - 1)
Þ
Þ
10 ( y ¢ - 1)
2 y-x
2x ö
æ
-ç5 ÷ = 5 ( y ¢ - 1)
5 ø
è
2x
-5 +
= 5 ( y ¢ - 1)
5
2
y ¢¢ =
25
38. g( x) = f -1 ( x)
Þ
Þ
f( g( x)) = x
f ¢( g( x)) g¢( x) = 1
Continuity, Differentiability and Differentiation
Þ
g¢(2) =
Þ
f(1) = 2
g(2) = 1
Þ
g¢(2) =
1
f ¢( g(2))
1
f ¢(1)
f ¢( x) = 3 x 2 + 4 x 3 +
Þ
1
x
f ¢(1) = 8
1
g¢(2) =
8
39. f( x) =| x|
=x
x Î ( -¥, - 1)
2
x Î [-1, 1)
= 2x - 1
x Î [1, ¥)
Function is not differentiable at x = -1.
40. g( x) = ( f( x)) 2 + ( f ¢( x)) 2 Þ g¢( x) = 2 f( x) f ¢( x) + 2 f ¢( x) f ¢¢( x)
or g¢( x) = 2 f( x) f ¢( x) - 2 f( x) f ¢( x) = 0 Þ g( x) = c Þ g(8) = 8
æ æ a ö ö
ç fç
÷-1 ÷
xø ÷
lim ç è
ç
÷
1
x ®¥
x
çç
÷÷
öö
x
è
ø
÷÷ = e
æ æ a
41. l = lim çç f çç
÷
x ®¥ è è x ø ÷
ø
Using L’ Hospital’s rule, we get
a2
a2
f ¢¢( 0)
l=e 2
=e 2
42.
d
d
d
fn ( x) = e fn-1 ( x )
fn -1 ( x) = fn ( x)
fn -1 ( x)
dx
dx
dx
= fn ( x) fn -1 ( x) ¼¼ f2 ( x) f1 ( x)
43. y = tan
-1
( x 1/ 3 ) - tan -1 ( a 1/ 3 )
44. f( x) is continuous at x = 0 then
45. Put x = sin q then y = tan -1 tan
4k - 1 4k + 1
=
3
5
q
2
e x cos x - ln(1 + x) - 1
x ®0
x
46. lim
( e x - 1)
ln(1 + x) æ cos x - 1 ö
cos x +ç
÷ =0
x ®0
x
x
x
è
ø
lim
49
50
Solution of Advanced Problems in Mathematics for JEE
y = 1–cos x
2
y = cos x
1
y = sin x
47.
p
p/2
2p
3p/2
Clearly 3 sharp points.
48. g( x) = f -1 ( x)
f( 4) = 2 Þ g(2) = 4
1
1
Þ g¢(2) = 16
G( x) =
f ¢( 4) =
g( x)
16
-1
-1
-1
G ¢( x) =
× g¢( x) Þ G ¢(2) =
× g¢(2) =
× 16 = -1
2
2
16
( g( x))
( g( x))
y
49.
1/81
O
1
9
x
1
1ö 1
æ
f( x) = maximum ç x 4 , x 2 , ÷ =
81 ø 81
è
= x2
= x4
1
f( x) is non-differentiable at x = , 1
9
50. lim
x£
1
9
1
< x<1
9
x³1
ln( f(2 + h 2 )) - ln( f(2 - h 2 ))
h2
Apply L’ Hospital rule,
h ®0
2 hf ¢(2 + h 2 )
lim
h ®0
f (2 + h 2 )
+
2h
2 hf ¢(2 - h 2 )
f (2 - h 2 )
=4
pö
æ
51. f( x) = ( x 2 - 3 x + 2)|( x - 1)( x - 2)( x - 3)| + sin ç x + ÷
4ø
è
3p 7p
Not differentiable at x = 3 ,
,
4 4
52. h( x) = f(2 x g( x) + cos px - 3)
h¢( x) = f ¢(2 x g( x) + cos px - 3)[2 g( x) + 2 xg¢( x) - p sin px]
h¢(1) = f ¢(2 g(1) - 4)[2 g(1) + 2 g¢(1)] = 32
Continuity, Differentiability and Differentiation
53. f( x) =
( x + 1) 7 1 + x 2
( x 2 - x + 1) 6
( f(0) = 1)
1
ln(1 + x 2 ) - 6 ln( x 2 - x + 1)
2
f ¢( x)
6 (2 x - 1)
7
x
=
+
2
f( x) 1 + x 1 + x
x2 - x + 1
ln f( x) = 7 ln(1 + x) +
f ¢(0) = 13
ì
ï - sin 2 x ;
ï
54. f( x) í ln(1 + x) ;
ïln 2 - sin 2
;
ïî
2
f( x) ;
55. f( f( x)) é
ëê1 - f( x) ;
x>1
x<1 ;
f(1+ ) ¹ f(1- ) ¹ f(1)
x =1
if f( x) is rational
if f( x) is irrational
x;
if x is rational
é
êë1 - (1 - x) ; if x is irrational
x ; if x is rational
f( f( x)) é
êë x ; if x is irrational
f( f( x))
y = |x2–x–2|
y = x2–3x
56.
0
1
3
2
y = ln(–x)
1
y = ex
a
1
So, graph of f( x) is
2
ln(–x)
ex
a
1
2
Clearly, 3 non-differentiability points.
51
52
Solution of Advanced Problems in Mathematics for JEE
57. g( f( x)) = x
58.
ln(2 - cos 2 x)
lim
x ® 0- ln 2 (1 + sin 3 x )
lim
1 - cos 2 x
2
-
e sin 2 x - 1
x ® 0+ ln(1 + tan 9 x )
= K = lim
sin 2 x
tan
9x
x ®0
= K = lim
sin 3 x
dx
3
2 -3 - 2t
59.
==
4
dt
t
t3
t4
x ®0
+
dy -3 2 -3 - 2t
=
=
dt t 3 t 2
t3
dy
=t
dx
3
dy
æ1+ t ö 3
æ dy ö
- xç ÷ =t -ç
÷ × t = -1
dx
dx
è ø
è t3 ø
60. -
2
y3
( y ¢) 2
y6
( y ¢) 2
y6
y ¢ = 2 2( -2 sin 2 x)
æ 1
ö
= 8 - (2 2 cos x) = 8 - ç
- 1÷
ç y2
÷
è
ø
2
=
8 y 2 - (1 - y 2 ) 2
y4
( y ¢) 2 = 8 y 4 - y 2 (1 - y 2 ) 2 then diff.
61. f( x) = x satisfy the equation.
\ f(5) = 5
x£0
é x
62. f( x) ê x 2
0< x<1
ê2 x - 1
x³1
ë
x£0
é1
f ¢( x) ê2 x 0 < x < 1
x³1
êë 2
f( x) is not derivable at x = 0.
63. y = ( x + 1 + x 2 ) n
dy
ny
=
dx
1+ x2
2
53
Continuity, Differentiability and Differentiation
yx ù
é
2
ê 1 + x y¢ ú
d2 y
dy
d y
1+ x2 ú
=nê
Þ (1 + x 2 )
+x
=n2 y
2
2
2
ê
ú
dx
dx
1+ x
dx
ê
ú
ë
û
2
2 ö
2ö æ
æ
ö æ
x
÷ = ç 1 - æç x - 1 - x 2 ö÷ ÷ × ç x + 1 - x ÷
64. g¢( x) = f ¢( x - 1 - x 2 ) × ç 1 +
ç
÷ ç
÷
è
ø ÷ø ç
1- x2 ø è
1- x2 ø
è
è
= 2 x æç x + 1 - x 2 ö÷
è
ø
f( x + h) - f( x)
æ f( h) - 1 ö
= lim f( x) ç
÷ = f( x) × f ¢(0) = 3 f( x)
h ®0
h ®0
h
h
è
ø
66. f ¢( x) = lim
67. f( x) = lim
log e (2 + x) - x 2n sin x
1 + x 2n
n ®¥
ìln(2 + x) | x|< 1
ï- sin x
| x|> 1
ïïln 3 - sin 1
f( x) = í
x =1
2
ï
ïsin 1
x = -1
ïî 2
68.
lim
x - e x + 1 - {1 - cos 2 x}
x2
x® 0
Þ
lim
x - e x + cos 2 x
x® 0
x
2
x - e x + 1 - 1 + cos 2 x
Þ lim
x2
x® 0
= lim
x® 0
1+ x - ex
x
2
+
(cos 2 x - 1)
x
2
69.
y = |x2–10x+21|
(0,21)
(0,4)
0
71.
(3,0)
xy = const.
y + xy ¢ = 0 Þ y ¢ = -
y
x
72. f( x) = -1 + | x - 2| is a continuous function.
(7,0)
x
=-
5
2
(Q f ¢(0) = 3)
54
Solution of Advanced Problems in Mathematics for JEE
g( x) = 1 -| x| is a continuous function.
Þ
f( g( x)) is a continuous function.
f( k + h) - f( k)
h ®0
h
73. f ¢( K + ) = lim
= lim
h ®0
K tan( pk + ph) - k tan kp
h
æ tan ph ö
= lim k ç
÷ = kp
h ®0 è
h ø
74.
lim
ae sin x + be - sin x - c
x2
x ®0
=2
a + b- c =0
Applying L Hospital Rule,
ae sin x × cos x - be - sin x × cos x
=2 Þ a = b
x ®0
2x
lim
75. tan x = sec a × tan y
sec 2 x = sec a × sec 2 y × y ¢
æp pö
y ¢ = 1 at ç , ÷
è4 4ø
2 sec 2 x tan x = sec a (sec 2 y × y ¢¢ + 2 sec 2 y × tan y × ( y ¢) 2 ) Þ y ¢¢ = 0
y = ( x 2 - 9)( x 2 - 4)( x 2 - 1) x
76. We gave,
= { x 6 - 14 x 4 + x 2 ( 49) - 36} x
= x 7 - 14 x 5 + 49 x 3 - 36 x
dy
= 7 x 5 - 70 x 4 + 147 x 2 - 36
dx
Therefore,
d2 y
Thus,
dx
2
= 42 x 5 - 280 x 3 + 294 x
2
d y
dx 2 x =1
77. f ¢( x) = lim
h ®0
Þ
= 42 - 280 + 294 = 56
f( x + h) - f( x)
f( x) f( h) - f( x)
æ f( h) - 1 ö
= lim
= lim f( x) ç
÷
h ®0
h ®0
h
h
h
è
ø
f ¢( x) = f ¢(0), f( x) Þ f( x) = e kx
(where k = f ¢(0))
…(1)
55
Continuity, Differentiability and Differentiation
78. f( g( x)) = x ;
79.
f ¢( g( x)) g¢( x) = 1 Þ g¢(6) =
1
1
=
f ¢( g(6)) f ¢(0)
dy dy dx f ¢( x)
=
=
dz dz dx g¢( x)
d æ dy ö
ç ÷
d y
d æ dy ö dx è dz ø g¢f ¢¢ - f ¢g¢¢
=
=
=
ç
÷
dz
dz 2 dz è dz ø
( g¢) 3
dx
2
f(x)
80.
ì f( x) + 1 = x + 2 , x Î ( -¥, - 1) x = -1, 1non differentiable
ï
g( f( x)) = í( f( x) - 1) 2 = ( x + 1 - 1) 2 = x 2 , x Î ( -1, 0)
ï
2
î(| x - 1| - 1) , x ³ 0
81. f( x) = [sin x] + [cos x]
1
O
p/2
p
, x Î [0 , p]
ìcos x
82. g( x) = í
îsin x - 1 , x > p
g( p - ) = g( p) = g( p + ) = -1
but not differentiable at x = p.
f r (0) f(0) f ¢(0) f ¢¢(0)
=
+
+
+¼
r!
0!
1!
2!
r =0
¥
83.
å
= 4n +
n × 4 n -1 n(n - 1) × 4 n -2
+
+¼
1!
2!
= n C 0 4 n + n C 1 4 n -1 + n C 2 4 n -2 +¼
= ( 4 + 1) n = 5 n
3p/2
2p
56
Solution of Advanced Problems in Mathematics for JEE
84. f( x) =
x
1- x
x £ -1
=
x
1+ x
-1 < x < 0
=
x
1- x
0£ x<1
=
x
1+ x
x³1
Function is discontinuous at x = -1, 1
f( x) is not differentiable at x = -1, 1
85. f( g( x)) = x
æ -7 ö
f ¢( g( x)) g¢( x) = 1 Þ g¢ ç
÷=
è 6 ø
86. f( x) = 0
1
æ æ -7 ö ö
f ¢ç gç
÷÷
è è 6 øø
=
1
f ¢(1)
x³0
2
= 4 x (1 - 2 x)
2
x<0
Differentiable everywhere.
88. f( x) is discontinuous at x = 1, 2
x=1
Þ g( x) = x 2 - ax + b = 0
x=2
89. f -1 ( f( x)) = x
( f -1 ( f( x))) ¢ f ¢( x) = 1
( f -1 ( f(9))) ¢ f ¢( g) = 1
( f -1 (3)) ¢ =
1
1
=
¢
f ( 9) 5
90. f(0 + ) = lim h n sin
h ®0
1
=0 Þ n > 0
h
æ 1ö
f(0 - ) = lim ( -h) n sin ç - ÷ = 0 Þ n > 0
h ®0
è hø
f ¢( x) = n × x n -1 × sin
1
1
- x n -2 × cos = finite Þ n = 2
x
x
57
Continuity, Differentiability and Differentiation
Exercise-2 : One or More than One Answer is/are Correct
1. f( x) has exactly one point of discontinuity so that sgn ( x 2 - lx + 1) is equal to zero for some
values of l.
D =0
\
Þ
2
l - 4 =0
l = ±2
Þ
2. Answer from the graph.
y
2–x
x–2
–1
1
2
x
2(x+1)
3. (a) L.H.L. = lim
x(3 e1/ x + 4)
x ® 0-
R.H.L. = lim
2-e
1/ x
x(3 e1/ x + 4)
x ® 0+
2 - e 1/ x
æ4ö
=0ç ÷ =0
è2ø
æ 3 + 4 e -1/ x ö
÷ = 0 æç 3 ö÷ = 0
= lim x ç
ç
1
/
x
+
x ®0
è -1 ø
- 1 ÷ø
è 2e
f(0) = 0
\
f( x) is continuous at x = 0.
æ 3 e 1/ x + 4 ö
ç
÷
ç 2 - e 1/ x ÷
f ( x ) - f ( 0)
è
ø
+
(b) f ¢(0 ) = lim
= lim x
+
+
x
0
x
x ®0
x ®0
= lim
3 + 4 e -1/ x
x ® 0+ 2 e -1/ x - 1
= -3
f ( x ) - f ( 0)
3 e 1/ x + 4 4
= lim
= =2
x -0
2
x ® 0x ® 0- 2 - e 1/ x
f ¢(0 - ) = lim
f ¢(0 + ) ¹ f ¢(0 - )
(c) f ¢(0 + ) = -3
(d) f ¢(0 - ) = 2 exist
4. Given | f( x)| £ sin 2 x
Clearly | f(0)| £ 0 Þ f(0) = 0
58
Solution of Advanced Problems in Mathematics for JEE
lim | f( x)| = lim f( x) = 0
x® 0
x® 0
| f ¢(0)| = lim
x® 0
f ( x ) - f ( 0)
f( x)
= lim
£0
x
®
0
x
x
é
æ
æ
x 3 ö÷ ù
x 2 x 4 ö÷
a ê1 - x ç x ¼ ú + bç 1 +
¼ +5
ç
÷
ç
÷
3
!
2
!
4
!
ê
ú
è
ø
è
ø
ë
û
5. f(0 - ) = lim
= f ( 0)
2
x® 0
x
bö
æ
( a + b + 5) - ç a + ÷ x 2 +¼
2ø
è
= lim
=3
2
x® 0
x
a + b+ 5 =0
…(1)
bö
æ
-ç a + ÷ = 3
2ø
è
2a + b = 6
On solving (1) and (2),
a + b+ 5 =0
2a + b + 6 = 0
– – –
-a -1=0
a = -1
b = -4
a + b = -5
\
+
f ¢(0 ) is exist when c = 0
lim (1 + dx) 1/ x = 3
x® 0
lim
e
1
x®0 x
( dx )
=3
d
e =3
Þ
d = ln 3
2
7. (a) f( x) = 3 x | x| - 1 -| x|
But x 2| x| =| x|3
So, f( x) =| x| - 1 -| x| = -1 is every where differentiable.
So, no where non-differentiable.
…(2)
59
Continuity, Differentiability and Differentiation
(b) lim [ x(tan -1 ( x + 1) - x tan -1 ( x + 1))] + [5 tan -1 ( x + 1) - tan -1 ( x + 1)]
x ®¥
(c) f( - x) = sin æç ln æç - x +
è è
= sin æç - ln æç x +
è
è
æpö
= lim 4 tan -1 ( x + 1) = 4 ç ÷ = 2 p
x ®¥
è2ø
æ
ö
1
÷
x 2 + 1 ö÷ ö÷ = sin ç ln
ç
÷
2
øø
è x + x + 1ø
x 2 + 1 ö÷ ö÷ = - sin æç ln æç x + x 2 + 1 ö÷ ö÷
øø
øø
è è
= - f( x)
So, f( x) is an odd function.
(d) f( x) =
4 - x2
4x - x3
is discontinuous at where denominator is zero, 4 x - x 3 = 0
x = 0, x = ± 2
a, b, c only correct.
8. g¢( x) = ae ax + f ¢( x) Þ g¢(0) = a - 5
Þ
\
g¢¢( x) = a 2 e ax + f ¢¢( x)
g¢¢(0) = a 2 + 3
Þ
a 2 + a - 2 = 0 ; a = -2 , 1
10. f(0 + ) = f(0) = f(0 - ) = 0
11.
ò f ¢( x) dx = ò f ¢( - x) dx
Þ
f( x) + f( - x) = c
12. | f( x)| £ x 4n
Þ
f(0) = 0
lim ( -h 4n ) £ lim f(0 + h) £ lim ( h) 4n
h® 0
h® 0
h® 0
Þ f(0 + h) = 0
lim ( - ( -h) 4n ) £ lim f(0 - h) £ lim ( -h) 4n Þ f(0 - h) = 0
h® 0
h® 0
Þ
f( x) is continuous at x = 0.
h® 0
f(0 + h) - f(0)
=0
h® 0
h
f ¢(0 + ) = lim
Þ
f( x) is differentiable at x = 0.
13. g( x) = 0
=x
xÎI
2
xÏI
gof( x) = 0 for x Î R
é
f(0 + h) - f(0)
-h 4 n
h 4n ù
£ lim
£ lim
ê lim
ú
h® 0
h® 0 h
h
ëh ® 0 h
û
60
Solution of Advanced Problems in Mathematics for JEE
14. If f( x) is continuous at x = 2 then 3 p + 10 q = 4
f( x) is differentiable at x = 2 then 2 p + 11q = 4
16. f( x) = x 2
-2 £ x £ 0
=x
0< x<1
=x
3
1£ x £ 2
f( x + h)
-1
f( x + h) - f( x)
f( x)
17. f ¢( x) = lim
= f( x) lim
h® 0
h® 0
h
h
hö
æ
f ç1 + ÷ - 1
xø
è
= f( x) lim
h® 0
h
So, f ¢( x) =
f( x)
× f ¢(1)
x
ln( f( x)) = 3 ln x + ln c
f( x) = cx 3
f(1) = 1 so c = 1
f( x) = x 3
So, we can check options.
18. f( x) = ( x - 1)( x - 2)( x + 1)( x + 2) = ( x 2 - 1)( x 2 - 4)
f ¢( x) = ( x 2 - 1) 2 x + ( x 2 - 4)(2 x) = 2 x(2 x 2 - 5) = 0
x = 0, ±
5
2
19. If f( x) is continuous at x = 2 then 3 p + 10 q = 4
f( x) is differentiable at x = 2 then 2 p + 11q = 4
3
20. y = e x sin x + e x ln (tan x )
3
dy
1
æ
ö
= e x sin( x ) [ x cos( x 3 )3 x 2 + sin( x 3 )] + e x ln(tan x ) ç ln(tan x) + x
sec 2 x ÷
dx
tan x
è
ø
3
y ¢ = e x sin( x ) [3 x 3 cos( x 3 ) + sin( x 3 )] + (tan x) x (ln(tan x) + 2 x cosec 2 x)
21. f( x) = 1 - (1 - x) + (1 - x) x 2 + (1 - x)(1 - x 2 ) x 3 +¼ + (1 - x)(1 - x 2 )¼(1 - x n -1 ) x n
= 1 - (1 - x)(1 - x 2 )(1 - x 3 ) ¼(1 - x n ) = 1 -
n
Õ(1 - x r )
r =1
n
( f( x) - 1) = -
Õ(1 - x r )
r =1
61
Continuity, Differentiability and Differentiation
22. \
Þ
f and g must be continuous.
1+ a =2 + b
a =1+ b
3 + b = 1 Þ b = -2
a = -1
ax3+b ;
0£x£1
2cos px + tan–1x ;
1<x£2
23. f(x)
is must be continuous and differentiable at x = 1.
p
…(1)
\
a + b = -2 +
4
1
…(2)
3a = 0 +
2
We get, a and b
(continuity)
(By differentiable)
24. f( f( x)) = 2 + x
0£ x£1
=2 - x
1< x £ 2
=4-x
2< x£3
25. ln( f( x)) = ln( x + 1) + ln( x + 2) +¼¼ + ln( x + 100)
f ¢( x)
1
1
1
=
+
+¼¼ +
f( x) x + 1 x + 2
x + 100
f( x) f ¢¢( x) - ( f ¢( x)) 2
( f( x)) 2
if
æ
ö
1
1
1
1
÷
= -ç
+
+
+¼¼ +
ç ( x + 1) 2 ( x + 2) 2 ( x + 3) 2
( x + 100) 2 ÷ø
è
g( x) = f( x) f ¢¢( x) - ( f ¢( x)) 2 = 0
1
Þ
( x + 1)
2
+
1
( x + 2)
2
+¼¼ +
1
( x + 100) 2
=0
Þ g( x) = 0 has no solution.
26. h( x) = -1
x<1
=| x - 2| + a + 2 -| x|
1£ x < 2
=| x - 2| + a + 1 - b
x³2
if h( x) is continuous at x = 1, then a = -3
if h( x) is continuous at x = 2, then b = 1
27. lim f( x) = 1 = lim f( x)
x ® 0-
x ® 0+
Clearly, C = 1 and use L’ Hospital’s rule.
28. Differentiable w.r.t. ‘ x’
2 f( x) f ¢( x) + 2 y = 2 f( x + y) f ¢( x + y)
62
Solution of Advanced Problems in Mathematics for JEE
put x = 0
k + y = f ¢( y) f( y)
integrate on both sides,
y2
f 2 ( y)
=
+c
2
2
put x = y = 0 in given equation, we get
ky +
f 2 ( 0) = 2
f(0) = 2 as ( f( x) > 0)
put y = 0 in (1)
1 + c = 0 Þ c = -1
also put y = 2
k 2 + 1=
4
-1
2
k 2 =0
k =0
2
y
f 2 ( y)
=
-1
2
2
\
f 2 ( y) = y 2 + 2
f( y) =
y2 + 2
f( x) = x 2 + 2
Hence, we can answer.
f( x + h) - f( x)
f( x + h) - f( x)
30. f ¢( x) = lim
= lim
h ®0
h
®
0
h
h
f( h)
2
2
= lim
+ x = x + f ¢(0) ; f ¢( x) = x 2 - 1
h ®0 h
1
32. f(1- ) = f(1+ ) = f(1) =
2
f ¢( x) = x
0£ x<1
= 4x - 3
f ¢¢( x) = 1
=4
34. gof( x) = 0
fog( x) = 0
2
= [x ]
1£ x £ 2
0£ x<1
1£ x £ 2
xÎI
xÏI
…(1)
63
Continuity, Differentiability and Differentiation
35. f( g( x)) = x
f ¢( g( x)) g¢( x) = 1
1
g¢( x) =
f ¢( g( x))
1
1
1
=
=
f ¢( g( e)) f ¢(1) e + 1
g¢( e) =
g¢¢( x) =
g¢¢( e) =
-1
( f ¢( g( x))) 2
-1
( f ¢(1)) 2
f ¢¢( g( x)) × g¢( x)
f ¢¢(1) × g¢( e)
36. f(2 + ) = lim [ x - 1] = 1
x ® 2+
3x - x2
=1
2
x ® 2-
f(2 - ) = lim
f(3 - ) = lim [ x - 1] = 1
x ® 3-
f(3 ) = lim ( x 2 - 8 x + 17) = 2
+
x ® 3+
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
2. f( x) = lim n 2
n ®¥
f( x) =
tan(ln(sec( x n))) (ln(sec( x n) - 1) + 1) sec( x n) - 1 æ x ö
´
´
´ç ÷
ln(sec( x n))
sec( x n) - 1
ènø
( x n) 2
x2
2
||||
|
1
|||
g(x)
||||
||
||||||||||||||||
–1
1
2
Paragraph for Question Nos. 3 to 4
4. f ¢( x) = 2 x + g¢(1)
f ¢¢( x) = 2
f ¢(1) = 2 - 3 = -1
g¢(1) = 2 f(1) + 2 + f ¢(1)
Þ
f(1) = -2
2
64
Solution of Advanced Problems in Mathematics for JEE
f( x) = x 2 - 3 x
g¢¢(2) = 2( -2) + 2(2) = 0
-2 = 1 + g¢(1)
2
g( x) = -2 x + x(2 x - 3) + 2
g¢(1) = -3
= -3 x + 2
f(1) + g( -1) = -2 + (3 + 2) = 3
Paragraph for Question Nos. 5 to 6
5. Clearly, 3 is non-repeated root where as 1 is repeats and also ( x - 2) 1/ 3 is not diff. at x = 2.
\ at 3, 2 is non-diff. and sum is 5.
6. h( x) is continuous.
x -1= x2 - x -2
x2 - 2x - 1 = 0
( x - 1) 2 = 2
x =1± 2
3p
æpö
tan
= 1 + 2 , tan ç ÷ = 2 - 1
8
è8ø
tan
7p
=1- 2
8
2 - 1 is not differentiable.
Paragraph for Question Nos. 7 to 8
Sol.
–2
–1
1
2
Paragraph for Question Nos. 9 to 10
1
10.
0
–1
p/2
p
3p/2
2p
x
65
Continuity, Differentiability and Differentiation
Paragraph for Question Nos. 11 to 13
0£ x<1
é 0
x
1£ x < 2
ê
11. f( x) = 2( x - 1) 2 £ x < 3
ê
êë3( x - 1)
x =3
No. of values where f( x) is discontinuous = 2
12. f( x) is non-differentiable at x = 1, 2 , 3.
13. No. of integers in the range of f( x) = 5
Paragraph for Question Nos. 14 to 16
Sol. f ¢( x) = lim
h® 0
Þ
f( x) f( h) - f( x)
= f ¢(0) f( x)
h
f( x) = e 2 x
g( x) = x
( f ¢(0) = 2)
2
Paragraph for Question Nos. 17 to 18
Sol. g¢( x) = l sec 2 x + (1 - l) cos x - 1 =
(1 - cos x)(l - f( x))
f( x)
Paragraph for Question Nos. 19 to 21
Sol. f( x) =
x2
x2 + 1
=1
=2
x<0
f(x)
2
1
x =0
x>0
0
Paragraph for Question Nos. 22 to 24
Sol.
f( x) = g¢(1) sin x + ( g¢¢(2) - 1) x
Þ
æpö
f ¢( x) = g¢(1) cos x + g¢¢(2) - 1 Þ f ¢ ç ÷ = g¢¢(2) - 1
è2ø
æpö
f ¢¢( x) = - g¢(1) sin x
Þ f ¢¢ ç ÷ = - g¢(1)
è2ø
æpö
æ pö
g( x) = x 2 - f ¢ ç ÷ × x + f ¢¢ ç - ÷
è2ø
è 2ø
æpö
Þ g¢( x) = 2 x - f ¢ ç ÷
è2ø
g¢¢( x) = 2
Þ g¢¢(2) = 2
f( x) = sin x + x
and
2
g( x) = x - x + 1
x
66
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 25 to 26
f( x) + f( h)
- f( x)
f( x + h) - f( x)
1 + f( x) f( h)
¢
Sol. f ( x) = lim
= lim
h ®0
h ®0
h
h
f ¢( x) = lim f( h)
h ®0
(1 - ( f( x)) 2 )
h
f ¢( x) = lim f ¢(0)(1 - f( x) 2 )
( f ( 0) = 0)
h ®0
Þ
f( x) =
e2x - 1
e2x + 1
f ¢( x) ³ 0 " x Î R
x
lim ( f( x)) = e
lim
-2 x
x ® 0 ( e 2x + 1 )
x ®0
=1
Paragraph for Question Nos. 27 to 28
Sol. f( x) = 3( x + 6)( x + 1)( x - 2)( x - 3) + x 2 + 1 then k why!
3( x + 1)( x - 2)( x - 3)( x + 6)
6!
=x ®-6
x +6
2
1
28. g( x) =
-3( x + 6)( x + 1)( x - 2)( x - 3)
27.
lim
Paragraph for Question Nos. 29 to 30
Sol. f( x) = g( x)
x ln x = e 2 x
(ln x) 2 = 2 + ln x
1
x = a = , b = e2
e
f( x) - cb f ¢( x)
29. lim
=
=4
2
g¢( x)
x ® e g( x ) - b 2
c = e2
30. h¢(a ) =
g(a ) f ¢(a ) - g¢(a ) f(a )
g 2 (a )
=
e( -2 e 2 ) - e 2 - e
( e) 2
= -3e
67
Continuity, Differentiability and Differentiation
Exercise-4 : Matching Type Problems
p
1. (A) Let I =
log(sin x)
ò cos 2 x dx
0
p/ 2
=2
ò
p/ 2
é p/ 2
ù
é
ù
cos x
2
p/ 2
dx = 2 ê log(sin x) sec x dx ú = 2 êlog(sin x) tan x|0 tan x dx ú
sin x
cos 2 x
êë 0
úû
êë
úû
0
log(sin x)
0
ò
ò
p/ 2
= 2(0 - 0) - 2
æpö
ò dx = 0 - 2 çè 2 ÷ø = -p = -k
0
k=p
3k
= 3 > 0 , 1, 2
p
Þ
\
(B) e x + y + e y - x = 1
e x + e-x = e-y
e x - e - x = e - y ( - y ¢)
e x + e - x = e - y ( - y ¢¢) + e - y ( y ¢) 2
e - y = e - y ( - y ¢¢) + e - y ( y ¢) 2 Þ y ¢ - ( y ¢) 2 + 1 = 0
\
(C) Let f
k =1
-1
=g
g{ f( x)} = x Þ ( g¢f( x)) f ¢( x) = 1
g¢(2 ln 2) f ¢(2) = 1
1
g¢2(ln 2) =
1 + ln 2
2
2( f -1 ) ¢(ln 4) =
> 0, 1
1 + ln 2
1
2
(D) l = lim ( x ln x) x +1
x® ¥
ln l = lim
x® ¥
ln x + ln(ln x)
x2 + 1
1
1 1
+
×
= lim x ln x x = 0
x® ¥
2x
ln(l) = 0 Þ l = 1
sec 2,
–2 £ x < –1
sec 1,
–1 £ x < 0
sec x,
0£x£2
2. g(f(x))
68
Solution of Advanced Problems in Mathematics for JEE
2
f(g(x))
1
–4p/3
–p
–2p/3
–p/3
p/3
2p/3
4p/3
p
–1
–2
3. (A) f(1+ ) = f(1- ) = -1
3
3
(B)
3
æ x2
ö
-3
([ x] × { x} -| x|) × dx = (2( x - 2) - x) dx = ç
- 4x÷ =
ç 2
÷
è
ø2 2
2
2
ò
ò
(C) [ x] × { x} = -1
x£0
1
1
x = -3 + , - 2 +
3
2
(D) l = lim f( x) = lim ([ x]{ x} -| x|) = -4
x ® 4+
x ® 4+
2 x +1
æ x 2 + 2 x - 1 ö 2 x -1 1
÷
4. (A) lim ç
=
x ®¥ ç 2 x 2 - 3 x - 2 ÷
2
è
ø
2
log sec x / 2 cos x
æ ln cos x ö
÷ =2
(B) lim
= lim (log x cos x) 2 = lim çç
x
sec
x ®0
x ®0
x ® 0è ln sec x / 2 ÷
ø
2
log sec x cos
2
p
(C)
0
p/2
(D) sin x ¹
1 2 3
, ,
3 3 3
5. f(1+ ) = f(1- ) = f(1) Þ b = 0
f ( 3 - ) = f ( 3 + ) = f ( 3)
3 = 9 p + 3q + 2 Þ 3 p + q = 0
f ¢( x) = 2 ax - a
x<1
=1
1£ x < 3
= 2 px + q x > 3
3p/2
x
Continuity, Differentiability and Differentiation
69
f ¢(3 + ) = f ¢(3 - ) = f ¢(3)
6p + q = 1 Þ p =
1
, q = -1
3
f ¢(1+ ) ¹ f ¢(1- )
a¹1
Exercise-5 : Subjective Type Problems
1. f( x) is discontinuous at x = 1,| f( x)| is diff. every where
f(1) = - f(1+ ) = - f(1- )
3 = - (0 + b)
b = -3
Þ
Þ
f ¢(1+ ) = - f ¢(1- )
(as | f( x)| is differentiable every where)
\
1 = - (2 a - a) Þ a = -1
Continuous at x = 3,
So,
5 = 9 p + 3q + 2
Þ
3p + q = 1
f ¢( x) is continuous at x = 3
f ¢(3 - ) = f ¢(3 + )
So,
Þ
6p + q = 1
On solving (1) & (2) we get, p = 0 , q = 1
So, |a + b + p + q| =|- 1 - 3 + 0 + 1| = 3
2. sin -1 y = 8 sin -1 x
y¢
1- y
2
=
8
1- x2
(1 - x 2 )( y ¢) 2 = 64(1 - y 2 )
(1 - x 2 ) y ¢¢ - xy ¢ = -64 y
3. yy ¢ = 4 a
( y ¢) 2 + yy ¢¢ = 0
4. f( x) is discontinuous at x = - 3 , - 2 , 2 , 3 , 5 , 6 , 7 , 8
sin px = 0 at x = -2 , - 1, 0 , 1, 2 , 3
So, continuous at these points.
5. Let f ¢( x) = K
Þ
…(1)
f( x) = Kx + c
…(2)
70
Solution of Advanced Problems in Mathematics for JEE
Þ f(9) - f( -3) = 12 K
Maximum value of f(9) - f( -3) = 96
6. g( x) = sin x 3 - x 3 + 1
3
x³1
3
= sin x + x - 1
0£ x<1
= - sin x 3 - x 3 - 1
-1 £ x < 0
3
3
= - sin x + x + 1
x £ -1
Function is not differentiable at x = -1, 1
7. F ( x) = g( x)
f( x) + g( x)
=
2
x>1
x =1
= f( x)
f( x) + g( x)
=
2
-1 < x < 1
x = -1
= g( x)
If F ( x) is continuous at x = 1
x < -1
F (1+ ) = F (1) = F (1- )
b=a+3
If F ( x) is continuous at x = -1
F ( -1- ) = F ( -1) = F ( -1+ )
a + b=5
8. f -1 ( x) = 2 - x
2£ x£5
=2 + x
-2 < x < 2
9. f( x) + 2 f(1 - x) = x 2 + 2
f (1 - x) + 2 f ( x) = (1 - x) 2 + 2 Þ f ( x) =
( x - 2) 2
3
10. g( x) = x( x - 3)( x - 7)
f( g( x)) = sgn( x( x - 3)( x - 7))
11.
d2
dx 2
(sin 2 x - sin x + 1) = -4 sin 2 x + sin x + 2
12. f( x) = a cos( px) + b
f ¢( x) = -ap sin( px)
3/ 2
2a
2
ò f( x) dx = - p + b = p + 1 Þ a = -1, b = 1
1/ 2
13. a ¢( x) = f ¢( x) - 2 f ¢(2 x) ; b ¢( x) = f ¢( x) - 4 f ¢( 4 x)
a ¢(1) = f ¢(1) - 2 f ¢(2) = 5
Continuity, Differentiability and Differentiation
71
a ¢(2) = f ¢(2) - 2 f ¢( 4) = 7
b ¢(1) = f ¢(1) - 4 f ¢( 4) = a ¢(1) + 2a ¢(2) = 5 + (2 ´ 7) = 19
b ¢(1) - 10 = 19 - 10 = 9
14. g( f( x)) = x
g¢( f( x)) f ¢( x) = 1
f(1) = - 7 6
\ x =1
æ 7ö
g¢ ç - ÷ f ¢(1) = 1
è 6ø
1
æ 7ö
g¢ ç - ÷ =
¢
6
f
(1)
è
ø
1- x
æ 1ö
f ¢( x) = -4 × e 2 ç - ÷ + x 2 + x + 1
è 2ø
f ¢(1) = 2 + 1 + 1 + 1 = 5
h( x) = ax
-
5
1
4 + bx 4
-9
-3
5a 4
b
x + x 4
4
4
5a -9/ 4 b -3/ 4
h¢(5) = 0 Þ ×5
+ ×5
=0
4
4
h¢( x) = -
Þ
Þ
5a × 5 -3/ 2 = b
a
= 51/ 2
b
2
æ aö
ç ÷ =5
è bø
a2
5
=
=5
1
2 æ -7 ö
5
´
¢
5b g ç
÷
5
è 6 ø
16. Let
x
æ
ö
lim ç f( x) + f(t ) dt ÷ = l
÷
x® ¥ ç
0
è
ø
ò
æ x
ö
ç e x f(t ) dt ÷
ç
÷
è 0
ø
lim
=l
x
x® ¥
(e )
ò
…(1)
72
Solution of Advanced Problems in Mathematics for JEE
x
ex
0
lim
Þ
ò f(t) dt
ex
x® ¥
=l
x
lim
Þ
x® ¥
ò f(t) dt = l
0
From (1) and (2) we get, lim f( x) = 0
x® ¥
17. f(0) = 0 , f ¢(0) = 1, f ¢¢(0) = 1, f ¢¢¢(0) = 2
g( f( x)) = x
Þ g¢( f( x)) f ¢( x) = 1
- f ¢¢( x)
Þ g¢¢( f( x)) =
( f ¢( x)) 3
Þ
é( f ¢( x)) 3 × f ¢¢¢( x) - 3 ( f ¢¢( x)) 2 ( f ¢( x)) 2 ù
g¢¢¢( f( x)) f ¢( x) = - ê
ú
( f ¢( x)) 6
êë
úû
Put x = 0
é 1 ´ 2 - 3 ´ 1ù
g¢¢¢(0) 1 = - ê
úû = 1
1
ë
æ æ
h öö
f ç x ç 1 + ÷ ÷ - f( x)
x øø
f( x + h) - f( x)
è è
19. f ¢( x) = lim
= lim
h ®0
h ®0
h
h
hö
æ
f ç1 + ÷
f( x)
xø
hö
æ hö
æ
è
+
- f( x)
f( x) ç - ÷ f ç 1 + ÷
1+ h x
x
x
xø
è
ø
è
= lim
= lim
+
h ®0
h ®0
h
hx
hö
æ
hç1 + ÷
x
è
ø
hö
æ
f ç 1 + ÷ - f(1)
- f( x)
xø
è
(as f(1) = 0)
=
+ lim
h ®0
x
2æ h ö
x ç ÷
è xø
- f( x) f ¢(1)
f ¢( x) =
+
x
x2
1
xf ¢( x) + f( x) =
x
d
1
( xf( x)) =
dx
x
1
xf( x) =
dx
x
ò
xf( x) = ln x + k
…(2)
73
Continuity, Differentiability and Differentiation
Put x = 1, we get k = 0
ln x
\ f( x) =
x
1
x
H( x) =
=
f( x) ln x
H ¢( x) =
ln x × 1 - 1
(ln x)
–
H( x) ³ e
2
0
H( e) = e
é 1 ù
lim ê
ú =2
x ® e ë f( x) û
\
21. f ¢( x) = tan -1 ( x 2 ) +
2x2
1+ x
4
+ 4x3
æ (1 + x 4 ) × 2 x - x 2 ( 4 x 3 ) ö
÷ + 12 x 2
+ 2ç
ç
÷
4 2
1+ x4
(
1
+
x
)
è
ø
dy dy dq
3
22.
=
= -3 sin q cos q = - sin 2q
dx dx dq
2
f ¢¢( x) =
d2 y
dx
2
=
2x
-3 cos 2q
sin q
23. Let 8 x - 16 = t 2 Þ
24. f( x) = [ x]
= { x}
= x-
3
2
t 2 + 16 + 8t
t 2 + 16 - 8t |t + 4| + |t - 4|
+
=
8
8
2 2
0< x<1
5
1£ x <
4
5
£ x<2
4
No. of points where f( x) is non-differentiable are three.
5 3
x = 1, ,
4 2
❑❑❑
+
e
74
Solution of Advanced Problems in Mathematics for JEE
4
APPLICATION OF DERIVATIVES
Exercise-1 : Single Choice Problems
1. Maximum value of f( x) = 3
Minimum value of f( x) = -1
2. f ¢¢( x) = 6 x - 6
f ¢( x) = 3 x 2 - 6 x + 3
(Q f ¢(2) = 3)
f( x) = x 3 - 3 x 2 + 3 x - 1
(Q f(2) = 1)
4
4
p(10 + T ) 3 - p(10) 3
3
3
dV
dT
= 4 p(10 + T ) 2
dt
dt
dT
1
Þ
=
(QT = 5 cm)
dt 18p
(| x| - 1)(| x| - 2)
6. g( x) =
(| x| - 3)(| x| - 4)
5. V =
g( x) is an even function so there is an extrema at x = 0.
Also number of extrema for x > 0 will be equal to number of extrema for x < 0
for x > 0
( x - 1)( x - 2)
g( x) =
( x - 3)( x - 4)
Number of extrema = 2
Total extrema = 5
Þ
75
Application of Derivatives
2
45 ö
75 2
æ
7. A ¢B ¢ = ç 700 t÷ +
t
2 ø
4
è
m
/s
B
O
-
8. f(0 ) ³ f(0) Þ a ³ 3
3+k–x,
x£k
Þ f ¢( k + ) > f( k), f ¢( k - ) > f( k)
9. f(x)
a2–2+
sin(x–k) ,
x>k
x–k
So, lim ( a 2 - 2) +
x ®k
+
sin( x - k)
= a2 - 1> 3
( x - k)
2
a >4
|a| > 2
dy
10.
= 3 x 2 - 4 x + C1
dx
y = x 3 - 2 x 2 + C1 x + C 2
Also,
11. m 1 =
dy ù
= 0 and y]at x = 1 = 5
dx úû at x = 1
dy ù
= 2a + b
dx úû at (1, 2)
m 2 = g¢( x) =
Also,
12. 18 y
dy ù
1
= 2 Þ 2a + b = dx úû at ( -2, 2)
2
2 =a+ b+
7
2
dy
= 3x2
dx
dy 3 x 2
=
dx 18 y
Þ
a2
= 1 Þ a 2 = 6b
6b
Also, 9 b 2 = a 2
dy
13.
= 3x2 - 4x + c
dx
dy
at x = 1,
=0
Þ c =1
dx
B¢
5
( A ¢B ¢) min at t = 30 sec
60°
A¢
700 m
20 m/s
A
76
Solution of Advanced Problems in Mathematics for JEE
dy
= 3x2 - 4x + 1
dx
y = x3 - 2x2 + x + d
at x = 1, y = 5
Þ 5 =1-2 + 1+ d
d =5
Þ
14. A(0 , 2)
5a 2 (3 x 2 ) + 10a(2 x) + 1 + 2
Þ
dy
dy -15a 2 x 2 - 20ax - 1
=0 Þ
=
dx
dx
2
dy
1
at A = dx
2
Equation of normal at A is y = 2 x + 2
Let normal meets the curve at B
5a 2 x 3 + 10ax 2 + x + 4 x + 4 - 4 = 0
5 x(ax + 1) 2 = 0
x =-
1
a
æ -1 -2
ö
So, B ç
,
+ 2÷
a
a
è
ø
\
Slope of tangent at B =
15. f( x) = cos x +
-15 + 20 - 1
=2
2
1
1
cos 2 x - cos 3 x
2
3
f ¢( x) = - sin x - sin 2 x + sin 3 x = 2 sin x(2 cos x + 1)(cos x - 1) = 0
16. Closest distance exist always alone the normal
1 - x dy
\
´
= -1
2 - x dx
P(x, x )
1- x
1
´
= -1
2-x 2 x
Let
x =t
x=
A(2,1)
2+ 3
2
17. Let x = 2 sin q
æ 2 cos 2 q 2 ö
æ 2 + 2 cos q ö
÷ - 2 cos q
y = ln ç
÷ - 2 cos q = ln ç
ç 2 sin 2 q 2 ÷
è 2 - 2 cos q ø
è
ø
qö
æ
= 2 ln ç cot ÷ - 2 cos q
2ø
è
77
Application of Derivatives
dy
1 æ
2 qö
=
ç -cosec
÷ + 2 sin q
dq cot q 2 è
2ø
dy
-2
-2 cos 2 q
=
+ 2 sin q =
dq sin q
sin q
dy
dx
= 2 cos q;
= -cot q
dq
dx
æ
ö
qö
æ
çç y - 2 ln ç cot ÷ + 2 cos q ÷÷ = - cot q ( x - 2 sin q)
2ø
è
è
ø
\
æ
ö
qö
æ
T = çç 0 , 2 ln ç cot ÷ - 2 cos q + 2 cos q ÷÷
2ø
è
è
ø
q
æ
ö
P = ç 2 sin q , 2 ln cot - 2 cos q ÷
2
è
ø
PT 2 = ( 4 sin 2 q + 4 cos 2 q ) = 4
18. g¢( x) = (2 x 2 - ln x) f( x)
f ¢( x) =
1
3x2 -
1
2x
ln x
ln x 2
x2 - x
f ¢( x) =
ln x
x( x - 1)
f ¢( x) =
> 0 " x > 1; f( x) > f(1) Þ f( x) > 0 " x > 1
ln x
For g( x) is increasing
3
g¢( x) > 0 Þ 2 x 2 - ln x > 0 as ( f( x) > 0)
Let H( x) = 2 x 2 - ln x
1 4x2 - 1
=
> 0 when x > 1
x
x
H( x) > H(1) Þ H( x) > 2
\ g¢( x) > 0 " x Î (1, ¥)
\ g( x) is increasing on (1, ¥).
19. f ¢( x) = 3 x 2 + 12 x + a
H ¢( x) = 4 x -
f ¢( x) < 0 in ( -3 , - 1)
Product of the roots =
a
=3 Þ a =9
3
æ 1- x ö
20. f( x) = tan -1 ç
÷
è1+ x ø
f ¢( x) =
æ -2 ö
2
1
ç
÷=
=
>0
2 ç
2 ÷
2
2
æ 1 - x ö è (1 + x) ø 2 (1 + x ) 1 + x
1+ ç
÷
è1+ x ø
1
78
Solution of Advanced Problems in Mathematics for JEE
f ¢( x) is decreasing " x Î R
f(0) = tan -1 (1) =
So, in [0 , 1]
p
(max)
4
f(1) = 0 (min)
21. f ¢( x) = 3 x 2 + 2( a + 2) x + 3 a
\
D£0
a 2 - 5a + 4 £ 0
\ a Î[1, 4]
22. f ¢( x) = 0
cos 2 x - 3 x + x 1/ 3 cos 2 x =
cos x = ±
\
1
=0
2
1
2
1
2
total number is 12.
y
23.
– 2,0
3
–1, 0
2
O 1
,0
2
x
(1,0)
y
2
24.
O
–3
x
–2
b2 + 1 ³ 2
25. f( x) is continuous and differentiable in [-1, 1].
cos x 1
26.
= - sin x 1 Þ x 1 = -cot x 1
x1
Point ( x 1 , cos x 1 ) always lie on
1
y
2
=
1
x2
+1
79
Application of Derivatives
27. x +
a
x2
> 2 " x Î (0 , ¥)
f( x) = x 3 - 2 x 2 + a > 0 " x Î (0 , ¥)
4ö
æ
f ¢( x) = 3 x 2 - 4 x = 3 x ç x - ÷
3
è
ø
4
Minimum value at x =
3
64
32
æ 16 ö
-2ç ÷ + a > 0 Þ a >
27
27
è 9 ø
29. f ¢( x) = cos 2 x + cos x + 2 > 0
f( x) min = f(0) = 0
f( x) max = f(2 p) = 5p
31. f( x) = x 3 - 3 x + c = 0
f ¢( x) = 3 ( x 2 - 1)
Þ
–1
f(1) f( -1) < 0
( c - 2)(2 + c) < 0
32. f ¢( x) = e x ( x - 1)( x - 2) < 0
33.
dy
= 3 ax 2 + 2 bx + c = 0 has one root Þ D = b 2 - 3 ac = 0 Þ b 2 = 6
dx
34. Let x = tan q then y = cos 2 q
dy
=|2 sin q cos 3 q|
dx
dy
p
at q =
dx max
6
35. h( x) = f( x) - g( x) = 2 x - 3 sin x + x cos x
h(0) = 0
h¢( x) = 2 - 2 cos x - x sin x
h¢(0) = 0
h¢¢( x) = sin x - x cos x
h¢¢(0) = 0
æ pö
h¢¢¢( x) = x sin x > 0 " x Î ç 0 , ÷
è 2ø
36. f( x) = 2 tan -1 ( g( x))
= p - 2 tan
-1
g( x)
= - p - 2 tan -1 g( x)
| g( x)| £ 1
g( x) > 1
g( x) < -1
1
80
Solution of Advanced Problems in Mathematics for JEE
f ¢( x) =
2 g¢( x)
=37.
| g( x)| < 1
1 + ( g( x)) 2
2 g¢( x)
| g( x)| > 1
1 + ( g( x)) 2
é 7 é ln(1 + 7 f( x)) ù 1 æ sin f( x) ö ù
lim ê ê
ú - 3 çç f( x) ÷÷ ú = 2
a 3
7 f( x)
x® e ë ë
û
è
øû
38. If f( x) is strictly decreasing for all x,
f ¢( x) = log 1/ 3 (log 3 (sin x + a)) £ 0
Þ
sin x + a ³ 3 " x Î R
Þ
a³ 4
39. f( x) = a ln| x| + bx 2 + x
f ¢( x) =
a
2 bx 2 + x + a
+ 2 bx + 1 =
x
x
if x = 1 and x = 3 are point of extrema.
1
a
Þ = 4 and
=3
2b
2b
y
40.
–3p
–2p
–p
O
x
1
f( x) has local maximum at x = 0.
x
41. f( x) = (t - a) 2n (t - b) 2m +1 dt
ò
1
f ¢( x) = ( x - a) 2n ( x - b) 2m +1
No sign change of f ¢( x) about x = a.
f ¢( x) will change sign from negative to positive at x = b Þ Point of minima.
43. Let point P on the curve y 2 = x 3 is P(t 12 , t 13 ).
Equation of tangent at P(t 12 , t 13 ) is
y - t 13 =
3
t 1 ( x - t 12 )
2
If this intersect the curve again at Q (t 22 , t 23 )
81
Application of Derivatives
Þ
t
t2 = - 1
2
tan a (3t 1 2)
=
= -2
tan b (3t 2 2)
44. y 2 = ax 3 - b
if (2 , 3) is lie on the curve
…(1)
8a - b = 9
Slope of normal at (2 , 3)
1
1
- =4
2a
Þ a =2
45. Equation of tangent at (0 , 1) to the curve y - 1 = kx meet x-axis at ( a , 0) then
1
é1 ù
-2 £ - £ -1 Þ k Î ê , 1ú
k
ë2 û
x
46. f( x) =
ò
-u 2
e x du =
0
1
2
x e -t dt
ò
0
u
where t =
x
Þ f( x) = K x , K > 0
47. f ¢¢(a ) = 0 Þ x = a is the point where concavity changes.
a
b
c
48. f( x) = x 6 - x - 1
f ¢( x) = 6 x 5 - 1 > 0 " x Î [1, 2]
If f(1) = -1 < 0 and f(2) = 2 6 - 3 > 0 then f( x) has one root in [1, 2].
49. Every line passing from ( a , b) is normal to the circle ( x - a) 2 + ( y - b) 2 = k
50. f ¢( x) = cos x(3 sin 2 x - m ) = 0
sin 2 x =
m
3
Þ 0<
m
<1
3
0<m<3
51. Let y = x
1/ x
y ¢ = x (1/ x )-2 (1 - ln x)
f( x) is increasing
and f( x) is decreasing
(0 , e)
( e, ¥)
82
Solution of Advanced Problems in Mathematics for JEE
52. Let y = mx
Point of tangency be ( x 1 , y 1 )
Þ
Þ
mx 1 = x 13 + x 1 + 16 & m = 3 x 12 + 1
x 1 (3 x 12 + 1) = x 13 + x 1 + 16
x1 = 2
m = 13
53. y ¢ = 3 x 2 - 6 x + 6
y ¢¢ = 6 x - 6 = 0
x =1
y¢ = 3
54. Let H( x) = ln( f( x) + f ¢( x) +¼ f n ( x)) - x
H( a) = H( b)
H ¢( c) = 0
Þ
Þ
Þ
f ¢( c) + f ¢¢( c) +¼ + f
n +1
n
( c)
f( c) + f ¢( c) +¼ + f ( c)
Þ
Þ
Þ
Þ
Þ
Þ
-1=0
f n +1 ( c) = f( c)
Þ
55.
(by L.M.V.T.)
h( x) = g( x) + x
h¢( x) = g¢( x) + 1
g¢( x) = h¢( x) - 1
g¢¢( x) = h¢¢( x)
h¢¢( x) - 3( h¢( x) - 1) > 3
h¢¢( x) - 3 h¢( x) > 0
d -3 x
(e
h¢( x)) > 0
dx
Let P( x) = e -3 x h¢( x)
P ¢( x) > 0
P( x) is an increasing function.
P(0) = h¢(0) = 0
Þ P( x ) > 0 " x > 0
Þ h¢( x) > 0 " x > 0
Þ h( x) is an increasing function " x > 0
dy
c
56.
== -1 Þ ( x + 1) 2 = c
dx
( x + 1) 2
Þ
Þ
Point ( c - 1, c ) lie on the line x + y = 3 Þ c = 2
83
Application of Derivatives
57. |sin x| = e - x
2
y
–p/2
O
p/2
p
3p/2
x
60. x n + ax + b = 0
x is even.
nx n -1 + a = f ¢( x)
f¢(x)
–a 1/(n–1)
n
f(x)
61. f( b) = sin x +
2
+b
" xÎR
3 + sin x
max
sin x = t
2
t Î [-1, 1]
3+t
2
g¢(t ) = 1 >0
(3 + t ) 2
g(t ) = t +
(3 + t ) 2 - 2 > 0
(3 + t - 2 )(3 + t + 2 ) > 0
2
increasing " Î [-1, 1]
3+1
3
g(t ) max =
2
–3 – 2 –3 + 2
–1
1
g(t ) = t +
g(t ) min = 0
3
+b
f( b) = 2
-b
if b ³ -
3
4
3
4
æ -3 ö 3
min. of f( b) = - ç
÷=
è 4 ø 4
b< -
62. y =
x
1+ x2
Þ
dy
1- x2
=
dx (1 + x 2 ) 2
–3/2
0
84
Solution of Advanced Problems in Mathematics for JEE
æ xö
63. f -1 ( x) = 2 cos -1 ç ÷
è3ø
d -1
-2 æ 1 ö
f ( x) =
ç ÷
dx
x2 è3ø
19
64. f( x) = sin x + tan x - 2 x
f ¢( x) = cos x + sec 2 x - 2 = 0
cos 3 x - 2 cos 2 x + 1 = 0 Þ (cos x - 1)(cos 2 x - cos x - 1) = 0
1- 5
2
a + 2c 4
65.
+ =0
b + 3b 3
cos x = 1,
Û 3 a + 4 b + 6 c + 12 b = 0
Û
Consider f( x) =
1
b
a + + d =0
4
3
ax 4 bx 3 cx 2
+
+
+ dx
4
3
2
then
f(0) = 0 = f(1)
\ f( x) satisfies the conditions of Rolle’s theorem in [0, 1].
Hence, f ¢( x) = 0 has at least one solution in (0, 1).
66. f ¢( x) = f ( x) × ( x - 2) 2
f(2) > 0 Þ f ¢( x) > 0 Þ f( x) ­
f(2) < 0 Þ f ¢( x) < 0 Þ f( x) ¯
67. f(1) = f(3) Þ a + b - 5 = 3 a + b - 27 Þ a = 11
f ¢( c) = 3 c 2 - 12 c + a = 0 Þ b Î R
3 at
, y = at 3/ 2
2 2/ 3
æ 9 a 2t 2 ö
÷
2ç
ç 2 4/ 3 ÷
dy
è
ø
=
= - cot a Þ t = -2 1/ 3 cot a
2 3/ 2
dx
9a t
70. Let x =
æ 3 at
at 3/ 2 ö÷
and P = cos a ç
ç 2 2/ 3 - cot a ÷
è
ø
P
Þ
= cos a cot 2 a
a
85
Application of Derivatives
Exercise-2 : One or More than One Answer is/are Correct
1. Equation of tangent to y = x 3
y - x 13 = 3 x 12 ( x - x 1 )
Equation of tangent to y = x 1/ 3 is
y - x 12/ 3 =
1
3 x 22/ 3
(x - x2)
If these tangents represent same line
2
2/ 3
1 9 x1 x 2
=
1
1
=
-2 x 13
1
Þ x1 = ±
2 1/ 3
3
x
3 2
f ( 4) - f ( 0) 1
= ; C 1 Î(0 , 4)
4 -0
4
f ( 8) - f ( 0) 1
(c) f ¢(C 1 ) =
= ; C 1 Î(0 , 8)
8 -0
8
2. (a) f ¢(C 1 ) =
f ( C 2 ) = f ( 8) = 1
x3
(d) Let g( x) =
ò f(t) dt
0
8
Þ
g(0) = 0 , g(2) =
ò f(t) dt
0
g(2) - g(0)
a Î(0 , 2)
2
g(2) - g(0)
g¢(b) = 3b 2 f(b 3 ) =
b Î(0 , 2)
2
g¢(a ) = 3a 2 f(a 3 ) =
and
8
g¢(a ) + g¢(b) = g(2) - g(0) =
ò f(t) dt
0
4. f( x) = 2 x 4 + x 4 sin
1
x
x ¹0
=0
x =0
1
1
f ¢( x) = 8 x 3 + 4 x 3 sin - x 2 cos
x
x
¢¢
5. -1 £ f ( x) £ 1
- x £ f ¢( x) £ x
(Q f ¢(0) = 0)
-
x2
x2
£ f( x) <
2
2
(Q f(0) = 0)
86
Solution of Advanced Problems in Mathematics for JEE
6. f ¢¢( x) > 0 " x Î [-3 , 4]
Þ f ¢( x) is increasing for x Î [-3 , 4]
7. f ¢¢( x) > 0 " x Î [0 , 2]
Þ f ¢( x) ­
f(1) - f(0)
f(2) - f(1)
, C 2 Î(1, 2)
f ¢(C 1 ) =
, C 1 Î (0 , 1) and f ¢(C 2 ) =
1-0
2 -1
f ¢(C 1 ) < f ¢(C 2 ) Þ f(0) + f(2) > 2 f(1)
é 2ù
é2 ù
Similarly applying LMVT between ê0 , ú and ê , 2 ú
ë 3û
ë3 û
æ2ö
æ2ö
f ( 2) - f ç ÷ f ç ÷ - f ( 0)
æ2ö
è3ø
è3ø
>
Þ 2 f ( 0) + f ( 2) > 3 f ç ÷
4
2
è3ø
3
3
8. Let g¢¢( x) = a( x - 1)
ax 2
- ax + b
2
3a
g¢( -1) = 0 Þ b = 2
g¢( x) =
g( x) =
ax 3 ax 2
+ bx + c Þ g( x) = x 3 - 3 x 2 - 9 x + 5
6
2
9. f( x) = 2 x 3 - 3 (l + 2) x 2 + 2lx + 5
f ¢( x) = 6 x 2 - 6 (l + 2) x + 2l = 0 has two real roots, then
D > 0 Þ 3l2 + 8l + 12 > 0
Þ
lÎR
10. f( x) = 1 + x ln( x + 1 + x 2 ) - 1 + x 2
f ¢( x) = ln( x + 1 + x 2 )
f ¢( x) ³ 0 for " x Î [0 , ¥)
f ¢( x) £ 0 for " x Î ( -¥, 0]
11. f( x , y) = x m ( k - x) n
f ¢( x , y) = mx m -1 ( k - x) n - x m n × ( k - x) n -1 = 0
Þ
x=
mk
m +n
Maximum value =
k m+n × m m × n n
( m + n) m + n
(Q g( -1) = 10 , g(3) = -22)
87
Application of Derivatives
12. Let line is tangent at (3t 12 , 2t 13 ) and normal at (3t 22 , 2t 23 )
dy
Þ
= t1
dx 3t 2, 2t 3
1
1
So, slope of normal at (3t 22 , 2t 23 ) = Þ
1
t2
1
t2 = t1
t1 = -
2t 13 +
Þ
1
t2
t1 =
2
t 13
æ
1ö
3 ç t 12 - ÷
ç
t 12 ÷ø
è
Þ t 12 (t 14 - 3) = 2 or t 1 = ± 2
dy
13. m =
dx
|my| y
= then solve it.
x+ y x
14. First draw the graph f( x) =
1
2
x -4
–2
2
While drawing diff. possibilities of y = ax 2 + bx + c
We get possible intersections.
15. y ¢ = 3 x 12
3 x 12 =
x 13 - 8
x1 - 2
Þ x 1 = -1 or 2
y ¢ = 3 or y ¢ = 12
16. Let f( x) = x + cos x - a
f ¢( x) = 1 - sin x ³ 0 " x Î R
Þ
Þ
f( x) is increasing.
x + cos x - a = 0 for one positive value of x , a Î (1, ¥)
88
Solution of Advanced Problems in Mathematics for JEE
17. Let y = ax + b = f( x) a ¹ 0
x -b
f -1 ( x) =
a
1
(1) m 1 = a , m 2 =
a
-1
(2) m 1 = a , m 2 =
a
1
(3) m 1 = - a , m 2 =
a
-1
(4) m 1 = - a , m 2 =
a
Þ m 1m 2 = 1
Þ m 1 m 2 = -1
Þ m 1 m 2 = -1
Þ m 1m 2 = 1
18. f ¢( x) = e x ( x 2 - 1) x 2 ( x + 1) 2011 ( x - 2) 2012
= e x x 2 ( x + 1) 2012 ( x - 1)( x - 2) 2012
–
–1
–
0
–
+
1
+
2
x = -1, 0 , 2 are points of inflections and might be more points in (1, 2).
x = 1 is point of minima (Answer can be given either d, ad, bd or abd)
19. f( x) = sin x + ax + b
f ¢( x) = cos x + a
if a > 1 then f( x) is increasing.
So, only one real root, which is positive if b > 0 and negative if b > 0
if a < -1
f( x) is decreasing so only one real root, which is negative if b < 0.
20. f ¢¢( c) = 0 for c Î(0 , 1)
f ¢¢( x) > 0 for x Î(0 , c)
f ¢¢( x) < 0 for x Î( c , 1)
21. f ¢( x) = 5 sin x cos x(sin x - cos x)(1 + sin x cos x)
æp pö
Clearly, f ¢( x) > 0 " x Î ç , ÷
è4 2ø
æ pö
f ¢( x) < 0 " x Î ç 0 , ÷
è 4ø
æpö
f ( 0) = 0 , f ç ÷ = 0
è2ø
æ pö
By Rolle’s theorem $ c Î ç 0 , ÷ Þ f ¢( c) = 0
è 2ø
æpö
Clearly, f( x) ³ f ç ÷
è4ø
89
Application of Derivatives
5
æ 1 ö
æ 1 ö
1
æ pö
÷÷ - 1 = 2 çç
÷÷ - 1 =
f( x) ³ 2 çç
- 1 and f( x) < 0 " x Î ç 0 , ÷
è 2ø
2 2
è 2ø
è4 2ø
22. f( x) = x 2a +1 ln x
x>0
=0
x =0
f( x) is not continuous at a = 23. f ¢( x) =
1
,-1
2
cos x
x
p
pö
æ pö æ
f ¢( x) > 0 Þ x Î ç 0 , ÷ È ç ( 4n - 1) ,( 4n + 1) ÷ " n Î N
2
2
2
è
ø è
ø
p
pö
æ
f ¢( x) < 0 Þ x Î ç ( 4n + 1) ,( 4n + 3) ÷ " n Î N
2
2ø
è
Clearly,
and
p
" nÎ N
2
p
p
f( x) has a local maxima at x = and ( 4n + 1) " n Î N
2
2
x ( - sin x) - cos x
f ¢¢( x) =
= 0 Þ x tan x + 1 = 0
x2
f( x) has a local minima at x = ( 4n - 1)
\
and
Also,
All the points of inflection of f( x) lie on the curve x tan x + 1 = 0
p
Also,
f ¢( x) = 0 Þ x = (2n + 1) " n Î N
2
\
\
Number of values of x in (0 , 10p) in which f ¢( x) = 0 are 20.
24. | f( x)| £ 1
Applying L.M.V.T. in x Î(0 , 1)
Þ
| f ¢( x)| =| f(1) - f(0)|
| f(1) - f(0)| £ 2
Þ
| f ¢( x)| £ 2 for atleast one x in (0 , 1)
Similarly | f ¢( x)| £ 2 for atleast one x is ( -1, 0)
F ( x) = ( f( x)) 2 + ( f ¢( x)) 2
For atleast one x in (0 , 1) & ( -1, 0)
| f ¢( x)| £ 2 & | f( x)| £ 1
Þ
( f ¢( x)) 2 £ 4 & ( f( x)) 2 £ 1
Þ
( f ¢( x)) 2 + ( f( x)) 2 £ 5
Þ
F ( x) £ 5, for atleast one x in ( -1, 0) & (0 , 1)
90
Solution of Advanced Problems in Mathematics for JEE
æ p+ ö
æpö
÷ and
25. f ç ÷ > f ç
ç 2 ÷
è2ø
è
ø
æ p- ö
æpö
÷
fç ÷> fç
ç 2 ÷
è2ø
è
ø
Also, absolute maximum occurs at x = -1
26. Symmetric about y = x
dy
=1
dx
2x = 1 Þ x =
1
2
æ1 5ö
Point = ç , ÷
è2 4ø
27. f ¢( x) > 0
- (1 + x) e - x - e - x
Þ
(1 + x) 2
>0
for x < -2 increasing.
–2
– e2
x = –1
28. Point (1, 2) lies on y = mx + 5
Point (1, 2) lies on x 3 y 3 = ax 3 + by 3
2
3
3
2
1
2
2
3 x y + x × 3 y y = 3 ax + 3 by y
Þ
29.
a=
16
3
, b=
5
5
f( x) - 1
x4 + x2 + 1
x2 - x + 1
=
=
f( x) + 1 ( x 2 + x + 1) 2 x 2 + x + 1
Þ f( x) =
x2 + 1
x
1
Þ m = -3
…(1)
Þ 8b + a = 8
…(2)
Þ a - 12 b = -4
…(3)
91
Application of Derivatives
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
x +1
x -1
x( x + 1)
2
g( x) =
=x +2+
x -1
x -1
2
g¢( x) = 1 =0
( x - 1) 2
1. f( x) =
x =1+ 2,1- 2
4
g¢¢( x) =
( x - 1) 3
g¢¢(1 + 2 ) > 0
Minimum value of g( x) is 3 + 2 2 .
1
2
1
Þ 1= Þ x = 3, - 1
2
2
2
( x - 1)
2. g¢( x) =
Paragraph for Question Nos. 3 to 5
1
3. g(1) =
ò
0
1
t2
f(t ) dt = (1 - t ) dt = t 2
ò
0
1
=
0
4. For x Î(2 , 3]
1
g( x) =
2
ò f(t) dt + ò f(t) dt + ò f(t) dt
0
1
3
1 ( x - 2)
+
2
3
5 æ 5 ö 13
at x = , g ç ÷ =
2 è 2 ø 24
g( x) =
g¢( x) = f( x)
æ 5ö 1
g¢ ç ÷ =
è2ø 4
y-
x
13 1 æ
5ö
= çx- ÷
24 4 è
2ø
12 y = 3 x - 1
2
1
2
92
Solution of Advanced Problems in Mathematics for JEE
1
4
2
Slope of tangent at R =
3
5
tan q =
14
5. Slope of tangent at P =
Paragraph for Question Nos. 6 to 8
y
6.
–1
x
0
y
7.
–1
0
x
Paragraph for Question Nos. 9 to 11
9. By putting x = 1, 2 , - 1, 0 we get a , b, c , d clearly other roots product is 1.
10. P( x) + k = 0 has 4 distinct real roots.
P( x) = -k , where - k Î (1, 2) Þ k Î ( -2 , - 1)
\ pull the graph more than 1 and less than 2, now the graph intersect the x-axis
in ( -2 , - 1),( -1, 0),(0 , 1),(2 , 3)
\ - 2 + ( -1) + 0 + 2 = -1
11. P( x) = 0 has two roots.
P ¢( x) = 0 has three root
P( x) = 0 has atleast 5 roots.
( P ¢( x)) 2 + P( x)P ¢¢( x) = 0 Þ ( P ¢( x)P( x)) ¢ = 0 has atleast four roots.
Paragraph for Question Nos. 12 to 14
Sol. The equation of chord AB will be y - f( x 1 ) =
f( x 2 ) - f( x 1 )
( x - x1 )
x 2 - x1
This line passes through (0 , 2 x 1 x 2 )
f( x 2 ) - f( x 1 )
\
2 x 1 x 2 - f( x 1 ) =
( -x1 )
x 2 - x1
93
Application of Derivatives
( x 2 - x 1 ) f( x 1 ) - x 1 f( x 2 ) + x 1 f( x 1 )
x 2 - x1
Þ
2 x1 x 2 =
Þ
2 x 1 x 2 ( x 2 - x 1 ) = x 2 f( x 1 ) - x 1 f( x 2 )
f( x 1 ) f( x 2 )
f( x 1 )
f( x 2 )
= 2 ( x 2 - x1 ) Þ
+ 2 x1 =
+ 2x2 = k
x1
x2
x1
x2
Þ
\
f( x)
+ 2x = k
x
\
f( x) = kx - 2 x 2
Given that f(1) = -1
\
-1 = k - 2 Þ
\
2
f( x) = x - 2 x
k =1
1/ 2
é x2 2x3 ù
f( x) dx = ê
ú
0
3 úû
êë 2
0
1
13. f ¢( x) = 1 - 4 x ³ 0 Þ x £
4
12. \
ò
1/ 2
=
1 2 1 1 1
1
- × = =
8 3 8 8 12 24
14. F ( x) = f( x) + x = 2 x - 2 x 2
Clearly, F (0) = F (1) = 0
\ Rolle’s theorem is applicable in [0 , 1].
Paragraph for Question Nos. 15 to 16
1
1
f( x) = 1 + x e y f( y) dy + e x
15.
ò
0
0
1
1
A = e y f( y) dy ,
Let
ò
0
Þ
ò y f( y) dy
B=
ò y f( y) dy
0
1
f( x) = 1 + Ax + Be x Þ A = e x (1 + Ay + Be y ) dy
ò
0
B =-
2
3
,A=e+1
2
f ¢( x) + 3 > 0
Þ
-
3
2e x
3( e + 1)
é4 ù
+ 3 > 0 Þ ex <
Þ ê e x ú = [e + 1] = 3
2 ( e + 1)
4
ë3 û
16. Ax 1 + Be x1 = -
3 x1
- 2 Þ Be x1 = -2
2
f ¢( x 1 ) = A + Be x1 = m 1
94
Solution of Advanced Problems in Mathematics for JEE
3
7
-2 = 2
2
3
m2 = 2
8
tan q =
25
m1 = -
Exercise-4 : Matching Type Problems
3.
y = f(x)
y = g(x)
4
0
1
x
–2
–1
1
– 2
–1
2
y = h(x) = f(–x)
4
1
–2
4. (A) y =
x3
( x - a )( x - b)( x - g)
y=
8
(2 - a )(2 - b)(2 - g)
–1
0 1
x 3 - 3 x 2 + 2 x + 4 = ( x - a )( x - b)( x - g)
2
x
1
x
2
95
Application of Derivatives
Put x = 2
8 - 12 + 4 + 4 = (2 - a )(2 - b)(2 - g) = 4
8
\
y|at x = 2 = = 2
4
(B) x 3 + ax + 1 = 0 ,
x 4 + ax + 1 = 0
…(1)
4
…(2)
2
x + ax + x = 0
(2) – (1) gives
2
ax - ax + x - 1 = 0
ax( x - 1) + ( x - 1) = 0
( x - 1)( ax + 1) = 0
1
x = 1 or x = a
put x = 1 in (1) we get,
1 + a + 1 = 0 Þ a = -2
|a| = 2
2
(C) f( x) = x + 4 cos x + 5
x/2
f ¢( x) = 2 x - 4 sin x = 2 ( x - 2 sin x) = 0
x
sin x =
2
f ¢(x)
(>0)
when x > a
<0
when 0 < x < a
>0
when b < x < 0
<0
when x < b
–p –2 b
a2 p
O
–
+
a
0
b
x = 0 only a point of maxima
So, number of local maxima is 1.
(D) Let | x| = t Î [0 , 2]
\
f( x) = 2t 3 + 3t 2 - 12t + 1 = g(t )
2
2
g¢(t ) = 6t + 6t - 12t = 6 (t + t - 2) = 6 (t + 2)(t - 1)
g(1) is min. of f( x) i . e. , fmin = 2 + 3 - 12 + 1 = -6
g(0) = 1, g(2) = 16 + 12 - 24 + 1 = 5
max. of f( x) is 5 > 4 , 3 , 2 , 0
6. f ¢ =
1
- a + 2 x > 0 since x > 0
8x
1 - 8 ax + 16 x 2 > 0
–
–2
0
+
1
96
Solution of Advanced Problems in Mathematics for JEE
See D = 0
D<0
D>0
7. Let g( x) = ax 3 + bx 2 + cx + d
f( x) = g( x)
f( x) has local maxima and local maxima at x = -2 and x = 2.
Þ
Þ
g( x) has same local minima and maxima and x = -2 and 2.
a < 0 ; a = -2
f ¢( x) =
3 ax 2 + 2 bx + c
2 ax 3 + bx 2 + cx + d
=0
f ¢( -2) = 0 and f ¢(2) = 0
Þ b = 0 , c = 24
Also,
g(2) > 0 and g( -2) > 0
Þ
-16 + 48 + d > 0 and 16 - 48 + d > 0
d > -32 and d > 32
d > 32
Þ
æ
h 2 ö÷
8. (A) V = pr 2 h = ph × ç R 2 ç
4 ÷ø
è
2 ö
æ
çQ R 2 = r 2 + h ÷
ç
4 ÷ø
è
æ
dV
3 h 2 ö÷
2R
2
and r =
= pçR 2 =0 Þ h =
R
ç
÷
dh
4 ø
3
3
è
(B) V =
pr 2 h ph 2
=
{R - ( h - R ) 2 }
3
3
dV p
= ( 4 hR - 3 h 2 ) = 0
dh 3
(C) sin q =
r
R
=
2
h-r
R + h2
Volume of cone =
Þ h=
{Q R 2 = r 2 + ( h - R ) 2 }
4R
2 2R
and r =
3
3
Þ R2 =
h2r 2
h 2 - 2 hr
p 2
p æ h 2 r 2 ö÷
R h= ç
3
3 çè h - 2 r ÷ø
dV p 2 æç ( h - 2 r) 2 h - h 2 ö÷
= r
= 0 Þ h = 4r
÷
dh 3 çè
( h - 2 r) 2
ø
æ 2x 2x 2x ö æ 3 y 3 y 3 y 3 y ö
1/7
+
+
+
+
+
÷
ç
÷+ç
3
3 ø è 4
4
4
4 ø æç 8 x 3 81 y 4 ö÷
32
è 3
(D)
Þ x3 y4 £
³
×
ç 27 256 ÷
7
3
è
ø
97
Application of Derivatives
Exercise-5 : Subjective Type Problems
1.
A2
pr 2
2p
=
=
A1 æ 2 p - q ö
2p - q
2
ç
÷ × pr
è 2p ø
V =
pæ q ö
ç
÷
3 è 2p ø
2
æ q ö
12 - ç
÷
è 2p ø
O
1
2
q
dV
8
=0 Þ q =
p
dq
3
A2
3
=
=3 + 6
A1
3- 2
2. f( x) = x 2 ln x
Þ f ¢( x) = x (1 + 2 ln x)
and f ¢( x) > 0 for Î[1, e]
\ f( x) is continuously increasing on [1, e] with the least value zero at x = 1 and the greatest
value e 2 at x = e.
3. f( x) = px e - x -
x2
+x
2
f ¢( x) = (1 - x)[ pe - x + 1] £ 0
p £ -1
ì ax e ax + e ax ; x £ 0
4. f ¢( x) = í
2
î1 + 2 ax - 3 x ; x > 0
Þ
Clearly, f ¢( x) is continuous at x = 0
ax
ì 2 ax
Þ f ¢¢( x) = ía x e + 2 ae ; x £ 0 ;
2a - 6 x ;
x>0
î
f ¢( x) increasing if ( ax + 2) ae ax ³ 0 and 2 a - 6 x ³ 0
5. f( x) = x 2 - 2 bx + 1
Case I : b > 1
f(0) - f(1) = 4
1 - (2 - 2 b) = 4
5
Þ
b=
2
1
Case II : 0 < b <
2
Þ
Þ
Þ
Þ
f(1) - f( b) = 4
b = 3, - 1
(Not possible)
98
Solution of Advanced Problems in Mathematics for JEE
Case III :
1
< b< 1
2
f(0) - f( b) = 4
b = ±2
Case IV : b < 0
Þ
f(1) - f(0) = 4
3
Þ
b=2
Þ
(Not possible)
6. x 2 + 9 y 2 = 36
x 2 + y 2 = 12
12 - y 2 + 9 y 2 = 36
8 y 2 = 24 Þ y 2 = 3 Þ y = 3 , - 3
when y = ± 3 , x 2 = 12 - 3 = 9
x = ±3
\
point of intersections are ( ± 3 , ± 3 )
Let one of point of intersect is (3 , 3 )
2x 2 y
æ 2x ö æ 4 ö
÷÷
+
y¢ = 0 Þ y¢ = ç ÷ ´ çç
36
4
è 36 ø è 2 y ø
Now,
( y ¢) ( 3,
3)
=-
1
3 3
= (m 1 )
2x + 2 y y¢ = 0 Þ y¢ = -
tan q =
tan q =
7. f ¢( x) = 2 e
2x
m1 - m 2
=
1 + m 2m 1
2
3
-
x
Þ ( y ¢) ( 3,
y
1
3 3
+ 3
1
1+
3
æ 2 ö
÷÷
Þ q = tan -1 çç
è 3ø
- (l + 1) e x + 2 ³ 0 " x Î R
i . e. , 2 e 2 x + 2 - (l + 1) e x
l + 1 £ 2( e x + e - x )
l+1
£ (e x + e-x ) " x Î R
2
l+1
Þ
£ ( e x + e - x ) min " x Î R
2
=
3)
=-
8
4 3
=
3
3
= - 3 =m2
2
3
99
Application of Derivatives
So,
l+1
£2
2
l + 1£ 4
l£3
l Î ( -¥, 3]
k =3
\
9. f( x) = x 2 ,
Þ q = p2
8
x
8
s=r
g( x) = Þ
s-q
8
= 2p & 2p =
r-p
r2
Also,
pr 2 = 4
s-q
= 2p
r-p
Þ
Þ
8
- p2
8
r
= 2 p Þ - - p 2 = 2 pr - 2 p 2
r-p
r
8
p 2 = + 2 pr
r
p 2 r = 16
Multiply (1) & (2),
Þ
Þ
é| x + 2|, x ³ 0
10. f( x) = ê
ë| x - 2|, x < 0
pr = 4
r = 1, p = 4
Minimum value of f( x) is 2.
x
11. f( x) = [( a - 1)(t 2 + t + 1) 2 - ( a + 1)(t 4 + t 2 + 1)] dt
ò
0
x
2 (t 2 + t + 1)( at - t 2 - 1) dt
ò
0
f ¢( x) = 2 ( x 2 + x + 1)( ax - x 2 - 1) = 0
D < 0 Þ a2 - 4 < 0
12. f( x) = x 2013 + e 2014 x
f ¢( x) = 2013 x 2012 + 2014 e 2014 x > 0
Þ
f( x) is increasing function.
…(1)
…(2)
100
Solution of Advanced Problems in Mathematics for JEE
14. P = ( x 1 , x 13 - ax 1 )
Q = ( x 2 , x 23 - ax 2 )
Q
y = x 3 - ax
P
dy
= 3x2 - a
dx
Slope at P = slope of PQ
æ x 3 - ax 2 - x 13 + ax 1 ö
÷
(3 x 12 - a) = ç 2
ç
÷
x
x
2
1
è
ø
\
(Q x 1 ¹ x 2 )
( x 2 - x 1 )( x 2 + 2 x 1 ) = 0
…(1)
x 2 = -2 x 1
Þ
Slope at P ´ Slope at Q = -1
(3 x 12 - a)(3 x 22 - a) = -1
…(2)
Put (1) in (2),
36 x 14 - 15ax 12 + ( a 2 + 1) = 0
(Q x 1 Î R )
D³0
4
9 a ³ 16 Þ a ³
3
2
b
15. I(t ) = ( x 2 + 2 x - t 2 ) dx =
ò
a
I( t ) =
x3
+ x 2 - t 2 x|ba
3
(max)
b3 -a3
+ (b 2 - a 2 ) - t 2 (b - a )
3
O
I ¢(t ) = -2t (b - a ) = 0
I ( t ) £ I ( 0)
0
\
I ( 0) =
2
4
p
-2
16.
4
ò ( x + 2 x) dx = - 3 Þ q =|I(0)|= 3
dy
3y
when t = 0 , y = 0
=1dt
100 - 2t
dy æ
3
ö
+ç
÷ y =1
dt è 100 - 2t ø
y(100 - 2t ) -3/ 2 = + (100 - 2t ) -1/ 2 + c
as when t = 0 , y = 0
1
c=10
1
y = (100 - 2t ) (100 - 2t ) 3/ 2
10
101
Application of Derivatives
dy
1 æ3ö
1/ 2
= -2 ç ÷ (100 - 2t ) ( -2) = 0
dt
10 è 2 ø
t=
250
7
= 27 +
9
9
17. Let f ¢( x) = K
Þ f( x) = Kx + c
Þ f(9) - f( -3) = 12 K
Maximum value of f(9) - f( -3) = 96
-9 y 12 æ
3 3ö
æ3
ö
19. Equation of normal at P ç y 13 , y 1 ÷ is y - y 1 =
ç x - y1 ÷
4 è
4
è4
ø
ø
If it passes from (0, 1) then 27 y 15 + 16 y 1 - 16 = 0 has only one real root.
æ x2
ö
20. e - x ç
+ x + 1÷ = a
ç 2
÷
è
ø
æ x2
ö
Let f( x) = e - x ç
+ x + 1÷
ç 2
÷
è
ø
æ x2 ö
÷<0
f ¢( x) = e - x ç ç 2 ÷
è
ø
21.
f ¢( x) = a - 2 sin 2 x + cos x - sin x
Let g( x) = -2 sin 2 x + cos x - sin x
= -2 {(cos x - sin x) 2 - 1} + cos x - sin x
where cos x - sin x = t
-2t 2 + t + 2 " t Î [- 2 , 2 ]
-2 - 2 £ g ( x) £
22. Let
x = 6 cos 3 q ,
17
17
Þ a³
8
8
y = 6 sin 3 q
dy 6 (3 sin 2 q cos q)
=
= - tan q
dx -6 (3 cos 2 q sin q)
Equ. of tangent
y - 6 sin 3 q = - tan q ( x - 6 cos 3 q) Þ p1 = 6 sin q cos q
Equ. of normal
y - 6 sin 3 q = cot q ( x - 6 cos 3 q)
Þ p 2 = 6 (cos 2 q - sin 2 q)
4 p12 + p 22 = 6 4 sin 2 q cos 2 q + cos 4 q + sin 4 q - 2 sin 2 q cos 2 q = 6
❑❑❑
102
Solution of Advanced Problems in Mathematics for JEE
5
INDEFINITE AND DEFINITE
INTEGRATION
Exercise-1 : Single Choice Problems
1.
ò
é
ù
ê a x ln x + a x ln a × x (ln x - 1) ú dx
1
424
3 14243 ú
ê
ë
û
II
I
é
ù
é 1
ù
= a x × ln x dx + ê x (ln x - 1) a x - ê x × + (ln x - 1)ú a x ú dx
ë x
û
ë
û
ò
ò
= a x × ln x dx + [ x (ln x e)] a x - (ln x) a x dx
ò
1 n
n® ¥ n
r =1
2. lim
3.
1
1
å
1+
r
n
=
ò
dx
ò x + 1 = 2 ( 2 - 1)
0
sin x
ò sin( x - a) dx
Let x - a = t Þ dx = dt
sin(t + a )
dt = t cos a + sin a log sin t + C = x cos a + sin a log sin( x - a ) + C
sin t
ò
2
4.
log( x 2 + 2)
ò ( x + 2) 2
0
2
2
æ - log ( x 2 + 2) ö
2 x dx
ç
÷
dx =
+
2
ç
÷
x +2
è
ø 0 0 ( x + 2)( x + 2)
ò
=
2
5
1
tan -1 2 log 2 +
log 3
3
12
12
5. For 0 < x < 1
1+ x9 < 1+ x8 < 1+ x4 < 1+ x3
x
6. Let g( x) =
2
ò 1 - ( f( s)) ds
0
103
Indefinite and Definite Integration
æ g(t ) - g( x) ö
÷ = f( x)
lim çç
t ® x è f (t ) - f ( x) ÷
ø
y dy
ò 1 - y 2 = ò dx
1 n
n® ¥ n
r =1
1
r
Þ g¢( x) = f( x) f ¢( x) = 1 - ( f( x)) 2
æ
1
3ö
çQ f æç ö÷ =
÷
ç
è 2 ø 2 ÷ø
è
1- y2 =1- x
Þ
1
2
å f æçè n ö÷ø = ò f( x) dx = 3 ò 1 - cos 6 x - sin 6 x dx
7. lim
0
0
1
1 - cos 2
2
= sin 2 x dx =
ò
0
1 ö
æ
1
1 2ç1 - 3 ÷
6
3
(x - x )
1
1
è
x ø
8.
dx =
dx = 3
3
3
2
36
1 ö
0 (2 x + 1)
0æ
ç2x + 2 ÷
è
x ø
ò
1
æ
ö
ç Put 2 x + 2 = t ÷
è
ø
x
ò
1/ 2
9. 2
ò
ò
q cos q
dq sin q
0
p/ 4
2
1
sin -1 x
tan -1 x
dx dx
x
x
0
ò
0
p/ 4
ò
0
q sec 2 q
p
dq = - ln 2 - 2
tan q
4
p/ 4
ò
0
p/ 4
=-
p/ 4
ln sin q dq +
ò ln tan q dq
0
p
ò ln sin 2q dq = 4 ln 2
0
x
0
10. f( x) = x 2 + e -t f( x - t ) dt = x 2 - e - x eu f(u) du
ò
ò
0
x
0
f ¢( x) = 2 x + e - x eu f(u) du + f( x)
ò
Þ f ¢( x) = x 2 + 2 x
x
x3
+ x2
3
1
æ
ö
ç k = f( x) dx ÷
1
ç
÷
0
è
ø
Þ f( x) =
11. f ¢( x) = f( x) + k1
Þ
ò
y = ke x - k1
if f(0) = 1 Þ k - k1 = 1
1
k1 = ( ke x - k1 ) dx Þ 2 k1 = k ( e - 1) Þ k =
ò
0
2
e -1
and k1 =
3-e
3-e
(Let x - t = u)
104
Solution of Advanced Problems in Mathematics for JEE
1+ cos2 x
12. I 1 =
1+ cos2 x
1+ cos2 x
ò tf(t(2 - t)) dt = 2 ò f (t(2 - t)) dt - ò tf(t(2 - t)) dt
sin 2 x
sin 2 x
sin 2 x
I1 = 2I 2 - I1 Þ I1 = I 2
5 sin x dx
cos x + 2 sin x
13.
= dx + 2
dx = x + 2 ln|sin x - 2 cos x| + C
sin x - 2 cos x
sin x - 2 cos x
2
1
+
2
3/ 2
(2 + x ) dx
x
x
14.
=
dx
2
( x + 1 + x)2
æ1
ö
1
+ 1÷
ç +
x
èx
ø
ò
ò
ò
ò
Let
15.
ò
ò
1
1
1 ö
æ 1
+
+ 1=t Þ ç÷ dx = dt
2
x
x
è x
2 x 3/ 2 ø
æ3
2 öæ6
2 ö
ç x + 2 - x ÷ ç 1 - x 2 - x ÷ dx
è
øè
ø
3
1- x2
3
=
ò
æ 2 - x2 - x ö
÷
x + (2 - x 2 ) 3 ç
ç
÷
2
è
ø
3
1- x2
= 2 1/ 6 dx = 2 1/ 6 x + C
ò
16.
dx
ò 1 - tan 2 x
cos x dx
1
-1
ò 1 - 2 sin 2 x = 2 sin ( 2 sin x) + C
17. I =
dx
ò x 5/ 6 ×( x + 1)7 / 6
x
=t
x +1
I=
dx
( x + 1) 2
= dt
( x + 1) 2 dt
ò [t( x + 1)]5/ 6 ( x + 1)7 / 6 = ò t
-5/ 6
dt = 6t 1/ 6 + C
18. I n = sin n x dx = sin n -2 x(1 - cos 2 x) dx
ò
I n = I n -2 -
ò
n -2
× cos x × cos x dx
ò 1sin442x 44
3 123
II
I
æ cos x × sin n -1 x
sin n -1 x ö÷
I n = I n -2 - ç
- - sin x ×
dx
ç
÷
n -1
n -1
è
ø
ò
I n = I n -2 -
cos x × sin n -1 x
1
In
n -1
n -1
nI n - (n - 1) I n -2 = - cos x × sin n -1 x
dx
105
Indefinite and Definite Integration
19.
1
2
Let a + bx = t then dx =
ò x ( a + bx) 2 dx
ò
\
x2
2
1
1 dt
1
æt - aö
dx = ç
=
÷ × 2 ×
2
b b3
è b ø t
( a + bx)
ò
=
=
20.
dt
b
1
b3
ò
ò
æ t 2 - 2 at + a 2 ö
ç
÷ dt
ç
÷
t2
è
ø
2 ö
2ù
æ
é
ç 1 - 2 a + a ÷ dt = 1 êt - 2 a ln|t| - a ú + C
ç
t
t û
t 2 ÷ø
b3 ë
è
1 é
a2 ù
a + bx - 2 a ln|a + bx| ê
ú +C
a + bx û
b3 ë
8 x 43 + 13 x 38
ò ( x 13 + x 5 + 1) 4 dx
8 x -9 + 13 x -14
ò (1 + x -8 + x -13 ) 4 dx
Let 1 + x -8 + x -13 = t
( -8 x -9 - 13 x -14 ) dx = dt
21.
dt
ò t4
\
-
=+
1
3t 3
+C =
1
3 (1 + x -8 + x -13 ) 3
+C =
x 39
3 ( x 13 + x 5 + 1) 13
+C
æ (cos 6 x + cos 4 x) + 5 (cos 4 x + cos 2 x) + 10 (cos 2 x + 1) ö
÷ dx
ø
10 cos 2 x + 5 cos x cos 3 x + cos x cos 5 x
ò çè
=
ò
2 cos 5 x cos x + 5 (2 cos 3 x cos x) + 10 (2 cos 2 x)
10 cos 2 x + 5 cos x cos 3 x + cos x cos 5 x
dx
= 2 dx = 2 x + C
ò
f( x) = 2 x Þ f(10) = 20
22.
-1
ò (1 + x - x ) e
x + x -1
dx = e x + x
ò
= e x+ x
-1
= xe x + x
23.
ò
-1
-1
× 1 dx + ( x + x -1 ) e x + x dx
ò
-1 æ
1 ö
-1
x + x -1
× x - e x+ x ç 1 dx + C
÷ x dx + ( x - x ) e
2
è
x ø
ò
ò
-1
+C
é
ù
ê 2 tan x
ú
1
2 tan x
2
ù
ú dx = e x é
ex ê
+
ê 1 + tan x + (1 + sin 2 x) ú dx
ê 1 + tan x æ 1 - cos( p 2 + 2 x) ö ú
ë
û
ç
÷ú
ê
2
è
øû
ë
ò
é
ù
sin x
1
=2 ex ê
+
dt
2ú
ë sin x + cos x (sin x + cos x) û
ò
106
Solution of Advanced Problems in Mathematics for JEE
Let f( x) =
sin x
,
sin x + cos x
f ¢( x) =
1
(sin x + cos x) 2
sin x
æ
ö
=2 ×exç
÷+C
sin
x
+
cos
x
è
ø
æ 5p ö
gç
÷ =1
è 4 ø
24.
d
( x sin x + cos x) = x cos x
f ¢( x) = x cos x
dx
Let f( x) = x sin x + cos x
f ¢¢( x) = - x sin x + cos x
æ
f ¢¢( x) ö÷
f ¢¢( x)
e f ( x ) ç xf ¢( x) +
dx = xe f ( x ) f ¢( x) dx + e f ( x ) ×
dx
2 ÷
ç
( f ¢( x) ) ø
( f ¢( x) 2 )
è
ò
ò
ò
= xe f ( x ) - e f ( x ) dx + e f ( x )
ò
= xe f ( x ) -
æ -1 ö
-1
÷÷ dx
- e f ( x ) × f ¢ ( x) çç
f ¢ ( x)
è f ¢ ( x) ø
ò
æ
e f( x)
1 ö
÷+C
+ C = e f ( x ) çç x f ¢ ( x)
f ¢ ( x) ÷ø
è
1 ö
æ
= e x sin x + cos x ç x ÷+C
x
cos
xø
è
1
25.
0
1
ò
0
26.
1
æ
ö
ò çè x + x + 1 + x ÷ø dx
1
1
2
2
( x + ( 1 + x - x ) dx = ( 1 + x dx = (1 + x) 3/ 2 = (2 3/ 2 - 1)
3
3
0
ò
0
x2
ò x x(2 ln x + 1) dx
2
xx =t
x 2 ln x = ln t
1
æ 2 1
ö
+ (ln x) 2 x ÷ dx = dt
çx
x
t
è
ø
2
dt
\
t×
= dt = t + C = x x + C = ( x x ) x + C
t
27. = sec 2010 x cosec 2 x dx - 2010 sec 2010 x dx
ò
ò
ò
ò
2010
= ò sec
x( - cot x) - ò 2010 sec 2010 x × tan x × ( - cot x) - ò 2010 sec 2010 x dx
cot x
- cot x
=+ 2010 ò sec 2010 x dx - 2010 ò sec 2010 x dx + C =
+C
2010
(cos x)
(cos x) 2010
\
f( x)
1
=
= { x} no solution.
g( x) sin x
107
Indefinite and Definite Integration
28. Let x x ln x = t
æ
x x ö÷
Þ ç x x ln x (1 + ln x) +
dx = dt
ç
x ÷ø
è
1ö
æ
Þ x x ç ln x + (ln x) 2 + ÷ dx = dt
xø
è
x
ò dt = t + C = x ln x + C
1 ö
æ 1
ç 3 - 5 ÷ dx
èx
x ø
29. I =
2
1
2+
2
x
x4
2
1
1
Let 2 +
=t Þ I =
2
4
4
x
x
ò
dt
ò t
30. Put ln x = t
2
æ 1
ö
2t
æ t -1 ö
÷ dt
I = et ç
dt = et çç
÷
2
2
2
2 ÷
è t + 1ø
è t + 1 (t + 1) ø
dx
dx
31. I =
=
3/ 4
5/ 4
3/ 4
( x - 1) ( x + 2)
æ x -1ö
2
ç
÷ ( x + 2)
è x +2ø
x -1
3 dx
Let
= t Þ dt =
x +2
( x + 2) 2
ò
ò
ò
32.
ò
2 - (1 + x 7 )
ò x(1 + x 7 )
33. I =
ò
dx = -
ò
dx 2
+
x 7
7x6
(sin 4 x - cos 4 x)(sin 4 x + cos 4 x)
2
2
1 - 2 sin x cos x
34. I = 2 1/ 3
ò
2
7
ò x 7 (1 + x 7 ) dx = - ln| x| + 7 ln|1 + x | + C
dx = (sin 2 x - cos 2 x) dx = - cos 2 x dx
ò
ò
(tan x) 1/ 3 d ((tan x) 1/ 3 )
(tan x) 2/ 3 + 1
Let (tan x) 1/ 3 = t Þ d((tan x) 1/ 3 ) = dt
I=
35.
2 1/ 3
2
2t
ò t 2 + 1 dt
(2012) x
ò 1 - (2012) 2 x ×(2012)
sin -1 ( 2012)x
dx
-1
Let sin
-1
x
1
(2012) sin ( 2012)
(2012) = t Þ
(2012) t dt =
+C
ln 2012
ln 2 (2012)
x
ò
108
Solution of Advanced Problems in Mathematics for JEE
36. Let x + 1 = t 2
Þ dx = 2t dt
1ö
æ
ç 1 + 2 ÷ dt
è
t ø
(t 2 + 1) dt
2
ò t 4 + t 2 + 1 = 2ò æ
2
1ö
çt - ÷ + 3
tø
è
37.
æ f( x) g¢ ( x) - f ¢ ( x) g( x) ö æ g( x) ö
÷÷ ln çç
÷÷ dx
f( x) g( x)
ø è f( x) ø
ò ççè
Let
38.
g( x)
f( x) g¢( x) - g( x) f ¢( x)
=t Þ
dx = dt;
f( x)
( f( x)) 2
æ
xæ
2
1 ö
æ
xæ
1ö
ln t
ò t dt =
(ln t ) 2
+C
2
ö
ò ççè ò e çè ln x + x - x 2 ÷ø dx ÷÷ø dx
xæ 1
1 ö
ö
æ
1ö
xæ
ö
x
ò ççè ò e çè ln x + x ÷ø dx + ò e çè x - x 2 ÷ø dx ÷÷ø dx = ò çè e çè ln x + x ÷ø + C 1 ÷ø dx = e ln x + C 1 x + C 2
æ -t cos( x + pt ) 1 1 1 × cos( x + pt ) ö
39. f( x) = p 2 ç
+
dt ÷ = p cos x - 2 sin x
ç
÷
p
p
0
0
è
ø
ò
40.
é 2
ö
2
£ 5 Þ xÎê
, 1÷
x
ë 5 ø
1
\
2/ 5
f( x) dx =
ò
0
f( x) dx +
ò
0
1
1
1
ò
ò
æ 2
ö
2
f( x) dx £ 5 ç
- 0÷ +
dx
x
5
è
ø
2/ 5
2/ 5
1
\
ò0 f( x) dx £ 2 + 2 [ln x]2/ 5
\
é 5ù
a = 2 + 2 ln ê ú
ë 2 û
42. f(0) = 0 , f(2 p) = 2 p
2p
\
ò
0
Þ
2p
f( x) dx +
ò
2p
f -1 ( x) dx =
0
2p
éx2
ù
- cos x ú + I = 4 p 2
ê
ë 2
û0
1
1
ò
ò
ò 2 p dx = 4 p
2
0
2p
Þ
I=
-1
ò f ( x) dx = 2 p
2
0
1
é é
ù
é
ù
ù 1
4
4
4
4
4
43. = 2 ê2 e - x dx - 8 x 4 e - x dx ú = 2 ê2 ê( xe - x ) 10 + 4 x 4 e - x dx ú - 8 x 4 e - x dx ú
êë êë
úû
êë 0
úû
úû 0
0
0
4
=
e
ò
ò
109
Indefinite and Definite Integration
46. Put y - 2 = z
2
I=
4
47.
z2 +1
ò
-2 2 z
3
òxe
2
sin x 3
sin( z ) dz = 0
+3
dx
1
Let x 3 = t Þ 3 x 2 dx = dt
64 sin t
e
ò t dt = F(64) - F(1)
1
x
x
2
ò
2
51. lim x et - x dt = lim
ò
x® ¥
2
x × et dt
0
x ®¥
0
ex
2
Apply L’ Hospital’s rule,
x
2
2
x × ( e x ) + et dt × 1
2
x® ¥
e x ×2x
n
52. L =
ò
0
lim
1
2 ×r + n
x
æ
ö
2
ç
et dt ÷
ç1
÷ 1
÷=
= lim ç + 0
2
x ® ¥ç 2
2x ex ÷ 2
ç
÷
ç
÷
è
ø
ò
(2 x + 1) dx
å r 2 + n × r + n 2 ò x 2 + x + 1 = ln( x 2 + x + 1)|10
r =1
=
0
L = ln 3
3
53. Let
x 2 + 2 x = y = f( x)
x = -1 + ( y 3 + 1) 1/ 2
2
I = ( f -1 ( x) + f( x) + 1) dx
ò
0
Consider
2
2
ò
f -1 ( x) = tf ¢(t ) dt
0
ò
Let f -1 ( x) = t ; x = f(t ); dx = f ¢(t ) dt
0
2
= tf(t )|20 - dx = 6
ò
0
p/ 2
54. Put x = 2 tan q then I =
æ ln 2 + ln tan q ö
÷÷ 2 sec 2 q dq then solve it.
2
4
sec
q
ø
0
ò ççè
110
Solution of Advanced Problems in Mathematics for JEE
55. Put x - 5 = t
x = 0, t = - 5
x = 10 , t = 5
5
ò
t3
(t + t + t ) dt =
3
2
5
3
-5
56. Let
I=
=
-5
¥
250
3
dx
…(1)
ò (1 + x 9 )(1 + x 2 )
0
Put x =
1
t
dx = -
Þ
1
dt
t2
-
0
I=
dt
¥
t 9 dt
t2
=
9
2
2 ö
æ 9
öæ
¥ ç t + 1÷ ç 1 + t ÷
0 (t + 1)(1 + t )
ç t9 ÷ç t2 ÷
è
øè
ø
ò
ò
On adding (1) & (2),
¥
2I =
dt
ò (1 + t 2 ) = tan
-1
0
2I =
p
2
Þ
p/ 2
p/ 2
ò
ò
57. I =
I=
t
¥
0
p
4
1 + 3 sin x - 4 sin 3 x
æ 1 + sin 3 x ö
dx
ç
÷ dx =
1 + 2 sin x ø
1 + 2 sin x
è
0
0
p/ 2
=
ò
0
-1
(tan
1
x® ¥
58. lim
x)
2 x2 + 1
lim (tan -1 x) 2
x® ¥
(1 + 2 sin x)( -2 sin 2 x + sin x + 1)
p
æ1 pö
= -2 ç
÷ + 1+ =1
(1 + 2 sin x)
2
è2 2ø
2
2x
1+ x2 p2
=
x
4
2013
59. Let t =
Õ( x 2 + r 2 )
r =1
2013
ln t =
å ln( x 2 + r 2 )
r =1
2013
æ 2013 x
ö
1
2x
÷ t dx
dt =
dx Þ dt = 2 ç
ç
÷
2
2
2
2
t
x
+
r
x
+
r
r =1
è r =1
ø
å
å
…(2)
111
Indefinite and Definite Integration
1
2013
ö
ö
dt t 1 æç 2013 2
1 æ 2013
= =
( x + r 2 )÷ = ç
(1 + r 2 ) r2 ÷
ç
÷
ç
÷
2 2 2 è r =1
r =1
ø 0 2 è r =1
ø
Õ
ò
\
60. f ¢( x) = 2 -
=
p/ 2
61. I =
ò
0
1
1+ x
2
Õ
æ
ö
2x
1
ç1 +
÷ =2 - 1 2
2 ç
2 ÷
1+ x
x + 1+ x è
2 1+ x ø
1+ x2
1
-
2(1 + x 2 ) - 1 - 1 + x 2
1+ x2
sin x cos x dx
=
1 + sin x + cos x
=
7
62. I =
p/ 2
æç (1 + x 2 ) - 1 + x 2 ö÷ + x 2
è
ø
=
>0" xÎR
2
1+ x
xæ
x
xö
ò sin 2 çè cos 2 - sin 2 ÷ø dx
0
p/ 2
1
2
p
ò [sin x + cos x - 1] dx = 1 - 4
0
cos x 2 dx
ò cos x 2 + cos(10 - x) 2
3
7
I=
Õ
cos(10 - x) 2 dx
ò cos(10 - x) 2 + cos x 2
3
2I = 4 Þ I = 2
Þ
1
e2
ò
ò
ln x
63. dx +
x
-1
e
1
1
é(ln x) 2 ù
é(ln x) 2 ù
ln x
dx = - ê
+ê
ú
ú
x
ë 2 û -1 ë 2 û
e
e2
=2 +
1
1 5
=
2 2
2
cosec x
ò tg(t) dt
2
64. lim
x®
p
4
x2 n
65. lim
n® ¥
p2
16
n-k
cosec 2 x g(cosec 2 x) 2cosec x( -cosec x) cot x -16
=
g(2)
p
2x
p
x®
= lim
4
4k
å n 2 cos n
k =1
kö
æ
ç1 - ÷
1
nø
æ 4k ö
è
lim
cos ç
÷ = (1 - x) cos 4 x dx
n® ¥
n
è n ø
k =1
n
å
ò
0
1
= (1 - x)
sin 4 x
+
4 0
1
ò
0
1
sin 4 x
1
dx = cos 4 x
4
16
0
1
1
= - (cos 4 - 1) =
(1 - cos 4)
16
16
112
Solution of Advanced Problems in Mathematics for JEE
2/ n
1
é1/ n
p
2p
np ù
66. lim ê sin
dx +
sin
dx +¼¼ +
sin
dx ú
n® ¥ê
2n
2n
2n ú
1/ n
1-1/n
ë 0
û
1é
p
2p
np ù
lim
sin
+ sin
+¼¼ + sin ú
n® ¥ n ê
2n
2n û
ë 2n
n
p
æ
ö
sin ç
÷
æ (n + 1) p ö 2
è 4n ø
lim
sin ç
÷=
p
n® ¥
è 4n ø p
n sin
4n
ò
ò
ò
n +1
67.
1
1
1
1
n+3
ò min.{| x - 1|,| x - 2|,| x - 3|, ¼¼| x - n|} dx = 2 (1) + 2 ´ 2 ´ (n - 1) + 2 ´ (1) = 4
0
1
1/2
1
68. S k =
2
3
n–1
n
1
æ kp ö
k sin ç
÷
2
è 2n ø
1
1
n
1
kp
1
2
æ px ö
k sin
=
x sin ç ÷ dx =
n® ¥ n 2
2
2
n
2
2
è ø
p2
k =1
lim
å
ò
0
g( x)
71. f( x) =
ò
dt
1+ t
0
f ¢( x) = g¢( x) ×
g¢( x) = - sin x × (1 + sin(cos x)) 2
3
1
1 + ( g( x)) 3
p
æpö
æpö
f ¢ ç ÷ = g¢ ç ÷ = - sin = -1
2
è2ø
è2ø
x
72. x 2 f( x) = ( 4t 2 - 2 f ¢(t )) dt
ò
4
2
x f ¢( x) + 2 x f( x) = 4 x 2 - 2 f ¢( x)
16 f ¢( 4) + 8 f( 4) = 64 - 2 f ¢( 4)
18 f ¢( 4) = 64
9 f ¢( 4) = 32
n+1
113
Indefinite and Definite Integration
2n
1 2n
r2
å n 3 + r 3 = nlim
å
n® ¥
®¥ n
73. lim
0
2
2
x 2 dx
ò 1+ x3
x2
cos (cos x) dx = 2 cos (cos x) dx = 2 ×
2
p
r =1
2p
ò
2
=
r =1
74.
æ rö
ç ÷
ènø
æ rö
1+ ç ÷
ènø
3
=
0
p
-1
-1
ò
1
1
ln|1 + x 3| = ln 9
3
3
0
= p2
0
0
x
75. 2 f( x) = ( x 2 - 2 xt + t 2 ) g(t ) dt
ò
0
x
x
x
2 f( x) = x 2 g(t ) dt - 2 x t × g(t ) dt + t 2 g(t ) dt
ò
ò
0
ò
0
0
x
2 f ¢( x) = x 2 × g( x) +
ò
x
g(t ) dt × 2 x - 2 x( x g( x)) - t × g(t ) dt × 2 + x 2 g( x)
ò
0
0
x
x
2 f ¢( x) = 2 x g(t ) dt - 2 tg(t ) dt
ò
ò
0
0
x
f ¢¢( x) = x × g( x) +
ò g(t) dt - xg( x)
0
x
f ¢¢( x) = g(t ) dt
ò
0
f ¢¢¢( x) = g( x)
p
76. I =
x 3 cos 4 x sin 2 x
ò p 2 - 3 px + 3 x 2
0
p
I=
3
p/ 2
dx = l
0
4
2
( p - x) cos x sin x
ò
p 2 - 3 px + 3 x 2
0
2
ò sin x dx
p
dx Þ 2 I = p cos 4 x sin 2 x dx
ò
0
3
77.
3
1
p
p
æ 2x ö
× tan -1 ç
= tan -1 x
= -0 =
÷
2
0
2
3
3
è1- x ø 0
3
78.
3
[ x]
ò { x}
dx = ( x - [ x])
ò
0
1
79. I =
ò
0
1
[ x]
0
-1
tan
x
x = tan q
x
dx
2
3
dx = 1 × dx + ( x - 1) dx + ( x - 2) 2 dx
ò
0
ò
1
ò
2
114
Solution of Advanced Problems in Mathematics for JEE
p/ 4
I=
ò
0
80.
q
× sec 2 q dq =
tan q
4/ p
4/ p
ò
ò
p/ 4
ò
0
2q
1
dq =
sin 2q
2
p/ 2
t
ò sin t dt
0
-
p
1 æ -1 ö 3
×ç
÷ × x dx x è x2 ø
1
1 ö 32 2
æ
- lim ç x 3 sin ÷ =
x
®
0
xø
è
2
p3
3x
+1
28
t
æ 2
ö
2
+ 4 ÷ dt =
ç 8t +
3
è
ø
log ( x +1) x + 1
x
81.
×
3
ò
cos
0
I
64
4p
4/ p
1
1
1
3 x 2 sin dx x cos dx = sin × x 3
123
x
x
x
0
0
0
II 123
ò
-1
8t 3 14t 2
+
+ 4t
3
3
3x
+1
= 2
1
-1
2
x
8 x 3 14 x 2
æ -8 14
ö
+
+ 4x - ç
+
- 4÷ = 3x + 2
3
3
3
è 3
ø
8 x 3 + 14 x 2 + 12 x + 8 - 14 + 12 = 9 x + 6
8 x 3 + 14 x 2 + 3 x = 0
x(8 x 2 + 14 x + 3) = 0
x(2 x + 3)( 4 x + 1) = 0
x = 0, -
3
1
,2
4
But x > -1 & x ¹ 0
1
So, x = 4
4
x
4
85. f( x) = e| x -t|dt = e( x -t ) dt + e(t - x ) dt = e x + e 4- x - 2 ³ 2 e 2 - 2
ò
ò
0
1
86.
4
87.
¥
ò
0
æ
4 cos 3 x
1
dx =
x
4
x
ò
0
x öæ
x
¥
ò
0
cos 3 x + 3 cos x
1
dx =
x
4
x
xö
æ
¥
ò
0
2 x
cos 3 x
3
dx +
x
4
2 xö
¥
ò
0
cos x
dx
x
ò çè sin 2 + cos 2 ÷ø çè cos 2 - sin 2 ÷ø dx = ò çè cos 2 - sin 2 ÷ø dx = ò cos x dx
4/ p
1
ò x cos x dx
0
115
Indefinite and Definite Integration
p
88. I
n+
1 =
2
p
p
ò
ò
sin(2nx + x)
sin 2nx × cos x
dx =
dx +
sin 2 x
sin 2 x
ò
0
0
=
p
p
ò
ò
0
cos 2nx × sin x
dx
sin 2 x
1 sin 2nx
1 cos 2nx
dx +
dx
2
sin x
2
cos x
0
0
1
89. f ¢( x) = 1 + ln 2 x + 2 ln x = 0 Þ x =
e
1/ e
1
æ 1ö
f ç ÷ = 1 + + (ln 2 t + 2 ln t ) dt
e
è eø
1
ò
1/ e
Let I =
2
ò (ln t + 2 ln t) dt
1
-1
x
x
ln t = x Þ t = e ; dt = e dx = ( x 2 + 2 x) e x dx = [e x × x 2 ]-01 =
ò
0
x
90. f( x) = x 2 + e -t f( x - t ) dt
ò
0
x
x
x 2 + et - x f(t ) dt = x 2 + e - x et f(t ) dt
ò
ò
0
0
x
f ¢( x) = 2 x - e - x
Þ
t
ò e f(t) dt + f( x)
0
f ¢( x) = 2 x + x 2 Þ f( x) =
Þ
y=
Þ
p/ 2
91.
I=
ò
1
( -2 x 2 + 6 x - 1)
4
cos 2 x
-p/ 2 1 + 5
x
p/ 2
dx =
p/ 2
Þ
2I =
2
ò cos x dx
-p/ 2
p/ 2
Þ
I=
2
p
ò cos x dx = 4
0
x3
+ x2
3
ò
cos 2 x
-p/ 2 1 + 5
-x
dx
1
e
116
92.
Solution of Advanced Problems in Mathematics for JEE
æ x 2 - x + 1ö
ò ççè x 2 + 1 ÷÷ø e
cot -1 x
Let cot -1 x = 1 Þ
t
dx
-1
1+ x2
2
dx = dt
t
ò e (cot t - cosec t) dt = e × cot t + c
= x × e cot
-1
x
+c
1
1
2
é( 1 + x 2 ) 3 / 2 ù
1 n æç r n 2 + r 2 ö÷
1 n r
æ rö
= lim
1 + ç ÷ = x 1 + x 2 dx = ê
ú
ç
÷ x ®¥ n
x ®¥ n
3
ènø
n2
r =1 è
r =1 n
ë
û0
0
ø
å
93. lim
å
( x 3 - 1)
x3
ò
1
1
94.
ò ( x 4 + 1)( x + 1)
95.
(cos -1 cos x)( - sin x)
- x sin x
1
= lim
=2
+
+
2 - 2 cos 2 x
4
x ®0
x ® 0 4 sin x
dx =
4
ò x 4 + 1 dx - ò x + 1 dx = 4 ln(1 + x ) - ln(1 + x) + c
lim
96. f(x)
0
x > tan1
cos x
0 < x < tan1
cos x
2
x = tan1
tan 1
¥
ò
f( x) dx =
0
ò
¥
cos x +
0
ò 0 dx = sin(tan 1)
tan 1
æ
ö
k
ç
÷ 1
k
1 n ç
æ
ö
n
÷ = x dx
97. lim
ç 2
÷ = lim
1
2
k
ç
÷
n ®¥
n ®¥ n
è
ø
n
+
n
+
2
k
k =1
k =1 1 +
0
+
ç
2 ÷
n
è
n ø
n
å
å
ò
y
ò|t - 1|dt
98.
99.
lim
1
y ®1 tan( y - 1)
+
Þ lim
y ®1
y -1
+
2
sec ( y - 1)
(Applying L’ Hospital Rule)
=0
1
é
ù
x
5 1
dx
=ê
ú +
2
4
2
4
1
2 3
0 (1 + x )
êë 2 ( 4 - 1)(1 + x )
úû 0 6 0 (1 + x )
ò
1
dx
ò
1
æ 1
ö 5é
ù
x
5 3 1
dx
=ç
- 0÷ + ê
+ ×
2 2
ç 6 ( 2) 3
÷ 6 ê 2 (2)(1 + x 2 ) 2 úú
0
è
ø
ë
û 0 6 4 (1 + x )
ò
117
Indefinite and Definite Integration
1
ù
1 æ 5ö é 1
x
5 1 1 dx
ù 5é
=
+ç ÷ê
- 0ú + ê
ú + ´
2
48 è 6 ø ë 16
8 2 01+ x2
û 8 êë 2 (1)(1 + x ) úû
ò
0
=
1
5
5æ 1
ö 5
+
+ ç - 0÷ +
[tan -1 x]10
48 6 ´ 16 8 è 4
16
ø
=
7
5
5 ép
ù
+
+
- 0ú
6 ´ 16 8 ´ 4 16 êë 4
û
22
5p
+
6 ´ 16 64
11 5p
=
+
48 64
=
Alternate solution :
I=
dx
1
ò0 (1 + x 2 ) 4
Put x = tan q; therefore, dx = sec 2 q dq.
p / 4 sec 2 q dq
I=
ò0
I=
ò0 (cos q) dq
(sec q) 8
That is,
p/ 4
6
p / 4 æ 3 cos q + cos 3q ö
2
÷ dq
0
4
ø
9 p/ 4
1 p/ 4
3 p/4
=
cos 2 q dq +
(cos 3q) 2 dq +
cos q cos 3q dq
0
0
16
16
8 0
9 p/ 4 1 + cos 2q
1 p/ 4 1 + cos 6q
3 p/ 4 cos 4q + cos 2q
=
dq +
dq +
dq
16 0
2
16 0
2
8 0
2
=
ò
ç
è
ò
ò
ò
ò
p/ 4
=
ò
ò
p/ 4
9 é
sin 2q ù
q+
32 êë
2 úû 0
+
1 é
sin 6q ù
q+
16 ´ 2 êë
6 úû 0
p/ 4
+
3 é sin 4q sin 2q ù
+
8 ´ 2 êë 4
2 úû 0
1 é p 1ù
3 é
1
æ 9 ö é p 1ù
ù
=ç ÷ê + ú +
- ú+
0 + - 0ú
ê
ê
2
è 32 ø ë 4 2 û 16 ´ 2 ë 4 6 û 8 ´ 2 ë
û
5
11
=
+
64 p 48
p/ 4
100. We have,
I=
4
ò (sin x) dx
0
We know that,
1 - cos 2 x
sin 2 x =
2
…(1)
118
Solution of Advanced Problems in Mathematics for JEE
Therefore,
2
æ 1 - cos 2 x ö
sin 4 x = (sin x) 4 = ç
÷
2
è
ø
1
= [1 - 2 cos 2 x + (cos 2 x) 2 ]
4
1æ
1 + cos 4 x ö
= ç 1 - 2 cos 2 x +
÷
4è
2
ø
=
1æ3
cos 4 x ö
ç - 2 cos 2 x +
÷
4è2
2 ø
Substituting this value of sin 4 x in Eq. (1), we get
p/ 4
I=
æ3
1
1
ö
ò çè 8 - 2 cos 2 x + 8 cos 4 x ÷ø dx
0
p/ 4
é3 ù
= ê xú
ë8 û 0
-
1
1
[sin 2 x]p0/ 4 +
[sin 4 x]p0/ 4
4
32
1
æ3 pö 1
= ç × ÷ - ( 1 - 0) +
( 0 - 0)
32
è8 4ø 4
3p 1
=
32 4
Alternate solution : We have,
p/ 4
I=
4
ò (sin x) dx
0
which can be written as
J = (sin 2 x)(1 - cos 2 x) dx
ò
1
= ò sin 2 x dx - ò 4 sin 2 x cos 2 x dx
4
1 - cos 2 x
1
dx (sin 2 x) 2 dx
2
4
1
1
1 1 - cos 4 x
= x - sin 2 x dx
2
4
4
2
x sin 2 x 1
1
= - x+
sin 4 x + c
2
4
8
32
3
sin 2 x sin 4 x
= x+
+c
8
4
32
=
ò
ò
ò
119
Indefinite and Definite Integration
Using the given limits, the above equation becomes
p/ 4
p/ 4
p/ 4
é3 ù
é sin 2 x ù
é sin 4 x ù
I = [ J ]p0/ 4 = ê x ú
-ê
+ê
ú
ë8 û 0
ë 4 û0
ë 32 úû 0
3p 1
=
32 4
(cos 9 x + cos 6 x) sin 5 x
5x
3x
101.
dx = 2 cos
cos
= (cos 4 x + cos x)
sin 10 x - sin 5 x
2
2
sin 4 x
=
+ sin x + C
4
1
A = ,B =1
4
dx
2014
æ x 2013 ö
1
x
÷+C
102.
=
ln ç
1
2013 çè 1 + x 2013 ÷ø
1+
x 2013
1
2
2
2
1 1
1é
ù
103.
x × (2 x × e - x ) dx = êæç - xe - x ö÷ + e - x dx ú
0
0
ø
2
2 ëè
û
1é 1
ù
= ê - + aú
2ë e
û
ò
ò
ò
ò
ò
ò
2
3
4
5
é 1
ù
104. 2 ê f( x) dx +
f( x) dx + f( x) dx ú + f( x) dx + f( x) dx
1
2
3
4
ë 0
û
ò
ò
ò
ò
ò
é 0 2 12 2 2 ù 3 2 4 2 35
=2ê
+
+
+
=
ú+
2
2 û
2
2
2
ë 2
105.
1
3
3x2
ò x 6 (1 + x 3 ) 2 dx
Let 1 + x 3 = t Þ 3 x 2 dx = dt
1
dt
1 æç 2
1
2
1 ö÷
Þ
=
+
+
dt
ç
2
2
2
3 t (t - 1)
3 èt t
t - 1 (t - 1) 2 ÷ø
ò
3n
106. lim
n ®¥
2
å
r =1
ò
3
1
n 1+
r
n
=
1
3
ò 1 + x dx = (2 1 + x ) 0 = 2
0
2
2 2
2
éx2
ù
x
x2
107. x f( x) dx = ê
f ( x)ú f ¢( x) dx = 0 +
dx
3
ë 2
û0 0 2
0
0 2 1+ x
ò
ò
ò
120
Solution of Advanced Problems in Mathematics for JEE
p/ 3
108.
ò (ln(cos x + 3 sin x) - ln cos x) dx
0
p/ 3
ì æ
ü
p öö
æ
p
ò íîlnçè 2 cos çè x - 3 ÷ø ÷ø - ln cos x ýþ dx = 3 ln 2
=
0
100 1
109.
åò
1
f( r - 1 + x) dx =
r =1 0
ò
1
f( x) dx +
0
1
=
ò
0
2
f( x) dx +
0
n
æ (2 x)
k =0
è
ò
1
f( x + 1) dx +
ò
0
3
f( x) dx +
1
ò
f( x) dx +¼ +
2
k ö
1
ø
0
111.
òx
5
1 + x 3 dx
Let 1 + x 3 = t 2
3 x 2 dx = 2t dt
2 2 2
2 æ t 5 t 3 ö÷
t (t - 1) dt = ç
+c
3
3 çè 5
3 ÷ø
ò
112. f ¢( x) =
sin x
x
f ¢( x) > 0 " x Î (0 , p)
f ¢( x) < 0 " x Î ( p , 2 p)
113.
ò
x( x 2 + 1) + 3 ( x 2 + 3)
( x 2 + 1)( x 2 + 3)
x
æ
3
dx
ö
ò çè x 2 + 3 + x 2 + 1 ÷ø dx
1
ln| x 2 + 3| + 3 tan -1 x + c
2
114.
sec 5 x
ò sin 3 x
dx =
sec 4 x
ò tan 3 x dx
Let tan x = t 2
sec 2 x dx = 2t dt
ò
(1 + t 4 ) × 2t × dt
t
3
æ 1
ö
=2 ç
+ t 2 ÷ dt
2
èt
ø
ò
ò f( x + 99) dx
0
100
å x 2 çç k ! ÷÷ = x 2 × e 2 x Þ ò x 2 e 2 x dx =
n ®¥
110. lim
ò
1
f( x + 2) dx +¼ +
ò
99
2
e -1
4
100
f( x) dx =
ò f( x) dx = a
0
121
Indefinite and Definite Integration
115. Let tx = y Þ x dt = dy
x2
òe
lim
sin y
0
x2
x ®0
p/ 2
116.
ò
0
dy
2
e sin x 2 x
= lim
=1
x ®0
2x
p/ 2
p/ 2
cos 2 x
æ sin 2 x ö
dx = ç
÷
x
è 2x ø0
+2
sin 2 x
ò (2 x) 2
p
dx = -1 +
0
sin q
(Q Let 2 x = q)
ò q 2 dq
0
Exercise-2 : One or More than One Answer is/are Correct
1.
dx
Let x = t 2 Þ dx = 2t dt
ò (1 + x ) 8
é
2t dt
dt
é
ù
dt
1
1
ù
ò (1 + t) 8 = 2 êë ò (t + 1)7 - ò (t + 1) 8 úû = 2 êêë - 6 (1 + t) 6 + 7(1 + t)7 úúû + C
a
2.
ò
-a
3. I =
=
=
4.
a
é e x + cos x ln( x + 1 + x 2 )ù dx = 2 e x dx = 2 ( e a - 1) Þ e a > 7
ëê
ûú
4
ò
0
x
Let x 3/ 2 = a 3/ 2 cos q
ò a 3 - x 3 dx
x
ò ( a 3/ 2 ) 2 - ( x 3/ 2 ) 2
dx =
2
3
a 3/ 2 cos q
ò a 3 - a 3 sin 2 q dq
æ x 3/ 2 ö
2
2
2
÷+C
dq = q + C = sin -1 ç
ç a 3/ 2 ÷
3
3
3
è
ø
ò
æ
ö
ò x sin x sec x dx = ò {xI çç 1tan44x2sec443x ÷÷ dx = x
3
è
=x
2
II
ø
ò
f( x) = sec 2 x , g( x) = tan x
(a) Clear f( x) Ï( -1, 1)
(b) tan x = sin x
ò
tan 2 x
(sec 2 x - 1)
tan 2 x 1
dx = x
- (tan x - x) + C
2
2
2
2
1
= ( x sec 2 x - tan x) + C
2
\
tan 2 x
tan 2 x
dx
2
2
122
Solution of Advanced Problems in Mathematics for JEE
cos x = 1 Þ tan x is not defined.
Þ
no solution
(c) g¢( x) = f( x) " x Î R except (2n - 1)
p
2
(d) sec 2 x = tan x
1 + tan 2 x - tan x = 0 has no solution.
5.
ò (sin 3q + sin q) cos q e
sin q
dq = ( 4 sin q - 4 sin 3 q) e sin q cos q dq
ò
Let t = sin q
dt = cos q dq = 4 [-t 3 + 3t 2 - 5 t + 5] et + C
Compare it
A = -4 , B = -12 , C = -20
q
7. I =
q
2 x dx
0
Þ
2 (q - x) dx
ò (3q - 2 x)(q + 2 x) ò (3q - 2 x)(q + 2 x)
=
0
I=
q
2
q
dx
ò q 2 - ( x - q 2) 2
0
8. Let f( x) = a x , F ( x) = F ( - x)
9. J =
=
0
2
ò
ò
é
ù
æ
-1 æ 1 ö
-1
-1 æ 1 ö
-1 ö
êcot ç x ÷ + cot ( x)ú dx + ç cot ç x ÷ + cot x ÷ dx
è
ø
è
ø
û
ø
-1 ë
0 è
0
2
ò
ò
pö
p
æ
dx
ç p + ÷ dx +
2ø
2
è
-1
0
p
(As 2p is period)
K = dx = p
ò
0
11. l1 = lim
x® ¥
x - cos 2 x
=1
x + sin x
1
l 2 = lim
h dx
2 tan
ò h 2 + x 2 = hlim
®0
h® 0
-1 1
+
-1
13.
dx
sec 4 x
h
=p
ò (1 + sin 2 x) cos 2 x ò 1 + 2 tan 2 x dx
=
123
Indefinite and Definite Integration
14.
ò
(1 + tan 2 x) sec 2 x
1
1 sec 2 x dx
sec 2 x dx +
2
2 1 + 2 tan 2 x
ò
=
1
1
tan x +
tan -1 ( 2 tan x) + c
2
2 2
(1 + 2 tan 2 x)
(1 + sin 2015 x) - 1 + sin 4030 x
2 sin 2015 x
2014
dx =
=
1
ò 2 dx
ò
(Rationalise)
ò odd = 0
-2014
15. tan -1 (nx)|¥
a =
p
- tan -1 (na)
2
a > 0, a = 0, a < 0
16. Let
x = cos 2q
dx = -sin 4q dq
p/ 4
I=
p/ 4
ò cot q sin 4q dq and J = ò tan q sin 4q dq
0
0
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
2. f( x) = (2 x cos x + 6 x 2 sin x cos x - 2 x 3 sin 2 x) dx
ò
3
2
æ
ö
= ç 2 x 3 cos 2 x + 3 x 2 sin 2 x ÷ dx
4 4
3
ç 123 12
÷
è I
ø
II
ò
f( x) = x 3 sin 2 x + c
f ( p) = 0 + c = 0 Þ c = 0
f( x) = x 3 sin 2 x
Paragraph for Question Nos. 6 to 8
6. g( x) = x - A
1
A=
ò f(t) dt
0
ò
124
Solution of Advanced Problems in Mathematics for JEE
x
f( x) =
x3
x3
x3
+ 1 - x ( x - A) dx =
+ 1+ Ax 2
2
2
2
ò
0
2
f( x) = Ax + 1
1
A = Ax 2 + 1 Þ A =
ò
0
f( x) =
7.
3
2
3x2
+ 1 ; min. f( x) = 1
2
3 2
3
x + 1= x 2
2
3x2 - 2x + 5 = 0
D < 0, no solution
8. g( x) = x A=
3
2
1 3 3 9
´ ´ =
2 2 2 8
Paragraph for Question Nos. 9 to 11
a
9.
ò
1
f( x) dx -
0
ò f( x) dx = 2 f( a) + 3 a + b
a
Diff. w.r.t. ‘ a’ on both sides,
( f( a) - 0) - (0 - f( a)) = 2 f ¢( a) + 3
2 f( a) = 2 f ¢( a) + 3
(2 f( a) - 3) = 2 f ¢( a)
2 f ¢( a)
=1
2 f( a) - 3
2 f ¢( a)
ò 2 f( a) - 3 da = ò da
ln|2 f( a) - 3| = a + c
2 f( a) - 3 = e a + c
2 f( a) - 3 = ke a
2 f( a) = ke a + 3
…(1)
125
Indefinite and Definite Integration
Put a = 1 Þ 0 = ke + 3 Þ k = 2 f( a) = -
\
3
e
3 a
e +3
e
f( a) =
3 3 a
e
2 2e
f( x) =
3 3 x
e
2 2e
Put f( x) in (1)
(By taking limiting case)
a
1
æ3
3
xö
æ3
3
3
xö
a
ò çè 2 - 2 e e ÷ø dx - ò çè 2 - 2 e e ÷ø dx = 3 - e e + 3 a + b
0
a
éæ 3 a 3 a ö æ
3 ö ù éæ 3 3 ö æç 3 a 3 e a ö÷ ù
3 a
e
0
ç
÷
ç
÷ ú - êç - ÷ - ç
ú = 3 - e + 3a + b
ê 2 2e
÷
2 e ø û êëè 2 2 ø è 2
2 e ø úû
e
ø è
ëè
3
-3 = b
2e
10.
Length of subtangent =
y
( dy dx)
y = f( x) =
3 3 x
e
2 2e
dy
3 x
= f ¢( x) = 0 e
dx
2e
dy
3
=dx x =1/ 2
2 e
when x =
1
1 ö
æ 1 ö 3 3 1/ 2 3 æ
, y = fç ÷= e
= ç1 ÷
2
2
2
e
2
2
è ø
eø
è
3æ
1 ö
ç1 ÷
2è
eø
Length of subtangent =
=
3
2 e
11.
1
1
ò
ò
e -1 = e -1
1
3x 3 x
3 ö
æ3 3 x ö
æ 3 3e ö æ
f( x) dx = ç e ÷ dx =
e
= ç - ÷ - ç0 - ÷
2 2e ø
2 2e
2
2
e
2e
è
è
ø
è
ø
0
0
0
3
æ3 3ö 3
=ç - ÷ +
=
è 2 2 ø 2e 2e
126
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 12 to 13
12. f3¢¢¢( x) = f0 ( x) see options or 3 times by parts.
13. fn ( x) =
xn æ
1 1
1ö
ç ln x - 1 - - ¼ - ÷
|n è
2 3
nø
Paragraph for Question Nos. 14 to 15
2
1ö
3
æ
f( x) = a ç x - ÷ +
2ø
4
è
f(1) = 1
a =1
2
f( x) = x - x + 1
14. g( x) = 1 + x 2
15.
now integrate
ex
x
ò e2x - e x + 1 e = t
Paragraph for Question Nos. 16 to 17
1/ 2
æ 1
ö
( xe x ) 2 ç
+ 3÷
2
xe 2 x (1 + 3 x 2 ) 1/ 2
èx
ø
17. L = lim
= lim
x ®¥ C × ( xe x ) C -1 × ( e x + xe x )
x ®¥
ö
x C æ1
C × ( xe ) × ç + 1÷
èx
ø
Exercise-4 : Matching Type Problems
1
2. (A)
ò
dx
3
x
=
1 ö
2
æ
( x 2 + 1) x 2 + 2
ç1 + 2 ÷ 1 + 2
è
x ø
x
dx
Let 1 +
x
2
x +x
8
4
(C)
2
ò
=t2
1 ö
æ
ç x + 3 ÷ dx
è
x ø
ò (1 - x 4 ) 7 / 2 = ò æ 1
2ö
ç 2 -x ÷
èx
ø
Let
1
x2
- x2 =t2
7/2
127
Indefinite and Definite Integration
(D) Let
x = cos 2q Þ dx = -2 sin 4q dq
1- x
ò 1 + x dx = - 2 ò tan q sin 4q dq
3. (A) Let sin x = t
1
ò
0
Þ cos x dx = dt
dt
=
(1 + t )(2 + t )
1
1 ö
1
0
41p / 4
(B)
æ 1
ò çè t + 1 - t + 2 ÷ø dt = [ln(1 + t) - ln(t + 2)]0
p/ 4
p
ò|cos x|dx = 10 ò|cos x|dx + ò cos x dx
0
0
0
p
é p/ 2
ù p/ 4
= 10 ê cos x dx - cos x dx ú + cos x dx
êë 0
úû
p/ 2
0
ò
0
(C)
ò
1/ 2
[ x] dx +
ò
ò
-1/ 2
1/ 2
[ x] dx +
0
0
ò
1/ 2
-1 dx +
-1/ 2
p/ 2
æ1+ x ö
ò ln çè 1 - x ÷ø dx
-1/ 2
=
(D) I =
p/ 2
2 cos q
dq =
0
p/ 2
2I =
2
1
ò 0 dx = - 2
0
ò 3 ( cos q + sin q )
Þ
ò
2 sin q
ò 3 ( sin q + cos q ) dq
0
p
ò 3 dq = 3
0
I=
Þ
p
6
4. (A) Common root a = b - a Þ 3 ( b - a) 2 + a( b - a) + 1 = 0 Þ 2 a 2 + 3 b 2 - 5ab + 1 = 0
4
(B)
x +1
2x2
(C) y =
(D)
ò
= sin 2
px
Þ
2
x2 +
1
1
1
+
-2
2
x -1
( x - 1)
æ x ö
ç
÷
è1+ x ø
7/6
dx
x2
.
1
x 2 = sin 2 px Þ x = ±1
2
2
x ¹1 ;
1
¹ -2 , 1
x -1
128
Solution of Advanced Problems in Mathematics for JEE
Let
\
x +1 6
1
=t Þ dx = 6t 5 dt
2
x
x
I = 6 t -7 ( -t 5 ) dt =
ò
6
æ x ö
+ C = 6ç
÷
t
è x + 1ø
1/ 6
+C
5. (A) We have,
1. 5
ò
1
[ x 2 ] dx = [ x 2 ] dx +
ò
0
0
1
0
1. 5
[ x 2 ] dx +
ò [ x ] dx
ò
1
= 0 dx +
ò
2
2
2
2
1. 5
1 × dx +
ò 2 × dx = 0 + ( 2 - 1) + 2 (15. - 2 ) = 2 - 2
ò
1
2
(B) We have,
4
ò
1
{ x } dx =
0
ò
0
4
Aliter :
4
x dx + ( x - 1) dx =
ò
1
4
2 2
7
+ (8 - 1) - 3 =
3 3
3
4
ò { x } dx = ò x dx - ò [ x ] dx
0
0
0
(C) We have,
æp
ö
sin x + cos x = 2 sinç + x ÷
è4
ø
\
é
æp
öù
[sin x + cos x] = ê 2 sin ç + x ÷ ú
è4
øû
ë
é
æp
öù
The graph of y = ê 2 sin ç + x ÷ ú is obtained from the graph of y = [ 2 sin x] by
4
è
øû
ë
p
translating it by units in the direction of OX ¢. The graph so obtained is shown in figure.
4
It is evident from the graph of y = [ 2 sin( x + p 4)] that
ì1,
ï0 ,
ï
ï-1,
f( x) = [sin x + cos x] = í
ï-2 ,
ï-1,
ï
î0 ,
0£ x£p 2
p 2 < x £ 3p 4
3p 4 < x £ p
p < x < 3p 2
3p 2 £ x < 7p 4
7p 4 £ x < 2p
129
Indefinite and Definite Integration
y
(0,–1)
p/4 p/2 3p/4
x¢
p
5p/4 3p/2 7p/4 2p
x
(0,–1)
(0,–2)
y¢
2p
ò [sin x + cos x] dx
\
0
p/ 2
=
ò
0
=
1 × dx +
3p/ 4
ò
3p/ 2
p
0 dx +
p/ 2
ò
( -1) dx +
3p/ 4
ò
7 p/ 4
( -2) dx +
ò
2p
(-1) dx +
3p/ 2
p
ò 0 dx
7 p/ 4
p
p
p
p
+ 0 - - 2 ´ + ( -1) + 0 = -p
2
4
2
4
(D) We have,
p
ò||sin x| -|cos x||dx
0
p/ 4
=
3p/ 4
p
ò (|sin x| -|cos x|) dx + ò (|sin x| -|cos x|) dx + ò -(|sin x| -|cos x|) dx
0
p/ 4
3p/ 4
y
y = |sin x|
y = |cos x|
x¢
p/4
0
p/2 3p/2 p
x
y¢
p/ 4
= - (sin x - cos x) dx +
ò
0
p/ 2
ò
(sin x - cos x) dx +
p/ 4
3p/ 4
ò (sin x + cos x) dx
p/ 2
p
+
ò -(sin x + cos x) dx
3p/ 4
130
Solution of Advanced Problems in Mathematics for JEE
= - [- cos x - sin x]p0/ 4 + [- cos x - sin x]pp// 24 + [- cos x - sin x]3p/p2/ 4 - [- cos x + sin x]p3p/ 4
= - [- 2 + 1] + [-1 + 2 ] + [ 2 - 1] - [1 - 2 ] = 4 2 - 4 = 4 ( 2 - 1)
Exercise-5 : Subjective Type Problems
1.
x dx
ò 1 - 9x2 ò
+
I1 =
(cos -1 3 x) 2
1 - 9x2
dx = I 1 + I 2
x dx
I2 =
ò 1 - 9x2
Let 1 - 9 x 2 = t 2
¥
2. I =
ò
(cos -1 3 x) 2
1 - 9x2
dx
Let cos -1 3 x = k
x 3 dx
ò (a 2 + x 2 )5
0
I=
1
a
2I =
p/ 2
ò
6
sin 3 q cos 5 q dq =
0
p/ 2
1
8a
a
6
1
3
6
1
p/ 2
3
5
ò cos q sin q dq
0
p/ 2
ò sin 2q dq = 32 a 6 ò (3 sin 2q - sin 6q) dq
0
0
I=
Þ
1
24 a 6
2p
3.
ò g( x) dx
0
2p
(Q x = f(t ))
ò t f ¢(t) dt
3p/ 2
2p
=
2
2
3p
ò (t cos t - t sin t) dt = 4 p - 2 - 1
3p/ 2
4.
5
3
6
4
2
ò ( x + x + x) 2 x + 3 x + 6 x dx
Let 2 x 6 + 3 x 4 + 6 x 2 = t 2 Þ 12( x 5 + x 3 + x) dx = 2t dt
=
1
1
2t 2 dt =
(2 x 6 + 3 x 4 + 6 x 2 ) 3/ 2 + C
12
18
ò
(Let x = a tan q)
131
Indefinite and Definite Integration
5. Put x = sin q
6.
tan x
ò tan 2 x + tan x + 1 dx
Let tan x = t , sec 2 x dx = dt, dx =
t
dt
æ
dt
1+ t2
1
1
ö
dt
ò (1 + t + t 2 ) × 1 + t 2 = ò çè 1 + t 2 - 1 + t + t 2 ÷ø dt = ò 1 + t 2 - ò
1
= tan -1 (t ) -
æ 3ö
1ö
æ
÷
ç t + ÷ + çç
÷
2
è
ø
è 2 ø
2
æ 2t + 1 ö
tan -1 ç
÷+C
è 3 ø
æ 3ö
ç
÷
ç 2 ÷
è
ø
= tan -1 (tan x) -
dt
2
2
æ 2 tan x + 1 ö
tan -1 ç
÷+C
3
3
è
ø
7. Let x 4 = t
4 x 3 dx = dt
-2011
1ö
æ
1
ç1 + ÷
1
1
1
tø
1 + t 2010
1
1
(1 + t ) -2011
è
dt =
dt +
dt =
+
2012
2012
2012
-2011
2011
1ö
0
0 (1 + t )
0 (1 + t )
0 2æ
t ç1 + ÷
tø
è
ò
ò
1
ò
0
-1 æ 1
1 æ 1
-1 æ 1
1 ö
1
l
ö
ö
=
ç 2011 - 1÷ +
ç 2011 - 0 ÷ =
ç 2011 - 1 - 2011 ÷ =
2011 è 2
ø 2011 è 2
ø 2011 è 2
ø 2011 m
2
3
8.
2
3
2
( x 2 x x + 2 x 2 x × x ln x) dx =
ò
1
2x2
( x + 2 x ln x) dx
1
3
ò æçè x
x2 ö
( 3 )3
ò
1
2
2
Let x x = t Þ x 2 ln x = ln t ; (2 x ln x + x) dx =
÷ ( x + 2 x ln x) dx
ø
1
9.
òx
dt
t ×
=
t
2
( 3 )3
ò
1
t2
t dt =
2
33 / 2
=
1
dt
t
33 -1
= 13
2
(cos x - sin x) dx
dx
ò (cos x - sin x)(1 + sin x cos x) = 2 ò (cos x - sin x) 2 (2 + (sin x + cos x) 2 - 1)
=2
(cos x - sin x) dx
ò ((sin 2 x + cos 2 x) - 2 sin x cos x)(1 + (sin x + cos x) 2 )
132
Solution of Advanced Problems in Mathematics for JEE
(cos x - sin x) dx
=2
ò (2 - (sin x + cos x) 2 )(1 + (sin x + cos x) 2 )
=2
ò (2 - t 2 )(1 + t 2 ) where t = sin x + cos x
=2
=
dt
dt
2é
1
dt
æ 2 + t öù
2é
1
-1
÷ú + C
lnçç
êtan (t ) +
3 êë
2 2 è 2 - t ÷ø úû
2
1
, B=
3
3 2
\
A=
\
12 A + 9 2 B - 3 = 12 ×
10. x a × y = la ;
y=
2
1
+9 2
-3 =8
3
3 2
la
(0, y1–mx1)
xa
dy
xa × y
= - ala x -a -1 = - a
dx
x a +1
m=
Þ
(x1, y1)
- ay 1
x1
y
x1 – m1 , 0
y
1
1
A = | y 1 - mx 1| x 1 - 1 = y 1 x 1 (1 + a) 2
2
m
2
=
1 a 1- a
l × x 1 (1 + a) 2
2
For A to be constant 1 - a = 0
p
p
p
æ x 6 ( p - x) 9 ö
( p - x) 9
ç
÷
11. I ( 6, 8) = x ( p - x) dx = + 6x5
dx
ç
÷
9
9
è
ø0 0
0
ò
6
8
ò
p
I ( 6, 8) =
6 ! ´ 8 ! 15
6
6 5
4
3
2
1
I ( 5, 9) = ´
´
´
´
´
( p - x) 14 dx =
p
9
9 10 11 12 13 14
15 !
ò
0
100
14. I =
ò
0
I=
155
3
ù
ò (2 - t 2 )(1 + t 2 ) = 3 êë ò 1 + t 2 dt + ò 2 - t 2 úû
2
2
2
2
3
102
é1
ù
ê
x dx - 0 × dx + dx +¼¼ + 2 dx +¼¼ + 9 dx ú
ê
ú
12
22
92
ë0
û
ò
ò
ò
ò
133
Indefinite and Definite Integration
1
17. f(q) =
sin q dx
ò
-1( x - cos q)
2
+ sin 2 q
= tan
-1 æ x - cos q ö
ç
è
sin q
1
÷
ø -1
Clearly, f(q) is not defined when sin q = 0
q = 0, p,2p
20. f( x) =
x
x
x
x
ù
1
1é
( x - t ) 2 g(t ) dt = ê x 2 g(t ) dt - 2 x tg(t ) dt + t 2 g(t ) dt ú
2
2ê
úû
0
0
0
ë 0
ò
ò
ò
ò
x
é x
ù
f ¢( x) = ê x g(t ) dt - t g(t ) dt ú
êë 0
úû
0
ò
ò
x
f ¢¢( x) = g(t ) dt
ò
0
f ¢¢¢( x) = g( x)
2
22. f(2 - x) = f(2 + x), it means it symmetric about x = 2 Þ
ò f( x) dx = ò f( x) dx = 5
0
Let 2 - x = t ; f(t ) = f( 4 - t ) i . e. ,
4
2
f( x) = f( 4 - x) = f( 4 + x)
50
æ2
ö
f( x) dx = 25 ç f( x) dx ÷ = 25 ´ 5 = 125
ç
÷
0
è0
ø
ò
ò
1
é1 æ
æ
x2 x3
x 2n ö÷
x3 x5
x 2n +1 ö÷ ù
23. I n = | x|ç 1 + x +
+
+¼¼ +
dx = 2 ê ç x +
+
+¼¼ +
dx ú
ç
ç
2
3
2n ÷ø
2
4
2n ÷ø ú
ê
è
è
-1
ë0
û
ò
ò
1
é x2
ù
x4
x6
x 2n +1
=2ê
+
+
+¼¼ +
ú
2n × ( 2n + 2) û
ë1 × 2 2 × 4 4 × 6
0
1
1
1
é 1
ù
=2ê
+
+
+¼¼ +
2n × (2n + 2) úû
ë1 × 2 2 × 4 4 × 6
In =1+
1æ 1
1
1
1æ1
1 ö
ö
+
+¼¼ +
ç
÷ =1+ ç ÷
2 è 1×2 2 ×3
n(n + 1) ø
2 è 1 n + 1ø
lim I n = 1 +
n® ¥
1 3 p
= =
2 2 q
pq( p + q) = 3 ´ 2(5) = 30
b
25.
ò|sin x|dx = 8
a
Þ b - a = 4p
134
Solution of Advanced Problems in Mathematics for JEE
a+ b
ò|cos x|dx = 9
Þ a+ b=
0
1
b
1
9p
p
17 p
Þ a = ;b=
2
4
4
17 p / 4
ò x sin x × dx = 2 p ò x sin x dx = 2
2p a
p/ 4
x
28. f( x) = 0 Þ
òe
-y
f ¢( y) dy = x 2 - x + 1
0
Þ e - x f ¢( x) = 2 x - 1 Þ f( x) = (2 x - 3)e x
p
æ 1 - cos np ö
æp
ö
29. I n = 2 ç -| x|÷ cos nx dx = 2 ç
÷
è2
ø
è
ø
n2
0
ò
1 ö 40
æ
I1 + I 2 + I 3 + I 4 = 4 ç 1 + ÷ =
9ø 9
è
❑❑❑
135
Area Under Curves
6
AREA UNDER CURVES
Exercise-1 : Single Choice Problems
1.
O
2
1
3
4
–1
1£ x + 3 y < 2
Area of shaded region =
1
3
1
2. Area = 2 ( x - x 2 ) dx
ò
0
y
–1
1
3. y 2 = ( x 2 - 1) 2 = 4 (1 - x 2 ) dx =
ò
0
O
x
1
8
3
y
1
–1
ln 2
4. Area = 4
æ -x
ò çè e
0
-
1ö
÷ dx = 2 - 2 ln 2
2ø
x
136
Solution of Advanced Problems in Mathematics for JEE
y
(0, 1/2)
O
(–ln2, 0)
(ln2, 0)
x
(0, –1/2)
y
5. As given relations are inverse of each other so A lies on y = x
æ
ö
n
n
ç
÷
i . e. ,
,
ç
÷
2
2
è n + 1 n + 1ø
A
O
So, required area = 8 area (OACBO) = 8 ( DOAB + area BACB )
2
æ æ
ö
1
ö
ç 1ç
÷
n
÷
=8ç
+
n × 1 - x 2 dx ÷
ç
÷
ç 2 è n2 + 1ø
÷
n / n 2 +1
è
ø
BC
x
ò
3
6.
æ
x2 ö
ò ççè 12 x - 12 ÷÷ø dx
0
12
æ
x2 ö
y
ò ççè 12 x - 12 ÷÷ø dx
15
=
49
3
8.
12
3
O
x
dy
=r
dx (1, 1)
1
Normal at (1, 1) is Þ y = - ( x - 1) + 1
r
1
Required area = x r dx + ar ( DPAB ) =
ò
0
f ¢( r) = -
1
( r + 1)
2
+
P
1
1
+ r = f( r)
r+1 2
O
1
=0
2
é1
1ù 2
9. Area = 4 ê 1 - x dx - ú =
2ú 3
êë 0
û
ò
y
(–1,0)
O
(1,0)
x
(1,1)
A r B
((r+1),0)
nr
137
Area Under Curves
10. Area of DAOB =
1 -x2
xe
2
y
(0,1)
2
dA
= (1 - 2 x 2 ) e - x
dx
2
A(x,e–x )
Area is maximum at x =
1
2
(0,0) O
B(x, 0)
2
12.
ò g( x) dx
1
Let x = f(t ) Þ dx = f ¢(t ) dt
1
1
2
ò t(3t - 6 t + 3) dt = 4
0
13. x 2 y 2 + y 4 - x 2 - 5 y 2 + 4 = 0
( x 2 + y 2 - 4)( y 2 - 1) = 0
y
x)
–1 (
y
=
(5, 5)
f
14.
(1, 1)
)
f(x
=
y
y=x
15. Ar. of shaded region =
x
p3
1
Ar. of circle =
.
2
2
y
O
x
16. We have,
y 2 = x (1 - x 3 )
…(1)
For x > 1, y 2 is negative. Since the square of a real number cannot be negative, y does not exist
1
at x = 0 or at x = 1 ; y = 0. Let x = . Therefore, from Eq. (1), we get
2
y
1
1ö 7
æ
y 2 = ç1 - ÷ =
2è
8 ø 16
y =±
7
4
O
1
x
138
Solution of Advanced Problems in Mathematics for JEE
Also, for x < 0, y 2 is negative. Therefore, the required area is
1
2
ò
1
y dx = 2 ( +) x 1 - x 3 dx
ò
0
0
1
=2
4
ò x - x dx
0
17.
y
0
18. | x| + | y| ³ 2 and
x
x1 p x2 2p
y2
x2
+
=1
9
4
y
O
x
Ar. of ellipse - Ar. of square = p (2)(3) - 8 = 6 p - 8
Exercise-2 : One or More than One Answer is/are Correct
1. (a) f( x) = - ( x - a)( x - b)( x + c) = ( x - a)( x)( x + c)
Clearly option (a) is correct.
c
(b)
ò
a
b
f( x) dx =
ò
[Q b = 0]
c
f( x) dx +
a
ò f( x) dx > 0 (from graph)
b
which incorrect
b
(c)
ò
a
b
f( x) dx < 0 &
ò f( x) dx < 0 but second term is large negative value so option (c) is
c
incorrect.
(d) Clearly, (d) is incorrect.
139
Area Under Curves
2. T n =
Let
( r n)
1 3n -1 ( r n)
1 3n
, Sn =
2
n r = 2 n 1 + ( r n)
n r = 2 n + 1 1 + ( r n) 2
å
å
f( x) =
f ¢( x) =
\
x
1+ x2
–
(1 + x 2 ) - 2 x 2
=
(1 + x 2 ) 2
1- x2
+
–1
–
1
2
(1 + x 2 ) 2
f( x) is decreasing in (2, 3).
3
Tn >
ò
3
f( x) dx, S n <
2
2
ò f( x) dx
2
3.
…(1)
a + b=2
4
ò ( a x + bx) dx = 8
0
4. Normal y + x =
1/ 2
æ7
Þ
2a
+ b=1
3
…(2)
7
4
ö
2
ò çè 4 - x ÷ø - ( x + 1) dx
–3/2
-3/ 2
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
Sol. f( x) =
3
x+a
2
bx + cx + 2
f( -1) = 0 Þ a = 1
If y = 1 is asymptotes, then b = 0 and c = 1 Þ f( x) =
y = g(x)
y = f(x)
1
–2
–1 O
x +1
1 - 2x
and g( x) =
x +2
x -1
O
1
x
–1
–2
x
1/2
140
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 4 to 6
Sol. y = e - x sin x
dy
= e - x [cos x - sin x] = 0
dx
p
tan x = 1 Þ x = np +
4
Þ
0
So, (i . e. , 0 £ x £ p)
I = e - x sin x dx =
ò
- e-x
[sin x + cos x] + c
2
( j+ 1)p
Sj =
e - x sin x dx = -
ò
jp
4. Put j = 0 , S 0 =
5.
6.
S 2009
=
S 2010
S j+1
1+ e
2
e -2009p
e
-2010p
e-x
[sin x + cos x]
2
-p
= ep
= e -p
Sj
1 + e -p
S0
1 + ep
2
Sj =
=
=
1 - e -p 1 - e -p 2( e p - 1)
j= 0
¥
Q
å
Exercise-5 : Subjective Type Problems
1. f( x) = x 2
1
A = 2 ( 2 - x 2 - x 2 ) dx =
ò
0
p 1
+
2 3
2. f( x) = 2 ln x
1
A = ( - x 3 + 6 x 2 - 11x + 6 - 2 ln x) dx =
ò
0
4. At x = 0 , y = 0
x + 5y - y 5 = 0 Þ 1 + 5y¢ - 5y 4 y¢ = 0
at x = 0 , y = 0
y¢ = -
1
5
17
4
( j+ 1)p
=
jp
e - jp - p
( e + 1)
2
p
2p 3p
141
Area Under Curves
Equation of tangent : y = -
x
, equation of normal : y = 5 x
5
1
´ 5 ´ 26 = 65
2
Area =
[ x]2 = [ y]2
5.
y
[ y] = ± [ x]
[ y] = ± 1,
1£ x < 2
= ± 2,
2£ x<3
= ± 3,
3£ x<4
= ± 4,
4£ x<5
= ± 5,
x =5
Now, when, x Î[1, 2)
then
y Î [-1, 0) È [1, 2)
when
x Î[2 , 3]
\
5
4
3
2
1
0
–1
–2
–3
then
when
then
when
then
when
then
Hence,
y Î [-2 , - 1) È [2 , 3)
x Î[3 , 4)
y Î [-3 , - 2) È [3 , 4)
x Î[4 , 5)
y Î [-4 , - 3) È [4 , 5)
x =5
y Î [-5, - 4) È [5, 6)
required area = 2 ( 4) = 8 sq. unit
6. f( x) + f( z ) = f( x + z ) and f(0) = 0 and f ¢(0) = 4 Þ f( x) = 4 x
4
So, area bounded = D = ( 4 x - x 2 ) dx
ò
0
7. Required area = 4
1
1
ò
x 2 - x 6 dx = 4 x 1 - x 4 dx =
ò
0
0
p
2
y
–1
O
Put x 2 = sin q
1
1
5
8
æ
ö
8. Ar = 4 (1 - x 2/ 5 ) dx = 4 ç x - x 7 / 5 ÷ =
7
è
ø0 7
0
ò
1
x
–4
–5
1
2
3
4
5
x
142
Solution of Advanced Problems in Mathematics for JEE
7
DIFFERENTIAL EQUATIONS
Exercise-1 : Single Choice Problems
4.
dy
xy
=
2
dx x + y 2
Let
ò
5.
y
dy
dt
=t Þ
=t + x
x
dx
dx
dx
1+ t2
=dt ;
x
t3
ò
y
ò 1 - y 2 dy = ò dx
- 1- y2 = x + c
Þ ( x + c) 2 + y 2 = 1
6.
dy
dx
ò y = - ò ( x - 3) 2
Þ
ln y =
1
+c
x -3
7. Let f( x) = y
2
dy
ex
- 2 xy =
dx
( x + 1) 2
8. Let x 2 y 2 = t
2 xy 2 + 2 x 2 y
dy dt
=
dx dx
dt
= tan t
dx
ln y =
x2
2y
2
-
1
2
(Q y(1) = 1)
143
Differential Equation
9. y = (C 1 cos C 2 ) cos x + (C 5 - C 1 sin C 2 ) sin x - C 3 eC 4 e - x
y = A cos x + B sin x + Ce - x
Q
C 1 , C 2 , C 3 , C 4 are arbitrary constants.
10. y = e( a +1) x
y ¢ = e( a +1) x (a + 1)
y ¢¢ = e( a +1) x (a + 1) 2
12.
dy æ
f ¢( x) ö
÷ y = f( x)
- çç 1 +
dx è
f( x) ÷ø
æ
f ¢( x ) ö
÷ dx
- ò çç 1+
e-x
f ( x ) ÷ø
è
I.F. = e
=
f( x)
-x
ye
= e - x dx + C
f( x)
ò
ye - x
= -e - x + C
f( x)
Þ
æ t2 ö
13. Equation of tangent at ç t , ÷ is
ç 2 ÷
è
ø
y = tx -
t2
2
Þ t=
dy
dx
Differential equation is
2
dy
æ dy ö
+ 2y =0
ç ÷ - 2x
dx
è dx ø
2
2
dy dt
1 dt
x
e-x
14. Let x + y = t Þ 1 +
=
;
+
=x Þ
= e-x + C
3
2
2
dx dx
t dx t
( x + y)
15.
dy
- 2 y tan x = tan 2 x
dx
-2 tan x dx
I.F. = e ò
= cos 2 x
y cos 2 x = sin 2 x dx =
ò
1
(1 - cos 2 x) dx
2
ò
16. f( x) = 2 e x + 1
17. Let
dy
d 2 y dt
= t then
=
dx
dx 2 dx
dt
2tx
=
dx x 2 + 1
Þ
t=
dy
= 3( x 2 + 1)
dx
(Q y ¢(0) = 3)
144
Solution of Advanced Problems in Mathematics for JEE
y = x3 + 3x + 1
Þ
(Q y(0) = 1)
18. cy 2 = 2 x + c
2 cyy¢ = 2 Þ c =
1
yy ¢
y 2 = 2 x yy ¢ + 1
19. Let cosec y = t
- cosec y cot y dy = dt
dt t
1
- =dx x
x2
Þ
t
1
1
1
=
+c Þ
=
+c
x 2x2
x sin y 2 x 2
Þ
20.
xdy - ydx
x2
æ yö
dç ÷
è xø
ò
æ yö
1+ ç ÷
è xø
21. lim
t®x
Þ
22.
=
2
x2 + y2
=
dx
òx
t 3 f ( x) - x 3 f (t )
2
dx
x2
2
=
1
Þ 3 x f( x) - x 2 f ¢( x) = 1
2
t -x
dy 3 y -1
=
Þ
dx
x
x2
y=
1
3
+ x2
4x 4
(Q f(1) = 1)
2 dp(t )
= p(t ) - 900
dt
dp(t )
2
= dt
p(t ) - 900
ò
ò
2 ln|900 - p(t )| = t + c
p(t ) = 900 - 50 et / 2
(Q p(0) = 850)
p(t ) = 0 Þ t = 2 ln 18
sin y dy tan x
23.
+
= sec x
cos 2 y dx cos y
Let
sin y
1
=t Þ
dy = dt
cos y
cos 2 y
dt
+ t × tan x = sec x
dx
145
Differential Equation
t × sec x = sec 2 x × dx + C
ò
sec y × sec x = tan x + C
dy
24.
= ( 4 x + y + 1) 2
dx
Let 4 x + y + 1 = t
dy dt
dt
4+
=
Þ
=t2 + 4
dx dx
dx
1
ætö
tan -1 ç ÷ = x + C
2
è2ø
dy y
dx
25. tan q = dx x = y dy
dy
1+ ×
x dx
y
P
q
q
x
O
2
Þ
2 y dy
æ dy ö
=1
ç ÷ x dx
è dx ø
1
26. I.F. = e
ò x dx
= e ln x = x
y × x = x 3 dx =
ò
x4
+c
4
27. x 3 dy + 3 x 2 y dx = y 2 dx + 2 xy dy
d ( x 3 y) = d ( xy 2 )
3
2
3
ò d ( x y) = ò d ( xy ) Þ x y = xy
2
Exercise-2 : One or More than One Answer is/are Correct
1.
xdy - ydx
ò
x2 -2
dx
x2
x2
æ
2 ö
æ yö
dç ÷ = ç1 ÷ dx
x
è ø
è
x2 ø
=
ò
y = x2 - 2x + 2
(Q f(1) = 1)
146
Solution of Advanced Problems in Mathematics for JEE
2. I.F. = x sec x ; yx sec x = tan x + c
3. Put y = h
Þ
or
Þ
x [ f( x + h) - f( x - h)] - h [ f( x + h) + f( x - h)] = 2( x 2 h - h 3 )
lim x
h ®0
[ f( x + h) - f( x - h)]
- [ f( x + h) + f( x - h)] = lim 2( x 2 - h 2 )
h ®0
h
xf ¢( x) - f( x) = x 2
Þ f( x) = x 2 + x
4. L.D.E., I.F. = 1 + sin 2 x ; (1 + sin 2 x) f = sin x + C , C = 0
5. 2 ydx + 2 xdy + (2 x 2 y 3/ 2 dx + x 3 y 1/ 2 dy) = 0
2
d( x 3 ´ y 3/ 2 ) = 0
3
y
y
sin 2 x
1
6. I.F. =
;
=
dx ;
= 2 cot x × cosec x dx = -2 cosec x + c
3
3
3
sin x sin x
sin x
sin 3 x
æ
ö
æpö
y = -2 sin 2 x + 4 sin 3 x ççQ y ç ÷ = 2 ÷÷
2
è ø
è
ø
2 d( xy) +
ò
ò
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
x
x
Sol. x g(t ) dt + (1 - t ) g(t ) dt = x 4 + x 2
ò
0
ò
0
differentiate w.r.t. ‘ x’
x
x g( x) +
3
ò g(t) dt + (1 - x) g( x) = 4 x + 2 x
…(1)
0
1. From (1)
x
3
ò g(t) dt + g( x) = 4 x + 2 x
0
Let g( x) + g¢( x) = 12 x 2 + 2 Þ
dy
+ y = 12 x 2 + 2
dx
2. Put x = 0 in (1) we get g(0) = 0
Paragraph for Question Nos. 3 to 5
3. f( g( x)) = e -2 x
x × [ f( g( x))]¢ [ g( f( x))]¢
=
f( g( x))
g[ f( x)]
(Q y = g( x))
147
Differential Equation
Þ
g( f( x)) = e - x
2
2
H( x) = e -( x -1) +1
4. f( g(0)) + g( f(0)) = 2
5. H( x) max = e
Paragraph for Question Nos. 6 to 8
æ eh - 1ö
g( x + h) - g( x)
e x g( h)
÷
= lim
+ lim g( x) ç
ç h ÷
h ®0
h ®0
h ®0
h
h
è
ø
Sol. g¢( x) = lim
= 2e x + g( x)
dy
- y = 2e x
dx
Þ
Þ y = 2 x e x + ce x
y = 2x ex
(Q g(0) = 0)
Exercise-4 : Matching Type Problems
1. (A) y
(B) y
dy
dx
dx
dx
dx
= - (1 + x) Þ
=- x = y2
+ 1 Þ ( y 2 - y)
dy
dy
dy
1+ x
y( y - 1)
dx
dx æ 2 ö
+ çç ÷÷ x = 10 y 2
+ 2 x = 10 y 3 Þ
dy
dy è y ø
2
I.E. = e
ò y dy = e
2 ln y
=y 2
d( xy 2 ) = 10 y 4
(C)
dy
= y¢
dx
y ¢ y ¢¢¢ = (3 y ¢¢) 2
y ¢¢¢ 3 y ¢¢
then integrate it.
=
y ¢¢
y¢
(D) Put x 2 = t
dt t
1
+ =
dy y y 3
then solve it.
(Q g¢(0) = 2)
148
Solution of Advanced Problems in Mathematics for JEE
Exercise-5 : Subjective Type Problems
1. x a × y = la ;
y=
la
(0, y1–mx1)
xa
dy
xa × y
= -ala x -a -1 = -a
dx
x a +1
m=
Þ
(x1, y1)
-ay 1
x1
y
1
1
A = | y 1 - mx 1| x 1 - 1 = y 1 x 1 (1 + a) 2
2
m
2
=
1 a 1- a
l × x 1 (1 + a) 2
2
For A to be constant 1 - a = 0.
dy
2.
= xy(1 + y)
dx
dy
= x dx
(1 + y) y
ò
ò
x2
2y
=e 2
1+ y
f ( 2) =
Þ
(Q f(0) = 1)
e2
2 - e2
3. y 2 = cos 2 x + 2
2y
y
dy
= -sin 2 x
dx
æ dy ö
+ ç ÷ = -cos 2 x
2
è dx ø
dx
y4 + y3
4.
2
d2 y
é æ dy ö 2
ù
= (cos 2 x + 2) 2 + (cos 2 x + 2) ê - ç ÷ - cos 2 x ú = 6
êë è dx ø
úû
dx 2
d2 y
t 2 f( x + 1) - ( x + 1) 2 f(t )
=1
t ® x +1
f(t ) - f( x + 1)
lim
Þ
2tf( x + 1) - ( x + 1) 2 f ¢(t )
=1
t ® x +1
f ¢(t )
lim
Þ [ x + 1][2 f( x + 1) - ( x + 1) f ¢( x + 1)] = f ¢( x + 1)
y
x1 – m1 , 0
149
Differential Equation
2 xf( x)
Þ
f ¢( x) =
Þ
f( x) = x 2 + 1
Þ
ln( f( x)) - ln 2
f ¢( x)
= lim
=1
x ®1
x ®1 f ( x )
x -1
x2 + 1
lim
❑❑❑
150
Solution of Advanced Problems in Mathematics for JEE
8
QUADRATIC EQUATIONS
Exercise-1 : Single Choice Problems
1. Let 3 x / 2 = a , 2 y = b
a 2 - b 2 = 77 , a - b = 7 Þ a = 3 x / 2 = 9 Þ x = 4
b=2y =2 Þ y =1
2. f( x) =
3
3
i =1
i =1
Õ( x - ai ) + å ai - 3 x = ( x - a1 )( x - a 2 )( x - a 3 ) + ( a1 + a 2 + a 3 ) - 3 x
f ( a1 ) = a 2 + a 3 - 2 a1 > 0
f ( a 3 ) = a1 + a 2 - 2 a 3 < 0
( a1 < a 2 < a 3 )
a1
a3
3. x 4 - 2 ax 2 + x + a 2 - a = 0
a 2 - a(2 x 2 + 1) + x 4 + x = 0
a=
a = x 2 + x,
1
a³ - ,
4
4. x 3 - 3 x 2 - 4 x + 12 = 0
2 x 2 + 1 ± (2 x - 1)
2
a = x2 - x + 1
3
a³
(Q x Î R )
4
a
b
g
Equation whose roots are a - 3 , b - 3 , g - 3 is
( x + 3) 3 - 3 ( x + 3) 2 - 4( x + 3) + 12 = 0
f( x) = x 3 + 6 x 2 + 5 x = 0
a–3
b–3
g–3
x
151
Quadratic Equations
5. | K - 1| < 3
3
-3 < K - 1< 3
-2 < K < 4
5
3
–1 1
–1
2
6.
x
1
x - x -6
( x - 3)( x + 2)
£0 Þ
- £0 Þ
£0
x( x + 6)
x +6 x
x( x + 6)
–
+
–6
–
+
–2
+
0
3
x Î ( - 6 , - 2] È ( 0 , 3]
7. P( x) = x 4 - 8 x 2 + 15 + 2 x 3 - 6 x = ( x 2 - 3)( x 2 - 5) + 2 x( x 2 - 3)
= ( x 2 - 3)( x 2 + 2 x - 5)
Q ( x) = ( x + 2)( x 2 + 2 x - 5)
l
-5
-7
, b = 1, g =
,f=
,c =6
2
2
2
1
l 2 -5 2
5 10
l2
1
-7 2 = 0 Þ l = ,
2 3
-5 2 -7 2
6
8. a = 1, h =
10. \ Let f( x) =| x - a| + | x - b|
a+b–2x
Suppose a > b
2x–a–b
b–a
f(0) = f(1) = f( -1)
\
f( x) = const. in [b, a]
So,
b
a
b £ -1 < a £ 1
a - b£ 2
Minimum |a - b| = 2
\
2
12. y =
x + 2x + c
x 2 + 4 x + 3c
Þ ( y - 1) x 2 + 2 (2 y - 1) x + (3 cy - c) = 0
D ³ 0 " y Î R and D £ 0
But at c = 0 and 1 there will be common factors among numerator and denominator.
c( c - 1) < 0
Þ
2
13. f(t ) = t - mt + 2 = 0
f(2) < 0
Þ
4 - 2m + 2 < 0 Þ m > 3
( D ³ 0)
152
Solution of Advanced Problems in Mathematics for JEE
3| x|
But
2
9 + | x|
æ 3| x| ö
çç
÷
2 ÷
è9 + x ø
So,
=
3
3
£
9
6
+ | x|
| x|
m
£
1
2m
<1
(by A.M. G.M. in equality)
[Qm > 3]
éæ 3| x| ö m ù
êçç
÷÷ ú = 0
êëè 9 + x 2 ø úû
14. x 2 ( x 6 - 24 x 5 - 18 x 3 + 39) = - 3 ´ 5 ´ 7 ´ 11
If ‘ x’ is integer, then there is no value of ‘ x’ .
1
15.
m4 +
= 119
m4
1
Þ m2 +
= 11
m2
2
Þ
1ö
æ
çm - ÷ = 9
mø
è
1 öæ
1
ö
æ
= çm - ÷çm 2 +
+ 1÷ =|3 ´ 12| = 36
2
m øè
è
ø
m
m
a
16. ax 2 + 2 bx + c = 0
b
m3 -
1
3
ax 2 + 2 cx + b = 0
a
g
By condition of common root
1
a
Þ Common root a = and + b + c = 0
2
4
2c
2b
and g =
Þ
b=
a
a
Equation whose roots are b and g is
4 bc
æ 2c 2b ö
x2 -ç
+
÷ x + 2 =0
a ø
è a
a
2 a 2 x 2 + a 2 x + 8 bc = 0
17.
9 x 2 ( x - 1) - 1( x - 1) = 0
1
1
,3
3
1
1
cos a = 1, cos b = , cos g = 3
3
x = 1, x =
æa
ö
ç + b + c = 0÷
è4
ø
153
Quadratic Equations
\
\
18. y =
a = 0, b + g = p
(Sa , S cos a ) = ( p , 1) = centre
é
pö ù é æpö ù
-1 æ
ê2 sin ç tan 4 ÷ , 4 ú = ê2 ç 2 ÷ , 4 ú = ( p , 4) ® point lies on the circle.
è
ø û ë è ø û
ë
Radius is 3.
11x 2 - 12 x - 6
x2 + 4x + 2
( y - 11) x 2 + ( 4 y + 12) x + (2 y + 6) = 0 " x Î R
D³0
2
( 4 y + 12) - 4( y - 11)(2 y + 6) ³ 0
y 2 + 20 y + 51 ³ 0
( y + 17)( y + 3) ³ 0
y Î ( -¥, - 17] È [- 3 , ¥)
1
³0
x2 - x -2 x - 4
x +3
19.
( x 2 - x - 12) - ( x 2 - x - 2)
³0
( x 2 - x - 2)( x - 4)
-10
³0
( x - 2)( x + 1)( x - 4)
Þ
20.
( x + 1)( x - 2)( x - 4) < 0
x Î ( -¥, - 1) È (2 , 4)
x = 4 + 3i
( x - 4) 2 = - 9 Þ x 2 - 8 x + 25 = 0
x 3 - 4 x 2 - 7 x + 12 = ( x 2 - 8 x + 25)( x + 4) - 88 = -88
21. f( x) =
x2 + x -1
x2 - x + 1
y
5/3
1
–1
0
–1
By graph : Min. = f(0); Max. = f(2)
5
If x Î [-1, 3]; y max. =
3
x
1
2
3
154
Solution of Advanced Problems in Mathematics for JEE
22. By graph min. = f(0); max. = f(1)
if x Î [-1, 1]; y Î [-1, 1]
1
1
1
23.
+
=
x+p x+q r
x 2 + x( p + q - 2 r) + pq - r( p + q) = 0
If one root is a. Then other root must be -a.
p+q
p + q - 2r = 0 Þ r =
2
Product of the roots = pq - r( p + q) = pq -
( p + q) 2
( p 2 + q2 )
=2
2
24. If a1 x 2 + b1 x + c1 = 0 has one root a.
1
.
a
Þ
a 2 x 2 + b2 x + c 2 = 0 has one root
Þ
c 2 x 2 + b2 x + a 2 = 0 has one root a
Condition of common root is
( a1 a 2 - c1 c 2 ) 2 = ( a1 b2 - b1 c 2 )( a 2 b1 - b2 c1 )
25. If a 2 - 5a + 3 = 0 and b 2 - 5 b + 3 = 0
Þ
x 2 - 5 x + 3 = 0 has two roots a and b.
Þ
a + b = 5, ab = 3
Sum of the roots =
a b (a + b) 2 - 2ab 19
+ =
=
b a
ab
3
Product of roots =
a b
´ =1
b a
Equation whose roots are
26.
b
a
and is 3 x 2 - 19 x + 3 = 0
b
a
|a - b| =|a 1 - b 1|
a 2 - 4b = b2 - 4a
a 2 - b 2 = 4 ( b - a)
( a - b)( a + b + 4) = 0
a ¹ b Þ a + b+ 4 =0
27. tan(q 1 + q 2 + q 3 + q 4 ) =
S1 - S 3
sin 2 b - cos b
cos b (2 sin b - 1)
=
=
= cot b
1 - S 2 + S 4 1 - cos 2 b - sin b sin b (2 sin b - 1)
28. ( a 2 + b 2 ) x 2 + 2 x( bd + ac) + ( c 2 + d 2 ) = 0
( a 2 x 2 + 2 acx + c 2 ) + ( b 2 x 2 + 2 bdx + d 2 ) = 0
155
Quadratic Equations
( ax + c) 2 + ( bx + d) 2 > 0
Þ
This equation has imaginary roots.
29. If a , b are roots of ax 2 + bx + c = 0
30.
Þ
a + 2 , b + 2 are roots of a( x - 2) 2 + b( x - 2) + c = 0
Þ
ax 2 + x( b - 4 a) + 4 a - 2 b + c = 0
a +b =1+ l
ab = l - 2
a + b - ab = 3
(a - 1)(b - 1) = - 2
Þ atleast one root is positive.
31. D ³ 0 Þ 3 k 2 + 8 k - 16 £ 0 Þ - 4 £ k £
4
3
a 2 + b 2 = (a + b) 2 - 2ab = k 2 - 2( k 2 + 2 k - 4) = -k 2 - 4 k + 8 = 12 - ( k + 2) 2
32.
P ( x ) = ( x - 2) Q 1 ( x ) + R ( x )
Q ( x ) = ( x - 2) Q 2 ( x ) + R ( x )
Þ P ( 2) = Q ( 2)
33. a + b = - a and ab = b
if b ¹ 0 , a = 1 and b = -2
2
1ö
9
æ
x 2 + ax + b = x 2 + x - 2 = ç x + ÷ 2ø
4
è
3
34.
æ böæ c ö
æ cö
x2 + ç ÷ç ÷ x + ç ÷ =0
è aøè aø
è aø
2
3 3
x - (a + b) × ab x + a b = 0
a2b
ab2
35. x 2 + 2 ( a + b + c) x + 6 k( ab + bc + ca) = 0
D³0
Þ
4( a + b + c) 2 - 24 k( ab + bc + ca) ³ 0
Þ
k£
also,
|a - b| < c , |b - c| < a , |c - a| < b
Þ
a 2 + b 2 + c 2 - 2( ab + bc + ca) < 0
Þ
a 2 + b2 + c 2
2
<2 Þ k<
ab + bc + ca
3
ö
1 æç a 2 + b 2 + c 2
+ 2÷
ç
÷
6 è ab + bc + ca
ø
156
Solution of Advanced Problems in Mathematics for JEE
9| x|2 -18| x| + 5 = 0
36.
Þ (3| x| - 1)(3| x| - 5) = 0
x =±
Þ
1
5
,±
3
3
and x 2 - x - 2 > 0 Þ ( x - 2)( x + 1) > 0 Þ x < -1 or x > 2
37. Difference of roots is same in both equation
b2 - c = B 2 - C
38. | x - p| + | x - 15| + | x - p - 15| = ( x - p) - ( x - 15) - ( x - P - 15) = 30 - x
min. = 15
a = –1/2
39. 4 p( q - r) x 2 - 2 q( r - p) x + r( p - q) = 0
b = –1/2
If x = -
1
is also the root of 4 x 2 - 2 x - m = 0
2
Þ m =2
40. Let cos x = t
Þ
t Î [-1, 1]
Þ
kt 2 - kt + 1 ³ 0 " t Î [-1, 1]
Case I : k ³ 0
1
x coordinate of vertex is .
2
æ 1ö
Þ
fç ÷³0
è2ø
k k
Þ
- + 1³ 0
4 2
Þ
Also,
–1
1/2
1
k£ 4
k³0
Þ
k Î[0 , 4]
Case II : k < 0
Þ
f(1) ³ 0
k - k + 1³ 0
and
and
Þ
1³ 0
and
Also,
Þ
k<0
é 1 ö
k Î ê- , 0 ÷
ë 2 ø
Þ
f( -1) ³ 0
k + k + 1³ 0
1
k³ 2
é 1 ù
k Î ê- , 4ú
ë 2 û
–1
1/2
1
157
Quadratic Equations
41.
y -1
1+ x
=y Þ x=
1- x
y +1
3
æ y - 1ö
æ y - 1ö
÷÷ - 2 çç
÷÷ + 5 = 0
H( y) = 3 çç
è y + 1ø
è y + 1ø
H( y) = 3 ( y - 1) 3 - 2 ( y - 1)( y + 1) 2 + 5 ( y + 1) 3 = 0
H( y) = 3 ( y 3 - 3 y 2 + 3 y - 1) - 2 ( y - 1)( y 2 + 2 y + 1) + 5 ( y 3 + 3 y 2 + 3 y + 1) = 0
H( y) = 3 ( y 3 - 3 y 2 + 3 y - 1) - 2 ( y 3 + y 2 - y - 1) + 5 ( y 3 + 3 y 2 + 3 y + 1) = 0
H( y) = 3 y 3 + 2 y 2 + 13 y + 2 = 0
H ¢( x) = 9 x 2 + 4 x + 13 Þ D < 0
H( x) > 0 " x > 0
Hence, it has one –ve real root.
42. (l2 + l - 2) x 2 + (l + 2) x - 1 < 0 " x Î R
l 2 + l - 2 < 0 Ç ( l + 2) 2 + 4 ( l 2 + l - 2) < 0
(l + 2)(l - 1) < 0 Ç 5l2 + 8l - 4 < 0
Þ
2ö
æ
l Î ( - 2 , 1) Ç l Î ç -2 , ÷
5ø
è
2ö
æ
l Î ç - 2, ÷
5ø
è
l = -2 is also the solution of this equation.
43. a = 1, b = 1, g = 1, d = 1 (as) (a - 1) 2 + (b - 1) 2 + ( g - 1) 2 + (d - 1) 2 = 0
\
\
The roots of given equation is equal to 1.
a
S2 = 2 = 6
a0
44. | x - 1| + | x - 2| + | x - 3| ³ 6
Case I :
x³3
3x - 6 ³ 6 Þ x ³ 4
Case II :
2< x<3
(Not possible)
x³6
Case III :
1£ x £ 2
4-x³6
(Not possible)
Þ
x £ -2
Case IV :
x<1
6 - 3x ³ 6
x£0
x Î ( -¥, 0] È [4 , ¥)
158
Solution of Advanced Problems in Mathematics for JEE
45.
x
0
x
3
0
Case-I : f(0) > 0 Ç f(3) £ 0
3
Case-II : f(3) > 0 Ç f(0) £ 0
a
46. x 3 + 3 px 2 + 3 qx + r = 0
b
g
2 1 1
= + ¼¼
b a g
(Q a , b , g are in H.P.)
Þ
3 1 1 1 ab + bg + ag
= + + =
b a b g
abg
Þ
b=-
r
q
which satisfy the given equation.
47. 4 y 2 + 4 xy + ( x + 6) = 0 " y Î R
D ³ 0 Þ x2 - x -6 ³ 0
48. log cos x 2 (3 - 2 x) < log cos x 2 (2 x - 1)
0 < cos x 2 < 1 Ç 3 - 2 x > 2 x - 1 Ç 3 - 2 x > 0 Ç 2 x - 1 > 0
49. px 2 + qx + r = 0
x<1
a
x<3 2
x>12
b
ab < 0
Þ
2
a( x - b) + b ( x - a ) 2 = (a + b) x 2 - 4abx + ab (a + b) = 0
Product of roots = ab < 0
D = 16a 2b 2 - 4ab (a + b) 2 = - 4ab (a - b) 2 > 0
a
3
2
b
50. x + 2 x - 4 x - 4 = 0
c
4x3 + 4x2 - 2x - 1 = 0
q = 1, r = -
1
1
,s=2
4
51. log 2 ( x 2 + 3 x) £ 2
0 < x2 + 3x £ 4
1/a
1/b
1/c
159
Quadratic Equations
52. k - 2 > 0 Ç D < 0
k > 2 Ç ( k + 6)( k - 4) > 0
Þ
k> 4
53. ab < 0
3m - 8
8
<0 Þ 2<m<
m -2
3
æ x2 + x ö
÷>1
54. log 6 ç
ç x+4 ÷
è
ø
x2 + x
x 2 - 5 x - 24
( x - 8)( x + 3)
>6 Þ
>0 Þ
>0
x+4
x+4
x+4
a
55. ax 2 + c = 0
b
a + b = 0, ab =
c
a
a 3 + b 3 = (a + b)(a 2 + b 2 - ab) = 0
56. ( k - 1) x 2 - ( k + 1) x + ( k + 1) > 0 " x Î R
k - 1 > 0 Ç ( k + 1) 2 - 4( k - 1)( k + 1) < 0
k>1
Þ
Ç ( k + 1)(3 k - 5) > 0
5
k>
3
57. y = - 2 x 2 - 4 ax + k; abscissa corresponding to the vertex is now,
b
æ 4a ö
i . e. , ç
÷ = -2 Þ a = 2
2a
è -4 ø
y( - 2) = 7
7 = - 8 + 16 + k Þ k = -1
58. If a + b + c = 0
Sum of coefficient ( b + c - a) + ( c + a - b) + ( a + b - c) = a + b + c = 0
Þ
Þ
x = 1 is one root of the equation.
a+ b-c
other root =
b+ c-a
59. x 3 - ax 2 + bx - c = 0
a
–a
b
Sum of roots a - a + b = a Þ b = a
If b is root of the equation, then ab = c.
60. a ¢b ¢ = 2 q 2 - r = 2a 2b 2 - (a 4 + b 4 ) = - (a 2 - b 2 ) 2 < 0
160
Solution of Advanced Problems in Mathematics for JEE
62. In DABC ,
A
B
C
A
B
C
+ cot + cot = cot × cot × cot
2
2
2
2
2
2
A
B
C
A
C
B
If cot , cot , cot are in A.P. then cot + cot = 2 cot
2
2
2
2
2
2
A
C
Þ
cot cot = 3
2
2
cot
63. f( x) = x " x Î [-9 , 9]
(3| x| - 3) 2 =| x| + 7
64.
Þ (| x| - 2)(9| x| - 1) = 0
| x| = 2 ,
1
1
Þ x = ± 2, ±
9
9
y = x ( x - 4)
D f : ( -¥, 0] È [4 , ¥)
65. x 2 + 3| x| + 2 = 0 Þ (| x| + 2)(| x| + 1) = 0
a
66. x 2 - bx + c = 0
a+1
b-1
Sum of roots 2a = b - 1 Þ a =
2
2
æ b - 1ö
æ b - 1ö
2
If a is the root of equation, then ç
÷ - bç
÷ + c = 0 Þ b - 4c = 1
è 2 ø
è 2 ø
67. y =
3 x 2 + 9 x + 17
3x2 + 9x + 7
3 ( y - 1) x 2 + 9 ( y - 1) x + (7 y - 17) = 0
y - 1 ¹ 0 then D ³ 0
81( y - 1) 2 - 12 ( y - 1)(7 y - 17) ³ 0
( y - 1)( y - 41) £ 0
1 < y £ 41
68.
x2 + 2x + 7
-6 < 0" xÎR
2x + 3
x 2 - 10 x - 11
<0
( 2 x + 3)
( x - 11)( x + 1)
<0
2x + 3
3ö
æ
x Î ç -¥, - ÷ È ( - 1, 11)
2ø
è
161
Quadratic Equations
69. y =
70.
5y + 2
3x - 2
ì3 ü
Þ x=
Þ yÎR -í ý
7x + 5
3 -7y
î7 þ
x +2
£ 0 Þ x Î [ - 2 , 4)
x -4
a
2
x - ax - 4 £ 0
Þ
b
–2
4
f( -2) ³ 0 Ç f( 4) > 0
a³0 Ça<3
a Î[0 , 3)
71. P( x) = ( P - 3) x 2 - 2 Px + (3 P - 6) " x Î R
P -3 > 0
Ç
D =0
P>3
Ç
P 2 - ( P - 3)(3 P - 6) = 0
x
Þ P =6
72. Graph is downward Þ a < 0
Graph cut y-axis Þ c > 0
-b
x-coordinate of vertex
< 0 Þ b< 0
2a
y
O
73. ax 2 + bx + c = 0 does not have real roots.
D<0
a>0
a<0
x
y = x2–3x+5
x Î [–4, 1]
y
5
74.
11
4
–4
0
x
1 3/2
y Î[3 , 33]
75. 3 x 2 - 17 x + 10 = 0 Þ ( x - 5)(3 x - 2) = 0
If x = 5 is common root, then m = 0
2
26
If x = is common root, then m =
3
9
x
x
162
Solution of Advanced Problems in Mathematics for JEE
76. x 2 + ( y + 2) x - ( y 2 + y - 1) = 0
8ù
æ
D ³ 0 Þ ( y + 2) 2 + 4 ( y 2 + y - 1) ³ 0 Þ y Î ç -¥, - ú È [0 , ¥)
5û
è
77. If x = 3 is root of this equation, then k = -5
Þ
3 x 4 - 6 x 3 - 5 x 2 - 8 x - 12 = ( x - 3)(3 x 2 + 4)( x + 1)
78. a = -
( 4 + sin 4 x)
2
put sin 2 x = t Þ t Î[0 , 1]
sin x
æ4
ö
a = - ç + t ÷ = f (t )
èt
ø
4
Here, f ¢(t ) =
- 1> 0
t2
\ For atleast one real root, a Î ( -¥, - 5]
O
–5
–¥
79. ( rs) 2 + ( st ) 2 + (tr) 2 = ( rs + st + tr) 2 - 2 rst ( r + s + t ) = b 2 - 2 ( - c)( - a)
80. Let the roots be t , t + 1 and t + 2.
t + (t + 1) + (t + 2) = - a Þ 3 (t + 1) = - a
å t (t + 1) = b
Þ
b + 1 = 3 (t + 1) 2
a2
[3(t + 1)]2
=
=3
b + 1 3(t + 1) 2
81. (3 x 2 + kx + 3)( x 2 + kx - 1) = 0
D1 = k 2 - 36 and D 2 = k 2 + 4 > 0
2
82.
3
1
1 1
æ rö
æ rö
æ rö
= +
Þ ç ÷ + ç ÷ + 1=0 Þ ç ÷ =1
r+s r s
è sø
è sø
è sø
84. If
x Î ( -¥, - 2] È [3 , ¥)
x 2 - 2 x - 8 = 0 Þ x = -2 , 4
if
x Î ( -2 , 3)
x2 = 4 Þ x = ±2
85. 5 x 2 + 12 x + 3 = 0 has D < 0
Þ
86.
Both roots common.
a +b + g =6
ab + bg + ag = 5
abg = 1
Þ
a 2 + b 2 + g 2 = 26
a 2b 2 + b 2 g 2 + a 2 g 2 = 13
1
y = f(t)
163
Quadratic Equations
a + 5i
2
a – 5i
2
2
87. 2 x - 6 x + k = 0
Sum of roots = a = 3
Product of roots =
a 2 + 25 k
= Þ k = 17
4
2
88. x 12 + x 22 = ( k - 2) 2 - 2 ( k 2 + 3 k + 5) = - ( k 2 + 10 k + 6) £ 18
89. a( x 2 - x + 1) - ( x 2 + x + 1) ³ 0
a³
Þ
x2 + x + 1
x2 - x + 1
90. f(1) = l - 13 > 0 Þ l > 13
f(2) = l - 18 < 0 Þ l < 18
f(3) = l - 15 > 0 Þ l > 15
Þ l Î (15, 18)
94. D = ( b - c) 2 + 4 a(2 b + a + c) = ( b - c) 2 + ( 4 ac - 4 b 2 ) + (2 a + 2 b) 2 > 0
a
3
95. x - x + 1 = 0
b
c
1
a+1
1
(1 - x) 3 - x 2 (1 - x) + x 3 = 0
b+1
1
c+1
Þ
x3 + 2x2 - 3x + 1 = 0
1
1
1
Sum of roots =
+
+
= -2
1+ a 1+ b 1+ c
96. x 2 - 2 ( 4 k - 1) x + 15k 2 - 2 k - 7 ³ 0 " x Î R
D£0
Þ
k2 - 6 k - 8 £ 0 Þ 2 £ k £ 4
97. f( x) =
x3 -1
( x - 1)( x 2 - x + 1)
=
x2 + x + 1
(Q x ¹ 1)
x2 - x + 1
y
3
1
–1
1/3
0
x
1
164
98.
Solution of Advanced Problems in Mathematics for JEE
2x2 + 2
x 2 + mx + 4
> 0" xÎR
Þ x 2 + mx + 4 > 0 " x Î R
Þ D < 0 Þ m 2 - 16 < 0
99. x 2 - 2|a + 1| x + 1 = 0
D ³ 0 Þ 4 ( a + 1) 2 - 4 ³ 0 Þ a Î ( -¥, - 2] È [0 , ¥)
100. P( x) = a1 x 2 + 2 b1 x + c1 > 0 ; D1 = 4 ( b12 - a1 c1 ) < 0 , a1 > 0 , c1 > 0
Q ( x) = a 2 x 2 + 2 b2 x + c 2 > 0 ; D 2 = 4 ( b22 - a 2 c 2 ) < 0 , a 2 > 0 , c 2 > 0
f( x) = a1 a 2 x 2 + b1 b2 x + c1 c 2
D = b12 b22 - 4 a1 a 2 c1 c 2 < 0
101. x 2 - 2 x + 4 = -3 cos( ax + b)
( x - 1) 2 + 3 = -3 cos( ax + b)
Þ
x = 1 and ax + b = p
102. a + b = a + a × r = 4
g + d = a × r 2 + ar 3 = 36 Þ r = 3 , a = 1
A = 3 , B = 243 = 3 5
103.
y = |2 – |x||
(0,2)
y = 3–|x|
(0,1)
(–2,0)
104. We have 4 x 2 - 16 x + 15 < 0 Þ
(2,0)
3
5
< x < Þ cot a = 2 , the integral solution of the given
2
2
inequality and sin b = tan 45° = 1
\
sin(a + b) sin(a - b) = sin 2 a - sin 2 b =
1
2
1 + cot a
-1=
1
4
-1= 1+ 4
5
Þ 2 + log e x = x
Þ log e x = ( x - 2)
Clearly graphs intersect once in (0,1).
Now check
Þ g( x) = 2 + ln x - x
g( e) > 0
g( e 2 ) < 0
Þ
one root between ( e, e 2 )
(1,0)
105. f1 ( x) = f2 ( x)
(0,–2)
(2,0)
165
Quadratic Equations
106. x 4 - 3 x 3 - 2 x 2 - 3 x + 1 = 0
Þ x2 - 3x - 2 -
3
1
+
=0
x x2
2
1 ö
1ö
1ö
1ö
æ 2
æ
æ
æ
ç x + 2 ÷ - 3 ç x + ÷ - 2 = 0Þ ç x + ÷ - 3 ç x + ÷ - 4 = 0
xø
xø
xø
è
è
è
è
x ø
1
Þ t 2 - 3t - 4 = 0 (where x + = t )
x
Þ (t - 4)(t + 1) = 0 Þ t = 4 or t = -1
1
1
Þ x + = 4 or x + = -1
x
x
1
Real solutions are from x + = 4 Þ x 2 + 1 = 4 x Þ x 2 - 4 x + 1 = 0
x
Hence, sum of roots = 4.
Þ
107. f( x) = x 2 - ( k + 4) + k 2 - 12
f( 4) = 16 - 4( k + 4) + k 2 - 12 < 0
Þ
-2 < k < 6
2
108. a + b 2 = (a + b) 2 - 2ab = k 2 - 2( k 2 + 2 k - 4) = -k 2 - 4 k + 8
Maximum value = 12
109. f( x) = a x - x ln a
f ¢( x) = ( a x - 1) × ln a
110. As a , b and c are the roots of x 3 + 2 x 2 + 1 = 0 , we have
a + b + c = -2
ab + bc + ca = 0
a b c
Now, for finding the value of b c a , evaluating using first row, we get
c a b
a( bc - a 2 ) - b( b 2 - ac) + c( ab - c 2 ) = abc - a 3 - b 3 + abc + abc - c 3
= 3 abc - a 3 - b 3 - c 3
= - ( a 3 + b 3 + c 3 - 3 abc)
= - ( a + b + c)( a 2 + b 2 + c 2 - ab - bc - ca)
= - ( -2)[( -2) 2 - 3(0)] = 8
111. x 2 + px + q = 0 , p , q Î R , q ¹ 0 a , b real roots.
g( x) = 0
a+
1
1
,b +
a
b
166
Solution of Advanced Problems in Mathematics for JEE
a +b +
1 1 æ
1 öæ
1ö
+ = ç a + ÷ çç b + ÷÷
a b è
a øè
bø
-p +
-p
a b
1
= ab + + +
q
b a ab
-p +
-p
p 2 - 2q 1
=q+
+
q
q
q
- pq - p = q 2 + p 2 - 2 q + 1
p 2 + p( p + 1) + q 2 - 2 q + 1 = 0
( q + 1) 2 - 4 ( q 2 - 2 q + 1) ³ 0
q2 + 2q + 1 - 4q2 + 8q - 4 ³ 0
-3 q 2 + 10 q - 3 ³ 0
3q2 - 2q - q + 3 £ 0
3 q ( q - 3) - ( q - 3) £ 0
é1 ù
êë 3 , 3 úû
1/3
3
112. ln( x 2 + 5 x) = ln( x + a + 3) Þ x 2 + 5 x = x + a + 3 > 0
y =x +a+3 Þ
113. f( x) = x 2 +
Let x +
1
x2
a+3>0
a > -3
-5 + a + 3 £ 0
a£2
-3 < a £ 2
–5
2
- 6x -
6
1ö
1ö
æ
æ
+ 2 =ç x + ÷ -6ç x + ÷
x
xø
xø
è
è
1
=t
x
f( x) = t 2 - 6t " t Î ( -¥, - 2] È [2 , ¥)
min. value = -9 at t = 3
cö
æ
114. x 3 + 2 x 2 + 2 x + c = ( x 2 + bx + b) ç x + ÷
b
è
ø
c
Þ b + = 2 and b + c = 2 Þ b = c = 1
b
115. ab + bg + ag = 0 Þ (ab) 3 + (bg) 3 + (ag) 3 = 3 (ab)(ag)(bg)
O
167
Quadratic Equations
¥
118.
å(a r + b r ) = (a + a 2 + a 3 +¼ ) + (b + b 2 + b 3 +¼ )
r =1
b
a
+
1 - a 1 -b
a
4x2 + 2x - 1 = 0
b
=
a
1– a
b
1– b
2
æ x ö
æ x ö
2
4ç
÷ + 2ç
÷ - 1 = 0 Þ 5x - 1 = 0
è1+ x ø
è1+ x ø
x
x
x
æ 2011 ö
æ 2012 ö
æ 2013 ö
119. ç
÷ +ç
÷ +ç
÷ =1
è 2014 ø
è 2014 ø
è 2014 ø
x
x
æ 2011 ö
æ 2012 ö
æ 2013 ö
Let f( x) = ç
÷ +ç
÷ +ç
÷
2014
2014
è
ø
è
ø
è 2014 ø
x
x 2 + ax + 12 = 0
123.
Þ f( x) is a decreasing function for x Î R .
…(1)
2
…(2)
2
…(3)
x + bx + 15 = 0
x + ( a + b) x + 36 = 0
Common roots
(1) + (2) – (3)
a2 =9 Þ a = ±3
positive root a = 3
124. e sin x = t
t 2 - 4t - 1 = 0 Þ t = 2 ± 5 Þ e sin x = 2 ± 5 (Not possible)
125. Maximum value of f( x) = 3
Minimum value of f( x) = -1
126. f(1) = l - 2 < 0
127. 2 x 2 + 5 x + 7 = 0 has non-real roots Þ
a b c
= =
2 5 7
Min. value of a + b + c = 2 + 5 + 7 = 14
Max. value of a + b + c = 28 + 70 + 98 = 196
128. Distance = ( x B - x A ) 2 + ( y B - y A ) 2 = (1 - 2t ) 2 + t 2 = 5t 2 - 4t + 1
Min. distance =
129.
a –3
af(1) < 0
2
1
5
at t =
2
5
b
–3
and
2
af( -1) < 0
2
;
f( -3) f(2) > 0
We have the equation ax + bx + c = 0 has two roots a and b such that a < -3 and b > 2.
168
Solution of Advanced Problems in Mathematics for JEE
If a > 0, then we have the following graphical representation :
y
–3
a
2
x
b
Then, for all x Î [-3 , 2], f( x) < 0 , we have the following graphical representation :
This implies that
y
f( -1) < 0 and f(1) < 0
Þ
a - b + c < 0 and a + b + c < 0
Þ
a( a +|b| + c) < 0
If a < 0, then for all x Î [-3 , 2], f( x) > 0. This imply that
Þ
f( -1) > 0 and f(1) > 0
Þ
a - b + c > 0 and a + b + c > 0
–3
a ( a +|b| + c) < 0
Þ
2
130. Let x + 5 x = t
t 2 - 2t - 24 = 0 = (t - 6)(t + 4)
x 2 + 5 x - 6 = 0 = ( x + 6)( x - 1)
x 2 + 5 x + 4 = 0 = ( x + 4)( x + 1)
131. Case-1: x ³ 2
3( x - 2) - (1 - 5 x) + 4(3 x + 1) = 13 Þ x =
Case-2:
4
5
(Not possible)
1
£ x<2
5
- 3( x - 2) - (1 - 5 x) + 4(3 x + 1) = 13 Þ x =
Case-3: -
2
(Possible)
7
1
1
£ x<
3
5
- 3( x - 2) + (1 - 5 x) + 4(3 x + 1) = 13 Þ x =
Case-4: x < -
1
(Not possible)
2
1
3
- 3( x - 2) + (1 - 5 x) - 4 (3 x + 1) = 13 Þ x = -
b
a
1
(Possible)
2
2
x
169
Quadratic Equations
132. log cos x sin x ³ 2 Þ sin x £ cos 2 x
sin 2 x + sin x - 1 £ 0
0 < sin x £
133. Minimum value
5 -1
2
-D
= -5 Þ D = 20
4
|a - b| =
134.
(sin x > 0)
D
= 20
1
| x - 3| + | x + 5| = 7 x
2x + 2 = 7x
x³3
- ( x - 3) + ( x + 5) = 7 x
- ( x - 3) - ( x + 5) = 7 x
-5 < x < 3
x £ -5
136. ( a + b + c) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ac) Þ ab + bc + ac = - 4
a 3 + b 3 + c 3 - 3 abc = ( a + b + c)( a 2 + b 2 + c 2 - ab - bc - ac) Þ abc = - 4
140. x 2 - 3 x + 4 < x 2 + 3 x + 4
x>0
Þ
a
2
142. x + 4 x + 3 = 0
b
a = -3 , b = -1
143. a 3 + b 3 + c 3 = 3 abc
Þ
a + b+ c =0
Þ
ax 2 + bx + c = 0 has one root x = 1
145. x 1 + x 2 + x 1 x 2 = a
x1 x 2 + x1 x 2( x1 + x 2 ) = b
x 12 x 22 = c Þ b + c = x 1 x 2 ( a + 1)
147. (| x| - 2)(| x| - 1) = 0 Þ x = ± 1, ± 2
149. a 2 + b 2 = (a + b) 2 - 2ab = 4 (1 - sin 2q) 2 + 4 cos 2 2q
= 4(2 - 2 sin 2q)
2
150. sin x + sin x = -b " x Î [0 , p]
0 £ -b £ 2
-2 £ b £ 0
152. x 2 + px - r = 0 = ( x - g)( x - d)
a 2 + pa - r = (a - g)(a - d) = -q - r
153. 2 x + 2 - 4 x £ 9 Ç 2 x + 2 - 4 x > 0; 2 x ( 4 - 2 x ) > 0
170
Solution of Advanced Problems in Mathematics for JEE
Exercise-2 : One or More than One Answer is/are Correct
2x - 1
>0
x( x + 1)(2 x + 1)
1.
–
+
–1
–
+
–1/2
0
+
1/2
æ 1 ö æ1 ö
x Î ( -¥, - 1) È ç - , 0 ÷ È ç , ¥ ÷
è 2 ø è2
ø
Þ
y
6
3.
(0,2)
–2
y=
x
0
3
x
+2
4. Apply D ³ 0 Ç f(2) > 0 Ç f( -2) > 0 Ç -2 <
a
<2
2
5. f( x) = x - 3
x>4
=5 - x
2< x<4
=x +1
1< x < 2
=3 - x
x<1
1 1
6. a - b = Þ ab = 1
(Q a ¹ b)
b a
a
1
Þ a - = a2 Þ a3 - a2 + 1 = 0
a-b=
b
a
7. If f(2 + x) = f(2 - x) and D > 0
D ö
æ
th
Vertex of parabola is ç 2 , ÷ lies in IV quadrant.
4a ø
è
y
3
2
1
x
1234
f(0) > f(1) > f(2)
8. If f(2 + x) = f(2 - x) and D < 0
f( -2) = 4 a - 2 b + c > 0
If log f ( 2) f(3) is not defined then f(2) = 1
f( x) ³ 1
-b
If
= 2 Þ a and b are opposite sign.
2a
9. Case-I : f( -1) ³ 0 Ç f(1) < 0 Ç f(2) ³ 0
3
a£0 Ça<0 Ça³ 2
é 3 ö
Þ
a Î ê- , 0 ÷
ë 2 ø
Þ
O1
–1
2
171
Quadratic Equations
Case-II: f(1) ³ 0 Ç f( -1) < 0 Ç f( -2) ³ 0
3
a³0 Ça>0 Ça£
2
3
æ
ù
Þ
a Î ç 0, ú
è 2û
y
–1 O
1
–2
x
10. As expression taking minimum value
So, a > 0
-b
-D
< 0;
<0
2a
4a
a > 0, b > 0, D > 0
Þ
2
11. ax + bx + c > 0 " x Î R
a > 0, D < 0
f ( 0) = c > 0
f( -3) + f( -2) = 13 a - 5b + 2 c > 0
f( -3) + f(2) = 13 a - b + 2 c > 0
12. D > 0
Þ k> 5
f(1) < 0 Þ k > 3
21
f(2) > 0 Þ k <
4
Þ
3< k<
21
4
x 2 + px + q = 0
13.
Sum of the roots = -13
Product of the roots = 30
Þ
Þ
x 2 + 13 x + 30 = 0 = ( x + 10)( x + 3)
Correct roots are x = -10 , - 3
2
14. x - 3 x + 2 > 0
( x - 2)( x - 1) > 0 Þ x Î ( -¥, 1) È (2 , ¥)
x2 - 3x - 4 £ 0
( x - 4)( x + 1) £ 0 Þ x Î [-1, 4]
then x Î [-1, 1) È (2 , 4]
15. 5 x + (2 3 ) 2 x - 169 £ 0
5 x + 12 x - 169 £ 0
y
0
1
2
x
172
Solution of Advanced Problems in Mathematics for JEE
if
Þ
x =2
5 2 + 12 2 = 169
x>2
5 x + 12 x > 169
x<2
5 x + 12 x < 169
x Î ( -¥, 2]
16. f( x) = x 2 + ax + b
D1 : a 2 - 4 b
g( x) = x 2 + cx + d
D2 : c 2 - 4d
D1 + D 2 = a 2 + c 2 - 4 ( b + d) = ( a - c) 2 > 0 Þ atleast one of them is positive.
17. Let x - 1 = t 2
1
1
+
x + 2 x -1
=
x -2 x -1
1
+
2
1
2
t + 2t + 1
t - 2t + 1
1
1
1
1
=
+
=
+
|t + 1| |t - 1| |1 + x - 1| | x - 1 - 1|
If 1 < x < 2, then 0 <
1
|1 +
If x > 2, then
x -1< 1
1
1
1
2
+
=
+
=
x - 1| | x - 1 - 1| 1 + x - 1 1 - x - 1 2 - x
x -1> 1
1
|1 +
+
1
x - 1| | x - 1 - 1|
2
=
1
x -1 + 1
+
1
x -1 -1
=
2 x -1
x -2
2
18. log 1/ 3 ( x + 2 px + p + 1) ³ 0
Þ ( x + p) 2 + 1 £ 1 Þ ( x + p) 2 £ 0 Þ x = - p
kp 2 - kp - k 2 £ 0 " k Î R
k2 + (p - p2) ³ 0 " k Î R
D£0
19. (a)
a +b =a2 +b2
and ab = a 2b 2 Þ ab(ab - 1) = 0 Þ a = 0 or b = 0 or ab = 1
sin 5q
np
= 0 Þ sin 5q = 0 Þ q =
cos 2q cos 3q
5
2
3
æ 2 x 1 128 x 3
x 2 ö÷
ç
+
+
2
ç x2
x2
4 x 1 x 32 ÷ø
è
(c)
³4
(Q AM ³ GM)
3
(b) tan 2q + tan 3q =
173
Quadratic Equations
(d) Equation of chord with mid-point ( h, k) is T = S 1
Þ
( h - 1) x + ( k - 3) y + ( h + 3 k - h 2 - k 2 ) = 0
If it is passes from (0 , 0).
Then, h 2 + k 2 - h - 3 k = 0
20. - 2 < a < 2
Þ a 2 Î[0 , 4)
x 2 - 4x - a2 = 0 Þ x = 2 ± 4 + a2
21. If
Þ
a + 2b = 0
ab < 0 Þ - 2b 2 < 0 Þ q < 0
a + b = -b = p
ab = - 2 b 2 = q Þ 2 p 2 + q = 0
22. f( x) = ( a + b - 2 c) x 2 + ( b + c - 2 a) x + ( c + a - 2 b) = 0 Þ f(1) = 0
(a) if a > b > c > 0
Þ
y = f(x)
a + b > 2c
f ( 0) = a + c - 2 b < 0
a
(c) g( x) = ax 2 + 2 bx + c = 0
1
x
b
g(0) = c > 0
g( -1) = a - 2 b + c < 0
(d) cx 2 + 2 bx + a = 0
0
–1
y = g(x)
1/a
1/b
–1
0
23. f( x) = 4 x 2 - 8 ax + a
D = 48 a 2 ³ 0
(a) If f( x) is non-negative " x Î R , then a = 0
(b) If a < 0, then f(0) < 0
(c) If f( x) = 0 has two distinct solutions in (0 , 1), then
f(0) > 0 Þ a > 0
4
f(1) > 0 Þ a <
7
-b
0<
<1 Þ 0< a<1
2a
24. ax 2 + bx + c = 0 has no real roots, then D < 0
æ 1ö
f ç - ÷ = a - 2b + 4c > 0 Þ a > 0
è 2ø
x
174
Solution of Advanced Problems in Mathematics for JEE
f ( 2)
4a + 2b + c
=
>0
a + 3b + 9c
æ 1ö
fç ÷
è3ø
25.
x = –1 x = 0 x = 1
x=3 x=4 x=5
x=2
26. ax 2 + bx + c = 0
x=
-b ± b 2 - 4 ac -b ± i 4 ac - b 2
=
2a
2a
2
4 ac - b
æ -b ö
|a| =|b|= ç
÷ +
è 2a ø
4a 2
27. 3 x - 6 > 6
2
=
c
a
x>3
2£ x£3
1£ x < 2
x<1
x>6
4-x>6
6 - 3x > 6
28. f( x) = ax 2 + x + b - a
D < 0 and f(1) = b + 1 > 0
f ( 0) = b - a > 0
f ( 1 2) = 4 b + 2 - 3 a > 0
29.
a 2 + b 2 = ( a + b) 2 - 2 ab = 7
…(1)
a 3 + b 3 = ( a + b)(7 - ab) = 10
…(2)
2
Þ ( a + b)[21 - ( a + b) ] = 20
Let
( a + b) = x
3
x - 21x + 20 = 0
( x - 1)( x + 5)( x - 4) = 0
31. a + b + g + d = 0
1
1
1
1
Root = - , - , - , d
g
b
a
Put x ® -
1
x
175
Quadratic Equations
32. D ³ 0 Ç
f( -1) > 0 Ç
f(1) > 0
1
-2 < K £
4
–1
y
33. (a)
y
a<0
c<0
b<0
x
y
x
a<0
c<0
b>0
x
a<0
c<0
b<0
y
a<0
c>0
b>0
x
(c)
(b)
(d)
1
æ 1ö
(b) f(1) f ç - ÷ > 0
è 2ø
34. (a) f(1) f( -1) > 0
(d) b 2 - 4 ac < 0
(c) f( -1) f( -2) > 0
but a can be +ve or –ve.
b
c
35. a + b = - , ab =
a
a
a 2 + b 2 = (a + b) 2 - 2ab =
1
a2
1
a3
+
+
1
b2
1
b3
=
=
(a + b) 2 - 2ab
a 2b 2
b2
a2
=
-
2 c b 2 - 2 ac
=
a
a2
b 2 - 2 ac
a2
(a + b)(a 2 + b 2 - ab)
(ab) 3
=
-b( b 2 - 3 ac)
c3
36. l = sin 2 x + sin x - 1
-1 < l < 1
37. x 2 + 5 x = x + a + 3 " x Î ( -5, 0)
x 2 + 4 x - 3 = a " x Î ( -5, 0)
a Î ( -3 , 2]
Þ
2
2
39. x - 2 ax - a = 0
Þ
x>a
x = a (1 ± 2 )
x 2 + 2 ax - 5a 2 = 0
43. (a + b) + ( g + d) = 12
x<a
Þ a +b = g + d =6
ab ( g + d) + gd (a + b) = 54 Þ
…(1)
ab + gd = 9
…(2)
(ab)( gd) = 14
…(3)
176
Solution of Advanced Problems in Mathematics for JEE
Þ
ab = 7 , gd = 2
æ K3 ö
æ 2
ö
÷ + mç K -3÷ + n =0
44. l ç
ç K - 1÷
ç K -1 ÷
è
ø
è
ø
K=a
K=b
K=c
lK 3 + mK 2 + nK - (3m + n) = 0
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1. f( -1 - x) = f( -1 + x) " x Î R
Þ graph of f( x) is symmetric about x = -1.
b
= -1 Þ b = 2 a
2a
a = f( -2) = 4 a - 2 b + c
b = f ( 3) = 9 a + 3 b + c
g = f( -3) = 9 a - 3 b + c
Using graph f(3) > f( -3) > f( -2) Þ b > g > a
2. p = b - 4 a = - 2 a < 0
q = 2a + b = 4a > 0
Þ
p ´ q< 0
x = –3 x = –2
Paragraph for Question Nos. 3 to 4
Sol. ( k + 1) x 2 - (20 k + 14) x + 91k + 40 = 0
f( 4) = 27 k > 0 ù
® One root is lie (4, 7)
f(7) = - 9 < 0 úû
f(10) = - 9 k < 0 ù
® Other root is lie (10, 13)
f(13) = 27 > 0 úû
Paragraph for Question Nos. 5 to 7
5. f( x) = x 2 + bx + c " x Î R
-b
= -1 Þ b = 2
2
Graph of f( x) cuts y-axis, when x = 0
Least value at
Þ
c =2
x=0 x=1 x=3
x = –1
x axis
177
Quadratic Equations
f( x) = x 2 + 2 x + 2
Þ
Least value of f( x) = 1
6. f( - 2) + f(0) + f(1) = 9
7. a Î (1, ¥)
Paragraph for Question Nos. 8 to 9
2
Sol. (log 2 x) - 4 (log 2 x) - m 2 - 2m - 13 = 0
8. D > 0 Þ m 2 + 2m + 17 > 0 " m Î R
9. m 2 + 2m - (log 2 x) 2 + 4 (log 2 x) + 13 = 0
D³0
æ 1ù
Þ (log 2 x - 6)(log 2 x + 2) ³ 0 Þ x Î ç 0 , ú È [64 , ¥)
è 4û
Paragraph for Question Nos. 10 to 11
Sol. x 4 - 2 x 3 - 3 x 2 + 4 x - 1 = 0 has four roots a ,
-1
1
, b,
a
b
1ö æ
1ö
æ
ça + ÷ + çb- ÷ =2
aø è
bø
è
1ö æ
1ö
æ
çb - ÷ -ça + ÷ = -4
b
a
è
ø è
ø
Paragraph for Question Nos. 12 to 14
Sol. f( x) - (6 - x) = 0 = ( x - 1)( x - 2)( x - 3)( x - 4)( x - 5)
f( x) = ( x - 1)( x - 2)( x - 3)( x - 4)( x - 5) + (6 - x)
Paragraph for Question Nos. 15 to 16
Sol. x 3 - x 2 (1 + sin q + cos q) + x (sin q + cos q + sin q cos q) - sin q cos q = 0
Þ
Roots are 1, sin q , cos q.
Paragraph for Question Nos. 17 to 18
Sol. 2 [1 + P( x)] = P( x - 1) + P( x + 1)
2 + 2 [ax 2 + bx + c] = a( x - 1) 2 + b( x - 1) + c + a( x + 1) 2 + b( x + 1) + c
Þ
a =1
P ( 0) = c = 8
P(2) = 4 a + 2 b + c = 32 Þ b = 10
…(1)
…(2)
178
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 19 to 21
3
2
2t – 9t + 30 = l
y
10
Sol.
0
–2
2
t
3
–22
Paragraph for Question Nos. 22 to 23
22. D > 0
(2t - 1) 2 - 4t (5t - 1) > 0
16t 2 - 1 < 0 Þ
-1
1
<t<
4
4
( t ¹ 0)
23. t > 0
Case-I: D < 0
y
1
(Q t > 0)
4
-b
1
1
Case-II: D ³ 0 Ç f(0) ³ 0 Ç
£0 Þ £t£
2a
5
4
(2t - 1) 2 - 4t (5t - 1) < 0 Þ t >
Exercise-4 : Matching Type Problems
1. (A)
(2 x - 1)
>0
x(2 x + 1)( x + 1)
–
+
–1
(B) 3 x 2 + 2 ( a 2 + 1) x + ( a 2 - 3 a + 2) = 0
ab < 0
a 2 - 3 a + 2 < 0 Þ ( a - 1)( a - 2) < 0
(C)
x + 3 - 4 x -1 +
x + 8 -6 x -1 =1
2
Let x - 1 = t ; |t - 2| + |t - 3| = 1
(D) A.M. =
a +b + g + d
=2
4
G.M. = (abgd) 1/ 4 = 2
A.M. = G.M. Þ a = b = g = d = 2
–
+
–1/2
0
+
1/2
O
x
179
Quadratic Equations
3. (A) x 4 - 8 x 2 - 9 = 0
( x 2 - 9)( x 2 + 1) = 0
Þ x = 3, - 3
(B) x 2/ 3 + x 1/ 3 - 2 = 0
( x 1/ 3 + 2)( x 1/ 3 - 1) = 0
Þ x = -8 , 1
(C) ( 3 x + 1) 2 = ( x - 1) 2
Þ
3x + 1 = x + 1 - 2 x
(D) (3 x - 9)(3 x - 1) = 0
4. (A) \
Þ 2 x = -2 x
(Not possible)
Þ x = 0, 2
( a + b) = -a & ab = b Þ (1, - 2) and (0, 0)
(B) P = O , Q = 8 cos
p
2p
4p
cos
cos
=1
9
9
9
(C) ar 6 = 2
Product = ( 2 ) 11 = 2 11/ 2
\
m = 11
n=4
(D) x = y = 3
Q
( x - y ) 2 + ( y - 3) 2 = 0
\
5x - 4 y = 3
Exercise-5 : Subjective Type Problems
pö
p
3p
3p
5p
p
5p
æ
1. f ç cos ÷ = sin sin
+ sin
sin
+ sin sin
7ø
7
7
7
7
7
7
è
p
p
= 2 cos 2 + cos - 1
7
7
Þ
f( x) = 2 x 2 + x - 1
2. ( r - a)( r - b)( r - c)( r - d) = ( - 1) ´ ( - 3) ´ (1) ´ (3)
Þ ( r - a) + ( r - b) + ( r - c) + ( r - d) = 0
é3
ö
3. Let x 2 + x + 1 = t
" t Î ê , ¥÷
ë4
ø
t 2 - ( m - 3) t + m = 0
180
Solution of Advanced Problems in Mathematics for JEE
æ3ö
fç ÷<0
è4ø
9 3
- ( m - 3) + m < 0
16 4
-45
m<
4
Case-I :
Þ
Case-II :
x
3/4
y
D = 0 Þ m = 1, 9
-b 3
9
> Þm>
2a 4
2
x
0
3/4
There is one positive integral value of m = 9.
4. t 2 - (m - 3) t + m = 0
y
t Î [3 4 , ¥) has four distinct real roots, then
D>0
Þ
m 2 - 10m + 9 > 0
Þ
m Î ( -¥, 1) È (9 , ¥)
-b 3
9
>
Þ m>
2a 4
2
0 3/4
- 45
æ3ö
fç ÷>0 Þ m>
Þ m Î (9 , ¥)
4
è4ø
5. f(t ) = (m 2 - 12) t 2 - 8 t - 4 = 0
( t ³ 0)
f(0) = - 4 < 0
m 2 - 12 £ 0 Þ m Î [-2 3 , 2 3 ]
Case-I :
D<0
Þ
m 2 - 8 < 0 Þ m Î ( -2 2 , 2 2 )
Case-II : D ³ 0 Þ m Î ( -¥, - 2 2 ] È [2 2 , ¥)
-b
4
=
< 0 Þ m Î ( -2 3 , 2 3 )
2
2 a m - 12
Þ
m Î [-2 3 , 2 3 ]
é æ
p öù
6. ( e x - 2) êsin ç x + ÷ ú ( x - ln 2)(sin x - cos x) < 0
4 øû
ë è
1
2
Þ
( x - ln 2) × (sin x + cos x)( x - ln 2)(sin x - cos x) < 0 =
cos 2 x > 0 , x ¹ ln 2
1
2
( x - ln 2) 2 (sin 2 x - cos 2 x) < 0
x
181
Quadratic Equations
é p ö æ 3p ù
x Î ê0 , ÷ È ç
, p ú - {ln 2}
ë 4ø è 4
û
Least positive integral value is 3.
7. x 2 + 17 x + 71 = l2
2
Þ lÎZ
2
x + 17 x + (71 - l ) = 0
D = perfect square = m 2 (say)
m 2 = 289 - 4 (71 - l2 )
(m - 2l)(m + 2l) = 1 ´ 5
m - 2l = 1
Þ
m + 2l = 5
4
8. P( x) = ( x - x 3 - x 2 - 1)( x 2 + 1) + ( x 2 - x + 1)
P(a ) + P(b) + P( g) + P(d) = (a 2 - a + 1) + (b 2 - b + 1) + ( g 2 - g + 1) + (d 2 - d + 1) = 6
9. If -
a
£1
2
f( x) max = f( 4) Þ 4 a + 18 = 6 Þ a = -3 (Not possible)
a
if - ³ 1
2
f( x) max = f( -2) Þ a = 0 (Not possible)
There is no real value of ‘ a’ .
10. x 2 - 8 x - (n 2 - 10n) = 0
D<0
Þ n 2 - 10n + 16 < 0
(n - 8)(n - 2) < 0
Þ 2 < n < 8 and n ¹ 10
2
11. x + 2 (m - 1) x + (m + 5) > 0 " ( x > 1)
Case-I :
D<0
m 2 - 3m - 4 < 0 Þ -1 < m < 4
Case-II : D ³ 0
Þ
m Î ( -¥, - 1] È [4 , ¥)
4
f(1) ³ 0 Þ m ³ 3
-b
<1 Þ m>0
2a
Þ
m Î ( -1, ¥)
x
1
182
Solution of Advanced Problems in Mathematics for JEE
12. ax 4 + bx 3 - x 2 + 2 x + 3 = ( x + 2)( x - 1) Q ( x) + ( 4 x + 3)
Put x = 1
x = -2
13. D > 0
Ç
a + b=3
b = 2a
-b
>4
2a
Ç
f( 4) ³ 0
k - 1> 0
Ç
4k > 4
Ç
k 2 - 3k + 2 ³ 0
k>0
Þ k³2
Ç
k>1
Ç
( k - 2)( k - 1) ³ 0
14. x 2 - 3 x + 2 = ( x - 1)( x - 2)
If ( x - 1) is a factor of x 4 - px 2 + q = 0. Then
…(1)
p - q=1
4
2
If ( x - 2) is a factor of x - px + q = 0. Then
…(2)
4 p - q = 16
p = 5, q = 4
p + q=9
Þ
Þ
15. x 2 + 2 xy + ky 2 + 2 x + k = 0
if it can be resolved into two linear factors, then
abc + 2 fgh - bg 2 - af 2 - ch 2 = 0
k2 - k - k = 0
k = 0, 2
2
16. ( a + 1) x + 2 = ax + 3 has exactly one solution.
Þ
D =0
a 2 + 4 ( a + 1) = 0
( a + 2) 2 = 0 Þ a = -2
17. y =
Þ a2 = 4
x 2 - ax + 1
x2 - 3x + 2
x 2 ( y - 1) - x (3 y - a) + 2 y - 1 = 0 " x Î R
D³0
(3 y - a) 2 - 4 ( y - 1)(2 y - 1) ³ 0 " y Î R
y 2 - 6 y ( a - 2) + a 2 - 4 ³ 0 " y Î R
D£0
36 ( a - 2) 2 - 4 ( a 2 - 4) £ 0
( a - 2)(2 a - 5) £ 0
183
Quadratic Equations
2£ a£
Þ
5
2
(x ¹ 1)
Integral value of a = 2
y
At a = 2
f( x) =
x2 - 2x + 1
x2 - 3x + 2
x -1
f( x) =
( x ¹ 1)
x -2
=
( x - 1) 2
( x - 2)( x - 1)
1/2
0
Range R - {0 , 1}
Þ
18. x
x=2
No integral values of ‘ a’ for which range is R.
100
at
x
1
= ( x 2 - 3 x + 2) × Q ( x) + ( ax + b)
x =1
at
x =2
Þ
a = 2 100 - 1, b = 2 - 2 100
Þ
…(1)
a + b=1
Þ
2a + b = 2
100
Remainder = (2 100 - 1) x + 2 (1 - 2 99 )
k = 99
Þ
19. x = 7
1/ 3
+ 7 2/ 3
x 3 = 7 + 49 + 3 ´ 7 ( x) Þ
x 3 - 21x - 56 = 0
Product of all roots = 56
21. Clearly P( x) is a second degree polynomial.
P( x) = ax 2 + bx + c
\
P ¢( x) = 2 ax + b
P( x) - P ¢( x) = ax 2 + ( b - 2 a) x + c - b = x 2 + 2 x + 1
a = 1, b - 2 a = 2 , c - b = 1
a = 1, b = 4 , c = 5
P( x ) = x 2 + 4 x + 5
P( -1) = 1 - 4 + 5 = 6 - 4 = 2
2
23. Let x = t
t 2 + kt + k = 0
D>0
Þ k Î ( -¥, 0) È ( 4 , ¥)
f(0) < 0 Þ k < 0
…(2)
184
Solution of Advanced Problems in Mathematics for JEE
4
3
8
f(3) < 0 Þ a < 7
25. f(1) < 0 Þ a < -
1
Integral values of ‘ a’ are - 5, - 4 , - 3 , - 2.
æpö
26. f(0) f ç ÷ £ 0
è2ø
- (n + 1)(2n + 1)(n - 3) £ 0 Þ n Î [3 , ¥)
27. f( x) = ax 2 + bx + c
a , b, c Î I
2
ax + bx + c = a( x - a )( x - b) + p
a,b Î I
ax 2 + bx + c - 2 p = a( x - a )( x - b) - p = 0
Not possible for integral values of x.
28. 9 x 2 + 2 x ( y - 46) + y 2 - 20 y + 244 = 0
D³0
Þ y 2 - 11 y + 10 £ 0
( y - 1)( y - 10) £ 0 Þ 1 £ y £ 10
2
y + 2 y ( x - 10) + 9 x 2 - 92 x + 244 = 0
D³0
Þ x 2 - 9 x + 18 £ 0
( x - 3)( x - 6) £ 0 Þ 3 £ x £ 6
29. a + b = 3 and a 3 + b 3 = 7 Þ a 3 + (3 - a) 3 = 7 Þ 9 a 2 - 27 a + 20 = 0
Sum of distinct values of ‘ a’ is 3.
30. ( y 2 - 3) 2 + ( x - 4) 2 = 1
Þ
x = 4 + cos q ,
y 2 = 3 + sin q
M = 36 , m = 1
31. x 1 + x 2 + x 1 x 2 = a
x1 x 2 + x1 x 2( x1 + x 2 ) = b
x 12 x 22 = c
If
b + c = 2( a + 1) Þ x 1 x 2 = 2
32. x 3 + 3 x 2 + 4 x + 5 = 0 Þ x = a is root
x 3 - 3 x 2 + 4 x - 5 = 0 Þ x = b is root
Þ
33. 5
a +b =0
(1) - (2)
( 2) - ( 3)
( x - z )( x + y + z ) = 1 and ( y - x)( x + y + z ) = -2
3
x
185
Quadratic Equations
Divide z =
x+ y
2
2
æ x + yö æ x + yö
2
2
y 2 + yç
÷+ç
÷ = 1 and x - 2 xy - 5 y = 0
è 2 ø è 2 ø
x = (1 + 6 ) y
y2 =
34.
9+3 6
put values
4 (1 - a - b) - ( a - b) 2 4 (1 + a + b) - ( a + b) 2
>
4
-4
Þ
35.
2
8 > ( a + b) 2 + ( a - b) 2 Þ a 2 + b 2 < 4
3
20 x + 3 20 x + 13 = 13
3
20 x + 3 20 x + 3 20 x + 3 20 x + 3 20 x +¼¥ = 13
Þ
3
20 x + 13 = 13
Þ 20 x = 2197 - 13
Þ
x=
2184 546
=
20
5
36. Let f( x) = x 2 - 2 ( a + 1) x + a ( a - 1)
f(1 - a) < 0
Ç
f(1 + a) < 0
4a - 3a - 1 < 0
Ç
3a + 1 > 0
1
- < a<1
4
Ç
a> -
2
Þ
1
3
æ 1 ö
a Î ç - , 1÷
è 4 ø
37. ( x - 8)( x - 2) < 0
2< x<8
-b
38. sin q + cos q =
a
c
sin q × cos q =
a
Þ
(sin q + cos q) 2 = 1 + 2 sin q cos q =
2
Þ
b -a
a
2
2
=
2c
a
b2
a
2
=1+
2c
a
186
Solution of Advanced Problems in Mathematics for JEE
39. cos 2 x + (1 - a) cos x - a 2 £ 0 " x Î R
Let cos = t " t Î [-1, 1]
–1
t 2 + (1 - a) t - a 2 £ 0 " t Î [-1, 1]
1
x
f( -1) £ 0
Þ
a2 - a ³ 0
Þ
a Î ( -¥, 0] È [1, ¥)
f(1) £ 0
Þ
a2 + a - 2 ³ 0
( a + 2)( a - 1) ³ 0 Þ a Î ( -¥, - 2] È [1, ¥)
a
40. 2 x 2 - 35 x + 2 = 0
b
2a - 35 = -
2
a
and 2b - 35 = -
2
b
42. xF ( x) - 1 = k( x - 1)( x - 2)( x - 3) ¼( x - 9)
k( x - 1)( x - 2)( x - 3) ¼( x - 9) + 1
Þ F ( x) =
x
Constant term = k( -9 !) + 1 = 0
1
Þ
k=
9!
44. cos A + cos B + cos C = -a
cos A cos B + cos B cos C + cos A cos C = b
cos A cos B cos C = -c
a 2 - 2 b - 2 c = cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cos C
=1
45. k > 0
D>0Ç
y
-b
>0
2a
k 2 - 10 k + 9 > 0 Ç
k -3
<0
k
O
k Î ( -¥, 1) È (9 , ¥) Ç k Î (0 , 3) Þ k Î(0 , 1)
❑❑❑
x
187
Sequence and Series
9
SEQUENCE AND SERIES
Exercise-1 : Single Choice Problems
1. AM ³ GM
3. 2 sec a = sec (a - 2 b) + sec (a + 2 b)
cos(a + 2 b) + cos(a - 2 b)
2
=
cos a
cos(a - 2 b) cos(a + 2 b)
cos 2a + cos 4 b = cos a (2 cos a cos 2 b)
2
2 cos a - 1 + 2 cos 2 2b - 1 = 2 cos 2 a cos 2 b
Þ
cos 2 a(1 - cos 2 b) + (cos 2 b + 1)(cos 2 b - 1) = 0
cos 2 a = 1 + cos 2 b
Þ
4. If a , b, c
A.P.
Þ
if c , d , e
H.P.
Þ
if b, c , d
G.P.
Þ
a+c
2
2 ec
d=
e+c
2
c = bd
b=
æ a + c ö æ 2 ec ö
c2 = ç
֍
÷
è 2 øè e + cø
Þ
c 2 = ae
5. ( a + nd) 2 = ( a + md)( a + rd)
Þ
a
mr - n 2
=
d 2n - m - r
if m , n , r in H.P., then n =
2mr
a -n
Þ =
m+r
d 2
4
=1
4
G.M. (a , b , g , d) = 1 Þ a = b = g = d = 1
So, equation is ( x - 1) 4 = 0
6. A.M. (a , b , g , d) =
188
Solution of Advanced Problems in Mathematics for JEE
7. S 3 = S 12 Þ
8. T r =
S 14 S 22 - S 22 S 32
S 12 + S 32
=
S 22 ( S 14 - S 32 )
S 12 + S 32
=0
r ×2r
( r + 2)!
Tr =
( r + 2 - 2) 2 r
1
1
=
2r 2 r +1
( r + 2)!
( r + 1)!
( r + 2)!
Sn =
2!
2 n +1
2 ! (n + 2)!
lim S n = S ¥ = 1
n® ¥
9. tan 2
é
ù
2 n +1
= 0ú
êas lim
ë n ® ¥ (n + 2)!
û
p
æ p
ö
æ p
ö
= tan ç
- x ÷ tan ç
+ x÷
12
è 12
ø
è 12
ø
p
tan
=
12
2
p
- tan 2 x
12
p
1 - tan 2
tan 2 x
12
tan 2
p
æ
ö
Þ tan 2 x ç tan 4
- 1÷ = 0 Þ tan x = 0
12
è
ø
x = 0 , p , 2 p , 3 p , ¼¼ 99 p
Sn
n n+1
10.
=
Sn - 1 n - 1 n + 2
n ö æ3 4 5
n
n + 1ö
æ2 3 4 5
Q n = ç ´ ´ ´ ´ ¼¼
´
÷ ´ ç ´ ´ ´ ¼¼ ´
÷
n - 1ø è 4 5 6
n +1 n +2ø
è1 2 3 4
3n
ænö æ 3 ö
Q n = ç ÷ ×ç
÷=
è 1ø è n + 2 ø n + 2
lim Q n = 3
n® ¥
11.
log l
log m
log n
p 1
log A + ( p - 1) log R
q 1 = log A + ( q - 1) log R
r 1
log A + ( r - 1) log R
log A
log A
log A
p 1
( p - 1) log R
q 1 + ( q - 1) log R
r 1
( r - 1) log R
p 1
q 1
r 1
p 1
q 1 =0
r 1
12. Numbers divisible by 6 ® 49
Numbers divisible by 18 ® 16
y+z
13.
= yz
Þ 1 - x ³ 2 yz
2
189
Sequence and Series
Thus,
(1 - x)(1 - y)(1 - z ) ³ 2 yz × 2 zx × 2 xy = 8 xyz
Þ
xyz
1
£
(1 - x)(1 - y)(1 - z ) 8
17. Clearly, both roots are lies in between -1 and 1.
n
n
n
æ
ö æ
ö
\
lim
(a r + b r ) = ç lim
a r ÷ + ç lim
br ÷
ç
÷
ç
÷
n® ¥
r =1
è n ® ¥ r =1
ø è n ® ¥ r =1 ø
b
a
1
=
+
=
1 - a 1 - b 12
å
18.
a
a
a
å
a
a
a
a
å
a
a
a
a
a
a
å aij = a12 + a13 + a14 + a12 + a 23 + a 24 + a13 + a 32 + a 34 + a14 + a 42 + a 43
1
æ
ö
çQ x + ³ 2 ÷
x
è
ø
³ 12
x 2 + 2 xy + 2 xy + 4 y 2 + z 2 + z 2 6 2
³ 2 ×4 × x4 × y4 ×z4
6
20. Let first term be ‘ a’ and difference be d.
19.
5 ( a + 4 d) = 8 ( a + 7 d)
a + 12 d = 0
25
S 25 = [2 a + 24 d]
2
Þ
Þ
S 25 = 25 ( a + 12 d) = 0
10 sin x = 5 ( 4 sin 2 x + 1)
21.
sin x ¹ 0
5 ±1
4
22. Let first term of G.P. be a and ratio be r.
Þ
a + ar + ar 2 = 70 and 10 ar = 4 a + 4 ar 2
1
Þ
a = 40
r=
2
a
40
S=
=
= 80
1
1- r
12
¥
k
k ¥ 1
k
23.
=
=
n+k
2 k n =1 2 n 2 k
n =1 2
Þ
sin x =
å
å
¥
k
1
2
3
4
å 2 k = 2 + 2 2 + 2 3 + 2 4 +¼¼ = 2
k =1
24. ( pqr) 1/ 3 ³
p+ q+ r
Þ p=q=r
3
if 3 p + 4 q + 5r = 12 Þ p = q = r = 1
190
25.
Solution of Advanced Problems in Mathematics for JEE
ù
1æ 1 1 1
1
1
1ö 1æ 1 1ö 1æ 1 1ö
ö 1 é1æ
+
+
+¼¼ ÷ = ê ç 1 + ÷ + ç + ÷ + ç + ÷ + ¼¼ú
ç + +
3 è 3 6 10 15 21
2ø 5è2 3ø 7 è3 4ø
ø 3 ë3 è
û
=
1é 1
1
1
ù 1
+
+
+ ¼¼ú =
3 êë 1 × 2 2 × 3 3 × 4
û 3
a
b
[2 A + ( a - 1) D]
[2 A + ( b - 1) D]
26. 2
=2
= c Þ D = 2 c, A = c
a2
b2
x r
x
27.
=4 Þ
= r - r2
1- r
4
1
if -1 < r < 1 then -2 < r - r 2 <
4
x 1
-2 < < Þ - 8 < x < 1
4 4
n+1
28. t 1 + t 3 + t 5 +¼¼ + t 2n +1 =
[2 a + n(2 d)] = 248
2
n
t 2 + t 4 + t 6 +¼¼ + t 2n = [2 ( a + d) + (n - 1) 2 d] = 217
2
Þ
t 2n +1 - t 1 = 2n × d = 56
n+1
n
[2 a + 56] = 248 and [2 a + 56] = 217
2
2
Þ
n = 7, a = 3
29. length of side A1 = 20
20
length of side A 2 =
2
20
length of side A 3 =
( 2)2
20
length of side A n =
( 2 ) n -1
400
Area of A n =
<1
2 n -1
¥
30. S k =
1
1
1
k+1
å ( k + 1)i = 1 + k + 1 + ( k + 1) 2 +¼¼ = k
i =0
n
n
n
n
k =1
k =1
k =1
å k S k = å( k + 1) = å k + å 1 =
k =1
2
31. T r =
n (n + 1)
n ( n + 3)
+n=
2
2
( r + 1) r -1 æ
1
2 ö r -1
×2
= ç1 + ÷2
r ( r + 1)
r r + 1ø
è
191
Sequence and Series
n
Sn =
å
n
Tr =
r =1
å
2 r -1 +
r =1
n
æ 2 r -1
é
2 r ö÷
2n ù æ n ö n
ç
= (2 n -1 ) + ê1 ÷2
ú =ç
ç
r + 1 ÷ø
n + 1û è n + 1 ø
r =1 è r
ë
å
29
32.
å(1.5) n = (1.5) 2 + (1.5) 3 +¼¼ + (1.5) 29
n =2
é(1.5) 28 - 1ù
2
= (1.5) 2 ê
ú = 2 k - 2 (1.5)
0.5
ë
û
33. 7 , A1 , A 2 , A 3 ,¼¼ , A n , 49 are in A.P.
æn +2ö
A1 + A 2 + A 3 +¼¼ + A n = ç
÷ (7 + 49) - (7 + 49)
è 2 ø
n
Þ
´ 56 = 364 Þ n = 13
2
2 2
34.
, , 2 , 2 r, 2 r 2
2 r
r
35. S n = 5n 2 + 4n
t n = S n - S n -1 = 10n - 1
é æ a 2 - b öù
÷ú
36. x 3 + y 3 = ( x + y)( x 2 + y 2 - xy) = a ê b - ç
ç
÷
êë è 2 ø úû
3 ab a 3
=
2
2
1
37. S 1 =
=3
2
13
3
S2 =
=5
2
15
M
M
2
2
2 ö
æ
çQ xy = ( x + y) - ( x + y ) ÷
ç
÷
2
è
ø
2n - 1
= 2n + 1
2
12n + 1
1
1
1
1
1
1
+
+
+¼¼ =
+
+
+¼¼
S1 S 2 S 3 S 2 S 3 S 4 S 3 S 4 S 5
3 × 5 × 7 5 × 7 × 9 7 × 9 × 11
Sn =
¥
S¥ =
å
r =1
¥
tr =
¥
1
1
1
é
ù
=
ê(2 r + 1)(2 r + 3) - (2 r + 3)(2 r + 5) ú
(
2
r
+
1
)(
2
r
+
3
)(
2
r
+
5
)
û
r =1
r =1 ë
å
å
38. ar 5 , 2 , 5, ar 13 are in G.P.
Þ ( ar 9 ) 2 = 10
t 1 t 2 t 3 ¼¼ t 19 = a 19 r 9´19 = ( ar 9 ) 19 = 10 19/ 2
192
Solution of Advanced Problems in Mathematics for JEE
39. A.M. ³ G.M.
1
1
1
1
A + + 1³ 3 ;
B + + 1³ 3 ;
C + + 1³ 3 ;
D + + 1³ 3
A
B
C
D
1
1
1
1
æ
öæ
öæ
öæ
ö
4
ç A + + 1÷ ç B + + 1 ÷ ç C + + 1÷ ç D + + 1÷ ³ 3
A
B
C
D
è
øè
øè
øè
ø
40. (Sr) 2 = Sr 2 + 2 Sr1 r2
S r1 r2 =
41.
a-b
2
2n
n
[2 a + (2n - 1)d] = x and
[2( a + 2nd) + (n - 1) d] = y
2
2
2y x
2y - x
Þ
- = 3nd Þ d =
n
n
3n 2
44. 2 , 6 , 2 ( k - 1) are in G.P.
Þ
6 2 = 2 ´ 2 ( k - 1)
Þ
k = 10
Þ
x 2 - x - 6 > 0 and | x| < 100
Þ x Î ( -100 , - 2) È (3 , 100)
Number of integers = 193
45.
n
n
r =1
r =1
3
1
1
3
å 1 + T r T r +1T r + 2T r + 3 = å 1 + æçè r - 2 ö÷ø æçè r - 2 ö÷ø æçè r + 2 ö÷ø æçè r + 2 ö÷ø
n
=
å
r =1
2
=
æ 2 5ö
çr - ÷
4ø
è
å
Tr -
n
å
n
=
5
5
å r2 - 4
r =1
n
5
1
n
n
r =2
r =2
5
å r2 - 4 + å r2 - 4 = 4 + å r2 - å 4
r =1
46. T r =
2
r =2
T r -1 = r 2 + r
n
n
2008
1
1 ö
n
æ1
= 2008
= (2008) ç ÷ = (2008)
Tr
r( r + 1)
r r + 1ø
n+1
r =1
r =1
r =1 è
å
å
å
(2008) n
= 2008
h®¥ n + 1
lim
n
48. P( x) =
1
1
1
å æçè x - r ö÷ø æçè x - r + 1 ö÷ø æçè x - r + 2 ö÷ø
r =1
n
Absolute term = -
1
1é n
1
1
ù
å r( r + 1)( r + 2) = - 2 êê å r( r + 1) - ( r + 1)( r + 2) úú
r =1
ë r =1
û
193
Sequence and Series
=lim -
n ®¥
50.
1 é1
1
ù
ê
2 ë 2 (n + 1)(n + 2) úû
1 é1
1
1
ù
=2 êë 2 (n + 1)(n + 2) úû
4
1 1 1
1
are in A.P.
,
,
,¼¼ ,
T1 T 2 T 3
Tk
1
+ 5d
T2 a
2
=
=9 Þ d = 1
T6
a
+d
a
1
+ 3d
T10 a
5
=
=
1
T4
17
+ 9d
a
1
1
1
1
æ
ö 2æ
ö 15
52. ç 1 +
+
+ ¼¼ ÷ + ç 1 +
+
+¼¼ ÷ =
2
4
2
4
3
è
ø
è
ø 8
3
3
3
3
53. ( x - 1)( x - 2)( x - 3)( x - 4) ¼¼( x - 10)
Coefficient of x 8 = sum of terms taken two at a time
1
= [(1 + 2 + 3 +¼¼ + 10) 2 - (12 + 2 2 +¼¼ + 10 2 )]
2
55.
AM = GM
a +b + g + d
1
= (abgd) 1/ 4 =
4
2
1
Þ
a =b = g = d =
2
56. Use AM ³ GM
¥
57.
å(a r + b r ) = (a + a 2 + a 3 +¼ ) + (b + b 2 + b 3 +¼ )
r =1
b
a
+
1 - a 1 -b
a
4x2 + 2x - 1 = 0
b
=
2
æ x ö
æ x ö
2
4ç
÷ + 2ç
÷ - 1 = 0 Þ 5x - 1 = 0
è1+ x ø
è1+ x ø
a
1– a
b
1– b
194
Solution of Advanced Problems in Mathematics for JEE
2
é 10 ´ 11ù
58. 2 2 [1 + 2 3 + 3 3 + 4 3 +¼ + 10 3 ] = 4 ê
= 12100
ë 2 úû
59. AM ³ HM
a a
b+ +
2 2³ 3
4 1
3
+
a b
60. 4 x - 15 = 4 2- x Þ 4 x = 16 Þ x = 2
pö 1
æ 2011p ö
æ
Common ratio = cos ç
÷ = cos ç 670 p + ÷ =
3ø 2
è 3 ø
è
61. AM ³ GM
c2 c2
1/ 4
+
4 4 4
2
2 ³ æç a b c ö÷
ç
÷
4
4
è
ø
a 4 + b4 +
62. x 2 + y 2 = x 2 +
63.
2
1
3
+
6
3
1 +2
3
+
1
x2
³2
12
3
+
20
+¼¥
1 +2 +3
1 + 2 + 33 + 43
1´ 2
2 ´3
3´4
=
+
+
+¼¥
3
3
3
3
1
1 +2
1 + 23 + 33
3
3
n
3
3
n(n + 1)
n
n(n + 1)
1
æ n(n + 1) ö
ç
÷
2
è
ø
å 13 + 2 3 +¼ + n 3 = nlim
å
n ®¥
®¥
= lim
1
n
= lim 4
n ®¥
2
n
1
1 ö
æ1
= 4 lim
ç ÷
n ®¥
n + 1ø
1 n(n + 1)
1 èn
å
å
1
1 ö
æ1 1 1 1
= 4 lim ç - + - +¼ + ÷
n ®¥ è 1 2
2 3
n n + 1ø
n
= 4 lim
=4
n ®¥ n + 1
64.
¥
æ
ö
ö
1
1
1
1 æ
1
çç
÷÷ =
çç
÷
( k - 1) n =1 è (n + 1)(n + 2) ¼(n + k - 1) (n + 2)(n + 3) ¼(n + k) ø ( k - 1) è 2 × 3 × 4 ¼ k ÷ø
å
3
6
and G - H =
2
5
As we know,
65. A - G =
G 2 = AH
6ö
æ3
öæ
Þ G2 ç + G÷çG - ÷
2
5ø
è
øè
Þ G = 6 and A =
15
2
195
Sequence and Series
ab = 36 and a + b = 15 Þ a = 12 and b = 3
ö
2+5
5+8
8 + 11
1 æ 5 2 - 2 2 8 2 - 5 2 112 - 8 2
66. S =
+
+
+¼ = ç
+
+
+¼ ÷
ç
÷
2
2
2
2
2
2
2
2
2
2
2
2
3 è 2 ×5
2 ×5
5 ×8
8 × 11
5 ×8
8 × 11
ø
Þ
10
67.
10
r
1æ 1
1
1
1
1
1
1
1 ö
ç
÷÷
+
+
+¼ +
2
3 çè 2 2 5 2 5 2 8 2 8 2 112
29
32 2 ø
=
1æ 1
1 ö
85
ç ÷=
3 è 4 32 2 ø 1024
r
å ( r 2 - 1) 2 - r 2 = å ( r 2 - r - 1)( r 2 + r - 1)
r =1
r =1
=
68.
=
¥
¥
r =1
r =1
¥
1 10 æ
1
1
ö
ç
÷
2
2
2 r =1 è r - r - 1 r + r - 1 ø
å
r
åt r = å r4 + r2 + 1
=
r
69.
r =1
1
1
1
æ
ö
ç
÷
2 r =1 è r 2 - r + 1 r 2 + r + 1 ø
å
S¥ = 1 +
4
7
10
+
+
+¼
5 52 53
1
1 4
7
× S¥ = +
+
+¼
2
5
5 5
53
Þ
Þ
r
å ( r 2 + 1) 2 - r 2 å ( r 2 - r + 1)( r 2 + r + 1)
r =1
¥
=
¥
=
4
3
3
3
7
S¥ = 1 + +
+
+¼ =
2
3
5
5 5
4
5
S¥ =
35
16
71. x 1 , x 2 , x 3 ¼ x 2n
2n
å( -1) r +1 x r2
r =1
x 12 - x 22 + x 32 +¼ - x 22n
( x 1 - x 2 )( x 1 + x 2 + x 3 +¼ x 2n )
- ( x 2 - x 1 )( x 1 + x 2 + x 3 +¼ x 2n )
( x - x1 ) 2 x
- 2n
[ x 1 + x 2n ]
2n - 1
2
x
( x 12 - x 22n )
2x - 1
196
72.
Solution of Advanced Problems in Mathematics for JEE
a +b
=9;
2
ab = 4
73. rms ³ AM
p 2 + q2 p + q
³
2
2
n
74. 150 ´ 9 + [300 + (n - 1)( -2)] = 4500 Þ n = 25
2
Total term = n + 9 = 34
20
75. S 20 =
[2 (1 - ad) + 19 d] = 20
2
76.
Þ
19 d - 2 ad = 0
¥
1
1 ¥ æ
n =3
1
n =3
78. 2 x + 2 2 x +1 +
2
x
= 2 x + 2 2x + 2 2x +
2 x + 2 2 x +1 +
Þ
¥
5
1
2
x
+
1
2
x
+
1
2
x
2 x + 2 2 x +1 + (5 2 x ) æ x
1 ö÷
³ ç 2 ´ (2 2 x ) 2 ´
ç
8
(2 x ) 5 ÷ø
è
Þ
79.
1
ö
å (n - 2)(n - 1) n(n + 1)(n + 2) = 4 å ççè (n - 2)(n - 1) n(n + 1) - (n - 1) n(n + 1)(n + 2) ÷÷ø
æ ( 4 r + 5) ö 1
5
2x
¥
+
1
2
x
+
1
2x
1/ 8
=1
³8
1
1
1
¥ æ
1
1
ö
1
å çè r(5r + 5) ÷ø × 5 r = å æçè r - 5r + 5 ö÷ø × 5 r = å çç r × 5 r - ( r + 1) × 5 r +1 ÷÷ = 5
r =1
r =1
r =1 è
Exercise-2 : One or More than One Answer is/are Correct
1. a =
Þ
a1 + a n
2 a1 a n
, b = a1 a n , c =
2
a1 + a n
a ³ b ³ c and b 2 = ac
2. D1 : b 2 - 4 ac < 0
D 2 : c 2 - 4 ab < 0
D 3 : a 2 - 4 bc < 0
D1 + D 2 + D 3 : a 2 + b 2 + c 2 < 4 ( ab + bc + ac)
1<
a 2 + b2 + c 2
<4
ab + bc + ac
ø
197
Sequence and Series
3. If a , b, c are in H.P.
A.M. > H.M.
a+c
Þ
>b Þ
2
a + c > 2b
a - b> b - c
1
1
or
<0
a-b b-c
G.M. > H.M.
Þ
ac > b 2
1
4. T p = a + ( p - 1) d =
q( p + q)
also
ac > b or
T q = a + ( q - 1) d =
1
Þ
p ( p + q)
a=d=
1
pq( p + q)
5. (a) a , H 1 , H 2 , H 3 , ¼¼ , H n , b are in H.P.
1 1
1
1
1 1
,
,
,
, ¼¼ ,
, are in A.P.
a H1 H 2 H 3
Hn b
Þ
1
1
1
1
n æ 1 1ö
+
+
+¼¼ +
= ç + ÷
H1 H 2 H 3
Hn 2 è a bø
(c) a , A1 , A 2 , A 3 , ¼¼ , A 2n , b are in A.P.
A1 + A 2n = A 2 + A 2n -1 = A 3 + A 2n -2 = ¼¼ = a + b
(d) 4 g 2 + 5 g 3 = 4 r + 5r 2
This is minimum at r = -
2
5
6. a , b, c are in H.P.
1 1 1
, , are in A.P.
a b c
a+ b+ c
a+ b+ c
a+ b+ c
(a)
- 2,
- 2,
- 2 are in A.P.
a
b
c
a+ b+ c
a+ b+ c
a+ b+ c
(b)
- 1,
- 1,
- 1 are in A.P.
c
b
c
2 1 1
2
(c)
Þ ac ³ b
= + ³
b a c
ac
a 5 + c 5 ³ 2 ( ac) 5/ 2 ³ b 5
(d) 2ac = ab + bc
7. Let the roots be a , ar , ar 2 , ar 3 and ar 4 .
a( r 5 - 1)
= 40
r -1
…(1)
198
Solution of Advanced Problems in Mathematics for JEE
æ 1
ö
- 1÷
ç
5
1èr
ø
= 10
a 1
-1
r
and
put
…(2)
r 5 - 1 40
in (2) we get ar 2 = ± 2
=
r -1
a
Now,
d = ( ar 2 ) 5 = ( ± 2) 5
8. (a) Q
2 a k +1 = a k + a k + 2
\
fk ( -1) = 0
-1 is a root.
\
Other is also real root.
(b) From (a) ( -1) is root for any ‘ k’ so any pair of equation has a common root.
(c) If one root is -1, other roots are - c a (form)
a k+ 2
a3 a4 a5
i . e. ,
,
,
¼¼ are not in A.P.
ak
a1 a 2 a 3
9. b =
a+c
2 ce
,d=
2
c+e
if c 2 = bd , then c 2 = 36
(Q a = 2 , e = 18)
10. If a , b, c are in A.P. then
a = b - d and c = b + d
a + b + c = 60 Þ b = 20
If ( a - 2), b,( c + 3) are in G.P., then
400 = (18 - d)(23 + d) Þ d = 2 , - 7
4
12.
4
81 + 144 a + 16 b + 9 c 4
³ 36 abc
4
Þ
A.M. = G.M.
Þ
81 = 144 a 4 = 16 b 4 = 9 c 4
13. x , y , z
A.P.
Let x = y - q and z = y + q
3q
2 × cos( y)
cos( y - q) + cos y + cos( y + q) = 1 =
q
sin
2
sin
199
Sequence and Series
3q
2 × sin( y) Þ cot y = 2
sin( y - q) + sin y + sin( y + q) =
=
2 sin q
2
3q
sin
2 = 3 = 3 - 4 sin 2 q Þ cos q = 3 - 2
q
2
2
2 2
sin
2
1
15.
10 n +1 + 1
10 n + 2 + 1
<
sin
10 m +1 + 1
10 m + 2 + 1
Þ
10 n +1 × 10 m + 2 + 10 n +1 + 10 m + 2 + 1 < 10 n + 2 × 10 m +1 + 10 n + 2 + 10 m +1 + 1
Þ
10 m +1 < 10 n +1
16. S r = r + S r
Þ S r2 - S r = r
17. 50, 48, 46, 44, ……… A.P.
T n = 50 + (n - 1)( -2) = 0
n = 26
Þ
18. S n = S n
n
n
r =1
r =1
2r + 1
n
1
1
1
å t r = å 12 + 2 2 + 3 2 +¼ + r 2 = å 6 æçè r - r + 1 ö÷ø = 6 æçè 1 - n + 1 ö÷ø
r =1
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
Sol. T1 = A + B = 0
T 2 = Aa + Bb = 1
Þ A = -B
Þ A(a - b) = 1
T 3 = Aa 2 + Bb 2 = 1
Þ A(a 2 - b 2 ) = 1
T 4 = Aa 3 + Bb 3 = 2
Þ A(a 3 - b 3 ) = 2
Þ
a + b = 1 and ab = -1
Paragraph for Question Nos. 3 to 4
Sol. Set A :
Set B :
Þ
5 - D , 5, 5 + D and
5 - d , 5, 5 + d
p 25 - D 2 7
=
=
q 25 - d 2 8
25 = 8 D 2 - 7 d 2 = d 2 + 16 d + 8
Þ
d = 1 and D = 2
Set A {3 , 5, 7} and set B {4 , 5, 6}
(Q D = 1 + d)
200
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 5 to 7
5.
( x - 3) + ( y + 1) + ( z + 5)
³ [( x - 3)( y + 1)( z + 5)]1/ 3
3
Þ ( x - 3)( y + 1)( z + 5) £ (21) 3
1ö æ
5ö
æ
6. Term is 6 ( x - 3) ç y + ÷ ç z + ÷ Þ
2
3
è
øè
ø
1
5
1/ 3
+z+
2
3 ³ é( x - 3) æ y + 1 ö æ z + 5 ö ù
ç
֍
÷
ê
3
2 øè
3 ø úû
è
ë
( x - 3) + y +
1öæ
5 ö (355) 3
æ
( x - 3) ç y + ÷ ç z + ÷ £
2 øè
3 ø 63 ´33
è
Maximum value =
7.
(355) 3
62 ´33
x+ y+z
³ ( xyz ) 1/ 3 ; xyz £ (20) 3
3
Paragraph for Question Nos. 8 to 10
Sol. Let removed number are A and A + 1.
n(n + 1)
105
- 2 A - 1 = ( n - 2)
2
4
2n 2 - 103n + 206 = 8 A
n = 50 , A = 7
Þ
Paragraph for Question Nos. 11 to 13
Sol. a n +1 - 1 = ( a n - 1)
2
a n - 1 = ( a n -1 - 1) 2
a n -1 - 1 = ( a n -2 - 1) 2
( a 2 - 1) = ( a1 - 1) 2
a1 - 1 = ( a 0 - 1) 2
2
Þ ( a n - 1)( a n -1 - 1) 2 ( a n -2 - 1) 2 ¼¼( a1 - 1) 2
Þ ( a n - 1) = 3 2
n
0
bn =
2 × (3 2 + 1)(3 2 + 1) ¼¼(3 2
(3
n
bn =
n-1
32 -1
n
32 + 1
2n
+ 1)
n-1
+ 1)
= ( a n -1 - 1) 2 ( a n -2 - 1) 2 ¼¼( a 0 - 1) 2
n
201
Sequence and Series
Paragraph for Question Nos. 14 to 15
n
f ( n) =
n
4
r =2
14. f(7) + f(8) =
15. x 2 +
æ
1
1
ö
æ 1
r =2
122
63
3
x + t =0
2
a
b
Paragraph for Question Nos. 16 to 17
Sol.
a1
a2
a3
a1005
1
=
=
= ¼¼
=
a1 + 1 a 2 + 3 a 3 + 5
a1005 + 2009 k
a1 =
1
ö
f ( n) = 1
å ( r - 1) r( r + 1) = 2 å ççè ( r - 1) r - r( r + 1) ÷÷ø = 2 ççè 1 × 2 - n (n + 1) ÷÷ø ; a = nlim
®¥
1
3
5
2009
, a2 =
, a3 =
, ¼¼ a1005 =
k -1
k -1
k -1
k -1
a1 + a 2 + a 3 +¼¼ + a1005 =
(1005) 2
1005
= 2010 Þ k - 1 =
k -1
2
Exercise-4 : Matching Type Problems
1. (A) a , b, c are in A.P.
b-a=c-b
b - a , c - b, a are in G.P.
c-b
a
=
Þ c-b=a
b-a c-b
(B) a , x , b are in A.P.
a+b
x=
2
(Q b - a = c - b)
a , y , z , b are in G.P.
y = a 2 / 3 b1 / 3 , z = a 1 / 3 b 2 / 3
(C) a , b = ar , c = ar 2
If c > 4 b - 3 a
r 2 - 4r + 3 > 0
(Q a > 0)
( r - 3)( r - 1) > 0
(D) 7 x 2 - 8 x + 9 < 0
a = 7 > 0, D = 64 - 252 < 0
No solution
202
Solution of Advanced Problems in Mathematics for JEE
2. (A) a + d = b + c = 20
(B) 2 , G1 , G 2 , G 3 , G 4 , G 5 , G 6 , 5 are in G.P.
G1 G 6 = G 2 G 5 = G 3 G 4 = 10
(C) a 4 h7 = a1 h10 = a10 h1 = 6
7ö
æ
(D) (2 x - 5) 2 = 2 ç 2 x - ÷ Þ (2 x - 8)(2 x - 4) = 0 Þ x = 3
2ø
è
2
3
3. (A) 2 × 2 x = 2 x + 2 x
Exponential series can’t be in A.P.
(B) If a1 , a 2 , a 3 ,¼¼ , a n are in A.P.
a 2 - a1 = a 3 - a 2 = a 4 - a 3 = ¼¼ = a n - a n -1 = d
æ
ö
1
1
1
÷
S = -dç
+
+ ¼¼ +
ç a + a
÷
a
+
a
a
+
a
2
2
3
n -1
n ø
è 1
æ a n - a1 ö
÷= a - a
= -d ç
1
n
ç
÷
d
è
ø
2n
[2 a + (2n - 1) d]
S 2n
= 2
=3
n
Sn
[2 a + (n - 1) d]
2
(C)
2 a = (n + 1) d
3n
[2 a + (3n - 1) d]
S 3n
= 2
=3
2n
2S n
[2 a + (n - 1) d]
2
Þ
(D)
t 1 + t 5 = t 2 + t 4 = 2t 3
4(t 1 - t 2 - t 4 ) + 6t 3 + t 5 3t 1 + (t 1 + t 5 ) - 4 (t 2 + t 4 ) + 3 (2t 3 )
=
=1
3t 1
3t 1
4. A ® Q; B ® P; C ® T; D ® S
1
1
5. (A) log 2 x + log 2 y = 5 and log 2 y + log 2 x = 7
3
3
Þ log 2 x = 6 and log 2 y = 3
Þ
x = 2 6 and y = 2 3
(B) ÐB = 60° and b 2 = ac
cos B =
a 2 + c 2 - b2 1
=
2 ac
2
Þ a 2 + c 2 = 2 ac
Þ
a=c
203
Sequence and Series
(C) AM ³ GM
b c a c a b
+ + + + +
a a b b c c ³1
6
(D) ( b + c) 2 - a 2 = lbc
Þ
Þ
b 2 + c 2 - a 2 = (l - 2) bc
b2 + c 2 - a 2 l - 2
=
2 bc
2
l -2
-1 <
<1
2
0<l<4
6. P(n) × ( f(n + 2) - f(n)) = q(n)
1 ö
æ 1
P ( n) × ç
+
÷ = q ( n)
èn +1 n +2ø
P ( n) × ( 2n + 3) = ( n 2 + 3n + 2) × q ( n)
Þ
P(n) = n 2 + 3n + 2 and q(n) = (2n + 3)
Exercise-5 : Subjective Type Problems
1. If a , b, c , d are in A.P. with common difference ‘ k’ , then
9 k 3 + ( x - 4) k 2 + 4 k = 0
k {9 k 2 + ( x - 4) k + 4} = 0
D ³ 0 Þ ( x - 4) 2 - 144 ³ 0
( x + 8)( x - 16) ³ 0
x Î ( -¥, - 8] È [16 , ¥)
Þ
2. S = 1 × 2 + 2 × 2 2 + 3 × 2 3 + 4 × 2 4 +¼¼ + n × 2 n
2 × S = 1 × 2 2 + 2 × 2 3 + 3 × 2 4 +¼¼ + (n - 1) × 2 n + n × 2 n +1
S = (n - 1) × 2 n +1 + 2 = 2 + 2 n +10
Þ
2(n - 1) = 2 10
Þ
n = 513
Þ
n
¥
r+2
é 1
ù
1
1
å 2 r +1 r ( r + 1) = å ê r × 2 r - ( r + 1) × 2 r +1 ú = 2
n ®¥
3. lim
r =1 ë
r =1
¥
4.
8r
¥
æ
1
û
1
ö
å 4 r 4 + 1 = 2 å çè 2 r 2 - 2 r + 1 - 2 r 2 + 2 r + 1 ÷ø = 2
r =1
r =1
204
Solution of Advanced Problems in Mathematics for JEE
5. Let three terms in A.P. a - d , a , a + d
If ( a - d) 2 , a 2 ,( a + d) 2 are in G.P. Þ d = ± 2 a
r=
6.
a2
( a - d)
=
2
1
(1 ± 2 ) 2
æ 10 n - 1 ö
æ n
ö
10 2n - 1
÷ = P ç 10 - 1 ÷ Þ P = 3
-2ç
ç
÷
ç
÷
9
9
9
è
ø
è
ø
7. a - d , a , a + d , a - d + 30
If last three terms are in G.P.
( a + d) 2 = a ( a - d + 30)
a=
Þ
8.
1
d2
30 - 3 d
n
å[k( k + 2)( k + 4)( k + 6) - ( k - 2) k ( k + 2)( k + 4)]
8n 4
k =1
1 é(n - 1)(n + 1)(n + 3)(n + 5) + n (n + 2)(n + 4)(n + 6) + 15 ù 1
ú = 4 (n ® ¥)
8 êë
û
n4
2
é n(n + 1) ù
9. Unit digit of ê
úû = 1
ë 2
n(n + 1)
Then unit digit of
is 1 because unit digit of n(n + 1) can not be 8.
2
10. 2 log b c = log c a + log a b
æ log a + 2 log r ö æ
ö æ log a + log r ö
log a
÷÷ = çç
÷÷ + çç
÷÷
2 çç
log a
è log a + log r ø è log a + 2 log r ø è
ø
Let A = log a and R = log r Þ 3 A 2 + 3 Ar - 2 R 2 = 0 Þ
A + 2R
A
3
=
A+R
A + 2R 2
3 3r
2
r
6r 3
11. 3 , ,
, 7s ;
= 1 + and
= + 7s
r s
r
s
s
r
d = log b c - log c a =
Þ
Þ
12.
7 r 3 - 6 r 2 + 21r - 18 = 0 Þ ( r 2 + 3)(7 r - 6) = 0
6
9
r = and s =
7
14
S=
12
31
+
22
32
+
32
33
+
42
34
+¼¼
S 12 2 2 3 2
=
+
+
+¼¼
3 32 33 34
A - 3 + 33
=
R
6
205
Sequence and Series
2S
S 1 3
5
7
=S - = +
+
+
+¼¼
3
3 3 32 33 34
2S
1
3
5
=
+
+
+¼¼
2
3
9 3
3
34
2S 2S 1 2
2
2
= +
+
+
+¼¼
2
3
3
9 3 3
3
34
4S 1 2 æ
1
1
ö
= +
+¼¼ ÷
ç1 + +
2
2
9
3 3 è
3 3
ø
4S 1 2 æ 1 ö 2
÷=
= + çç
9
3 9 è 2 3 ÷ø 3
13.
S ¥ = f( x) max
a - ar = f ¢(0) = 3
If
\
\ f( x) max = f(3) = 27 + 9 - 9 = 27
a
1- r
a(1 - r) = 3 Þ
\
3 p
=
2 q
x Î [-4 , 3]
f ¢( x) = 3 x 2 + 3 > 0
S ¥ = 27 =
Þ S=
1
a
=
1- r 3
æ aö
27 = a ç ÷
è3ø
a 2 = 81 Þ
a = ±9
a =9
1- r =
p 2
=
q 3
3
9
2
r=
3
If a = -9
1- r = r=
1
3
4
> 1 (rejected)
3
\ p + q=5
14. Total runs from 1 to 9 = 1350
Let, number of terms in A.P. be n.
n
Þ
[300 + (n - 1) ´ ( -1)] = 4500 - 1350 = 3150
2
Þ
n = 25 or 126, n = 126 (not possible)
Þ
n = 25, total matches = 34
15. x =
10 100 æ 1
1 ö 10 æ
1 1 1
1
1
1
1 ö
ç
÷=
ç1 + + + ÷
4 n = 3è n - 2 n + 2 ø 4 è
2 3 4 102 101 100 99 ø
å
206
Solution of Advanced Problems in Mathematics for JEE
16.
f ( n) =
Let
(2n + 1) + (2n - 1) + (2n + 1)(2n - 1)
2n + 1 + 2n - 1
2n + 1 = a and
2
f ( n) =
( a + b + ab) ( a - b) a 3 - b 3
=
( a + b)
( a - b) a 2 - b 2
f ( n) =
Þ
60
2n - 1 = b
2
(2n + 1) 3/ 2 - (2n - 1) 3/ 2
2
60
(2n + 1) 3/ 2 - (2n - 1) 3/ 2 (121) 3/ 2 - 1
=
= 665
2
2
n =1
å f ( n) = å
n =1
17. 3 0 {2 0 + 2 -1 + 2 -2 ¼¥} = 1{2}
1
{2}
3
1
3 -2 {2 0 + 2 -1 + 2 -2 ¼¥} = {2}
3
3 -1 {2 0 + 2 -1 + 2 -2 ¼¥} =
M
M
¥
Hence,
2 ´1
=3
1
13
18. 15 2 + (15 + d) 2 + (15 + 2 d) 2 +¼ + (15 + 9 d) 2 = 1185
Þ
19 d 2 + 90 d + 71 = 0
Þ
d = -1
S n ³ S n -1
n
æ n - 1ö
(31 - n) ³ ç
÷ (32 - n) Þ n £ 16
2
è 2 ø
a/r
3
2
19. 24 x - 14 x + kx + 3 = 0
a
ar
Product of roots a 3 = -
1
1
Þ a=8
2
Þ k = -7
If x = 7 lies between the roots, then
f(7) = 49 + 7a 2 - 112 < 0
a2 -9 < 0
20. 9 x 3 + 3 y 3 + 1 = 9 xy
(9 1/ 3 x) 3 + (3 1/ 3 y) 3 + 13 = 3 (9 1/ 3 x)(3 1/ 3 y) Þ 9 1/ 3 x = 3 1/ 3 y = 1
207
Sequence and Series
21. If a , x , y , z , b
A.P.
3a + b
a+b
a + 3b
and z =
x=
,y=
4
2
4
If a , x , y , z , b
H.P.
4 ab
2 ab
4 ab
and z =
x=
,y=
3b + a
a+b
3a + b
æ 3a + b ö æ a + b ö æ a + 3b ö
æ 4 ab ö æ 2 ab ö æ 4 ab ö 343
If ç
Þ ab = 7
֍
֍
÷ = 55 and ç
֍
֍
÷=
è 4 øè 2 øè 4 ø
è 3 b + a ø è a + b ø è 3 a + b ø 55
❑❑❑
208
Solution of Advanced Problems in Mathematics for JEE
10
DETERMINANTS
Exercise-1 : Single Choice Problems
1. Direct expansion.
k 1 1
2.
D = 1 k 1 =0
1 1 k
1
also D1 = k
k2
3.
a a2
b b2
c c2
Þ ( k - 1) 2 ( k + 2) = 0
1 1
k 1 ¹0
1 k
Þ k ¹ 1 Þ k = -2
1 + a3
1 a a2
3
1 + b = (1 + abc) 1 b b 2 = 0
1 + c3
1 c c2
1 2a a
2 1 1
4. D = 1 3 b b = 0 Þ = +
b a c
1 4c c
6.
and
and
2 x + ay + 6 z = 8
4 x + 2 ay + 6 z = 8
6 x + 12 y + 6 z = 30
y=
Þ
Þ 2 x + ay = 0
Þ 4 x + (12 - a) y = 22
22
12 - 3 a
a¹4
7. R 1 ® R 1 + R 2 + R 3
x2 - 4
2
x 2 - 13
x2 - 4
x 2 - 13
3
x2 - 4
1
2
2
= ( x - 4)
2
7
x 2 - 13
Þ ( x 2 - 4)( x 2 - 15)(20 - x 2 ) = 0
1
1
x - 13 2 = 0
3
7
2
209
Determinants
k
k + 1 k -1
8. D = k + 1
k
k + 2 =0
k -1 k + 2
k
R1 ® R1 - R 2
-1
1
-3
D= 2
-2
2 =0
k -1 k + 2 k
R2 ® R2 - R3
log a + (n - 1) log r
log a + (n + 1) log r log a + (n + 3) log r
9. D = log a + (n + 5) log r log a + (n + 7) log r log a + (n + 9) log r
log a + (n + 11) log r log a + (n + 13) log r log a + (n + 15) log r
C3 ® C3 -C2
C 2 ® C 2 - C1
log a + (n - 1) log r 2 log r 2 log r
Þ
log a + (n + 5) log r 2 log r 2 log r = 0
log a + (n + 11) log r 2 log r 2 log r
a1
10. D 2 = b1
c1
2a3
2 b3
2c3
5a 2
a1
5b2 = 10 b1
5c 2
c1
a3
b3
c3
a2
a1
b2 = -10 b1
c2
c1
a2
b2
c2
1 bc a
11. D 2 = 1 ac b
1 ab c
R 1 ® aR 1
R 2 ® bR 2
R 3 ® cR 3
2
a abc a
a 1 a2
1 a a2
1
D2 =
b abc b 2 = b 1 b 2 = - 1 b b 2 = -D 1
abc
c abc c 2
c 1 c2
1 c c2
0 0
-1
12. C 1 ® C 1 - C 2 + C 3 0 1
1- a =1
1 a 1+ a - b
13.
1 2
2 3
3 5
x
x 2 = 10 Þ x 2 + x - 12 = 0
2
Sum = -1
14. R 1 ® R 1 - R 2 , R 2 ® R 2 - R 3
a3
b3
c3
210
Solution of Advanced Problems in Mathematics for JEE
-1
-1
D = d-a+1 e-b+1
x+a
x+b
-1
f - c + 1 , C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3
x+c
On solving D does not depend on x.
15. R 1 ® R 1 + R 2 + R 3
1
1
D = ( x + y + z) 2 y y - z - x
2z
2z
0
D = ( x + y + z)
16.
3
0
1
2y
z-x-y
C 1 ® C 1 - C 2 and C 2 ® C 2 - C 3
1
1 -1
2y
0 1 z-x-y
Þ D = ( x + y + z) 3
ÐBOC = 60°
Þ BC = OB = OC = r
D
AB = 2 r cos 30° = 3 r
P
2
Area or rectangle
3r
3
=
=
2
Area of circle
p
pr
A
C
30°
O
60°
Q
B
17. C 1 ® C 1 - bC 3 , C 2 ® C 2 + aC 3
1 0
-2 b
2
2 2
(1 + a + b ) 0 1
2a
= (1 + a 2 + b 2 ) 3
2
2
b -a 1 - a - b
18.
2
a+ b+ c+ d
ab + cd
1
1
0 1 a + b ab
a+ b+ c+ d
2( a + b)( c + d)
ab( c + d) + cd( a + b) = c + d a + b 0 1 c + d cd
ab + cd
ab( c + d) + cd( a + b)
2 abcd
cd
ab 0 0
0
0
=0
l m n
19. |B| = p q r = [2 Ar( DABC )]2
1 1 1
C
(l, p)
3
A
1 2 1
20. D = 1 3 4 = 0
1 5 10
5
(m, q) 4
B
(n, r)
211
Determinants
1
D1 = K
K2
2 1
3 4 = 5 ( K 2 - 3 K + 2) = 5 ( K - 1)( K - 2)
5 10
1 1
D2 = 1 K
1 K2
1 2
D3 = 1 3
1 5
1
4 = -3 ( K 2 - 3 K + 2) = -3 ( K - 2)( K - 1)
10
1
K = K 2 - 3 K + 2 = ( K - 2)( K - 1)
K2
1 x + 1 ( x + 1) 2
21. ( x + 1)( x + 2)( x + 3) 1 x + 2 ( x + 2) 2 = 2( x + 1)( x + 2)( x + 3)
1 x + 3 ( x + 3) 2
-2
22.
cos C
cos B
cos C
-1 cos A
cos B cos A
-1
-2 (1 - cos 2 A) - cos C( - cos C - cos A cos B ) + cos B (cos C cos A + cos B)
-2 + cos 2 A +
1 + cos 2C 1 + cos 2 B
+
+ 2 cos A cos B cos C
2
2
cos 2 A + cos 2C + cos 2 B + 2 cos A cos B cos C
2 cos( A + B ) cos( A + B ) + 2 cos 2 C - 1 + 2 cos A cos B cos C
2 cos C [cos C - cos( A - B )]
-2 cos C cos A cos B - 1 + 2 cos A cos B cos C = -1
24. As a , b and c are the roots of x 3 + 2 x 2 + 1 = 0 , we have
a + b + c = -2
ab + bc + ca = 0
abc = -1
a b c
Now, for finding the value of b c a , evaluating using first row, we get
c a b
a( bc - a 2 ) - b( b 2 - ac) + c( ab - c 2 ) = abc - a 3 - b 3 + abc + abc - c 3
= 3 abc - a 3 - b 3 - c 3 = - ( a 3 + b 3 + c 3 - 3 abc)
= - ( a + b + c)( a 2 + b 2 + c 2 - ab - bc - ca)
= - ( -2)[( -2) 2 - 3 (0)] = 8
212
Solution of Advanced Problems in Mathematics for JEE
l
l + 1 l -1
25. For non-trivial solution,| A| or D = 0, That is, l + 1
l
l + 2 =0
l -1 l + 2
l
l l + 1 l -1
Now, R 2 ® R 2 - R 1 ; R 3 ® R 3 - R 1 gives 1
-1
3 =0
-1
1
1
Also, R 3 ® R 3 + R 2 gives
l l + 1 l -1
1 -1
3 =0
0
0
4
Evaluation using third row, we get
4( -l - l - 1) = 0 Þ l = -
1
2
which is exactly the real value of l.
Exercise-2 : One or More than One Answer is/are Correct
1. f( a , b) = a( a + b)( a + 2 b)
2. R 1 ® R 1 - R 2 and R 2 ® R 2 - R 3
1
-1
0
1
0
1
-1
= 0 Þ tan q = 3
cos 2 q sin 2 q 1 + 2 3 tan q
3. R 1 ® R 1 - R 2 , R 2 ® R 2 - R 3
-1
-1
3
2
D =d
-1
2
-1 = -d 2 (13 d + 12 a)
a + 2d a a + d
4.
1-l
3
-4
1
- ( 3 + l) 5 = 0
3
1
-l
x 2 + 4 x - 3 2 x + 4 13
5. D( x) = 2 x 2 + 5 x - 9 4 x + 5 26
8 x 2 - 6 x + 1 16 x - 6 104
C 3 ® C 3 - 4C 2 , C 2 ® C 2 - 2C 1
3x + 3
3
0
D( x) =
26 x - 37
26
0
8 x 2 - 6 x + 1 16 x - 6 104
213
Determinants
a 1 2
7. D = 1 2 1 = 0 Þ a 2 - a - 2 = 0 Þ ( a - 2)( a + 1) = 0
2 1 a
0 1 2
D1 = b 2 1
0 1 a
a 0 2
D2 = 1 b 1
2 0 a
a 1 0
D3 = 1 2 b
2 1 0
a = 2 infinite solution
a = -1, b ¹ 0 has no solution.
a 1 2
8. D = 1 2 1 = 0 Þ a 2 - a - 2 = 0
2 1 a
Þ ( a - 2)( a + 1) = 0
a 0 2
a 1 0
D= 1 b 1
D= 1 2 b
2 0 a
2 1 0
0 1 2
D= b 2 1
0 1 a
a = 2 infinite solution
a = -1, b ¹ 0 has no solution.
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
2 l 6
D = 1 2 m = (l - 2)(m - 3) ;
1 1 3
8 l 6
D1 = 5 2 m = (l - 2)( 4m - 15)
4 1 3
2 8 6
D2 = 1 5 m = 0 ;
1 4 3
2 l 8
D 3 = 1 2 5 = ( l - 2)
1 1 4
Exercise-4 : Subjective Type Problems
2. R 1 ® R 1 + R 2 + R 3
a1 + b1 + c1 a 2 + b2 + c 2
3
2 b1 + c1
2 c 1 + a1
2 b2 + c 2
2c2 + a2
a 3 + b3 + c 3
2 b3 + c 3
2c3 + a3
214
Solution of Advanced Problems in Mathematics for JEE
a1 + b1 + c1
R 2 ® R 2 + R1 - R 3 = 9
b1
2 c 1 + a1
a 2 + b2 + c 2
b2
2c2 + a2
a 3 + b3 + c 3
b3
2c3 + a3
Now, operate as R 3 ® R 3 - R 1 + R 2
then R 1 ® R 1 - R 2 - R 3
(1 + x) 2 (1 + x) 4
3. Let f( x) = (1 + x) 3 (1 + x) 6
(1 + x) 6
(1 + x) 9
(1 + x) 4 (1 + x) 8 (1 + x) 12
Coefficient of ‘ x’ is f ¢(0).
2(1 + x) 2
f ¢( x) = (1 + x) 3
(1 + x) 4
4(1 + x) 3
(1 + x) 6
(1 + x) 8
6(1 + x) 5
(1 + x) 2
(1 + x) 9 + 3(1 + x) 2
(1 + x) 12
(1 + x) 4
(1 + x) 4
6(1 + x) 5
(1 + x) 8
(1 + x) 2
+ (1 + x) 3
4(1 + x) 3
Put x = 0, f ¢(0) = 0
5. For non-zero solution,
(1 + x) 6
9(1 + x) 8
(1 + x) 12
(1 + x) 2
(1 + x) 6
8(1 + x) 7
D =0
2 3 -1
3 2 k =0 Þ
4 1 1
k =0
Now, let x = l
So,
Þ
6.
y =-
3l
5l
, z =2
2
Minimum positive integer value of z is at l = -2 ; z = 5
2 a -2 3
1
a 2 =0 Þ a =2
2
0 a
7. Let three terms be A - d , A , A + d.
Þ
A 4 = ( A - d) 2 ( A + d) 2 = A 4 + d 4 - 2 A 2 d 2
Þ
d = ± 2 A , r = 3 + 2 2 or r = 3 - 2 2
3 a1 + b1
8. D 3 =
3 b1
3 c1
3 a 2 + b2
3 b2
3c2
3 a 3 + b3
3 a1
3 b3
= 3 b1
3c3
3 c1
3a2
3 b2
3c2
3a3
a1
3 b3 = 27 b1
3c3
c1
a2
b2
c2
a3
b3
c3
(1 + x) 6
(1 + x) 9
12(1 + x) 11
215
Determinants
6 a1
D 2 = 3 b1
12 c1
2a2
b2
4c2
2a3
a1
b3 = 24 b1
4c3
c1
a2
b2
c2
a3
b3
c3
1
cos q
1
9. D = - cos q
1
cos q = 3 (1 + cos 2 q)
-1
- cos q
2
Its minimum value = 3
1 1 1
10. D = 1 2 3 = l - 8 = 0 Þ l = 8
2 5 l
1 1 6
D 3 = 1 2 14 = m - 36 = 0 Þ m = 36
2 5 m
æ
1
1
1 ö÷
11. n sin 2 p ç 1 + 1 +
+
¼
|n
ç
|2 |3 | N ÷ø
è
1
1
1
æ
ö
n sin 2 p ç 1 +
+
+¼
÷
n
+
1
(
n
+
1
)(
n
+
2
)
(
n
+
1
)(
n
+
2
)
¼
(
N
)
è
ø
Using sin(2np + q) = sin q
1
1
æ 1
ö
= n(2 p) ç
+
+¼
÷
(n + 1)(n + 2)¼ N ø
è n + 1 (n + 1)(n + 2)
Using lim
q®0
sin q
=1
q
= 2p
12.
cos q sin q cos q
sin q cos q sin q = 0 Þ -2 cos q cos 2q = 0
cos q sin q - cos q
❑❑❑
216
Solution of Advanced Problems in Mathematics for JEE
11
COMPLEX NUMBERS
Exercise-1 : Single Choice Problems
2. arg ( z - 2 - 7i ) = cot -1 (2) Þ
p
æ z - 5i ö
arg ç
÷=±
2
è z + 2 -i ø
y -7 1
=
x -2 2
Þ x( x + 2) + ( y - 5)( y - 1) = 0
4. z 12 + z 22 = z 1 z 2
1
5. Let w = reiq then z = ei ( p/ 2+ q )
r
1
z w = e -i ( p/ 2+ q ) r × eiq = e -i p/ 2
r
6. a
n
n
r =1
r =1
å rw r -1 + bå w r -1 = a(1 + 2w + 3w 2 +¼¼ + nw n -1 ) + b(1 + w + w 2 +¼¼ + w n -1 )
ìï1 + w + w 2 +¼¼ + w n -1 nw n üï
=aí
ý + b ( 0)
1-w
1 - w ïþ
ïî
n ö
æ
= aç0 ÷+0
1
wø
è
8. z 4 + z 3 + 2 = 0 has roots z 1 , z 2 , z 3 and z 4 .
Þ ( z - 1) 4 + 2 ( z - 1) 3 + 32 = 0 has roots (2 z 1 + 1),(2 z 2 + 1),(2 z 3 + 1) and (2 z 4 + 1)
9.
æ z - 6 - 3i ö p
arg ç
÷=
è z - 3 - 6i ø 4
Þ ( x - 6) 2 + ( y - 6) 2 = 9
11. |iz + z 1| =|i||z - iz 1| =|z - iz 1|
Maximum distance of iz 1 ( -3 + 5 i ) from z is 2 + 3 2 + (5 - 1) 2 = 7
217
Complex Numbers
z 1 - z 2 |z 1 - z 2| iq
=
e
z 3 - z 2 |z 3 - z 2|
12.
z1
z 1 - z 3 |z 1 - z 3| -iq
=
e
z 2 - z 3 |z 2 - z 3|
æ z - z 2 iq z 1 - z 3 -iq ö
æ z - z 2 z1 - z 3 ö
÷
÷÷ = arg ç 1
arg çç 1
+
ç z -z e - z -z e ÷
è z3 - z2 z3 - z2 ø
2
2
3
è 3
ø
p
=±
2
z 2 3 i p/ 3
13.
= e
z1 2
z2
q
q
O
(0,2)
A(1,3)
z3
pö
p
æ
3 ip/ 3
ç 2 + 3 cos ÷ + 3i sin
e
z1 + z 2
3ø
3
è
2
=
=
3 ip/ 3
p
p
z1 - z 2
1- e
2 - 3 cos - 3i sin
2
3
3
1+
2
2
2
2
æ3 3 ö
æ7ö
÷
ç ÷ + çç
÷
è2ø
è 2 ø
=
æ3 3 ö
æ 1ö
÷
ç ÷ + çç
÷
è2ø
è 2 ø
=
49 + 27
133
=
1 + 27
7
14.
z 1 z 2 z 3 = -c
1 =|c| Þ |c| = 1
|z 1 + z 2 + z 3| £ |z 1| + |z 2| + |z 3|
|a| £ 3
|b| =|z 1 z 2 + z 2 z 3 + z 3 z 1| £ |z 1 z 2| + |z 2 z 3| + |z 3 z 1|
Þ |b| £ 3
1
15.
£ |z| £ 4
2
Þ
1
z+ =
z
2
ææ
ö
ææ
ö
1ö
1ö
çç ç r + ÷ cos q ÷÷ + çç ç r - ÷ sin q ÷÷
rø
rø
èè
ø
èè
ø
2
= r2 +
16. |3 + i ( z - 1)| =|z - 1 - 3i|
Maximum distance of A from ( z ) = OA + r
= 1+ 1 + 2 =2 2
17. x 2 - ( 2 i ) x - 1 = 0
x=
2 i ± -2 + 4
1
=
( ± 1 + i)
2
2
1
r2
+ 2 (cos q - sin q)
2
218
Solution of Advanced Problems in Mathematics for JEE
p
3p
, cis
4
4
3p
p
x 2187 = cis
, cis
4
4
1
1
3p
p
æ - 3p ö
æ pö
= cis ç
= 2i sin
, 2i sin = 2 i
÷ , cis ç - ÷ Þ x 2187 2187
2187
4
4
è 4ø
è 4 ø
x
x
x = cis
18. 1 ×
(1 + z 9 )
= 0 , z ¹ -1
1+ z
Þ
z 9 = -1
Þ
iq
re
i ( 2n +1) p
9
=e
, n = 1, 2 , ¼¼ 8
19. Let P( rei a ) & Q ( rei b )
Point of intersection of tangents at ‘ a ’ , ‘ b’ to circle x 2 + y 2 = r 2 is
æ a +b ö
a +b
a +b ö
æ
iç
÷
r sin
ç cos
÷
è 2 ø
2w 1 w 2
2 -i
2 ÷ = re
çr×
=
a -b
a -b ÷
a - b w1 + w 2
ç
cos
ç cos
÷ cos
è
2
2 ø
2
® ®
® ®
® ®
20. |z 1 - z 2|2 + |z 2 - z 3|2 + |z 3 - z 1|2 = 2 ( 4 + 9 + 16) - 2 ( a × b + b × c + c × a)
® ® ®
where a , b , c are position vectors of points z 1 , z 2 , z 3
æ 1ö
Maximum value = 58 - 2 (6 + 12 + 8) ç - ÷ = 84
è 2ø
7+i
21. We have
Z=
3 + 4i
Þ
Simplifying (i . e. , rationalizing the denominator), we get
7 + i 3 - 4i 21 + 4 - 28i + 3i
´
=
3 + 4i 3 - 4i
9 + 16
25 - 25i
=
=1-i
25
Therefore,
æ 7+i ö
ç
÷
è 3 + 4i ø
14
= (1 - i ) 14
= [(1 - i ) 2 ]7 = (1 + i 2 - 2i ) 7
= ( +2 7 ) i
22. |Z - 4| + |Z + 4| = 10
PS + PS ¢ = 2 a
which implies that foci at 4 and -4 and a = 5 as shown in the following figure.
219
Complex Numbers
bi
–5
–4
0
4
5
–bi
Now,
b 2 = 25 (1 - e 2 ) = 25 - (5e) 2
= 25 - 16 = 9
Þ
b=3
Z lies on the ellipse circumference |Z| denotes the distance from the origin. Therefore,
|Z|max = 5
|Z|min = 3
Thus, the difference between the maximum and the minimum values of |Z| is
|Z|max -|Z|min = 5 - 3 = 2
Exercise-2 : One or More than One Answer is/are Correct
1. Let z 1 = rei q and z 2 = rei f
|z 1 + z 2| =|z 1|
Þ |ei q + ei f| =|ei q| = 1
Þ (cos q + cos f) 2 + (sin q + sin f) 2 = 1
Þ
cos(q - f) = -
1
2
Þ
q -f =
2p
2p
or 3
3
z1
= ei ( q -f) = ei 2p/ 3 or e -i 2p/ 3
z2
æz ö p
2. (a) If arg çç 1 ÷÷ = then z 1 and z 2 subtend right-angle at circumcentre origin.
è z2 ø 2
\ the chord joining z 1 and z 2 will subtend an angle q at ‘ z ’ such that
ìq = p 4 if |z| = 1
ï
íq < p 4 if |z| > 1
îïq > p 4 if |z| < 1
1
1
1
(b) |z 1 z 2 + z 2 z 3 + z 3 z 1| =|z 1| ×|z 2| ×|z 3|
+
+
z1 z 2 z 3
=|z 1 + z 2 + z 3|
æ ( z + z 2 )( z 2 + z 3 )( z 3 + z 1 ) ö æ ( z 1 + z 2 )( z 2 + z 3 )( z 3 + z 1 ) ö
÷
÷÷ = ç
(c) çç 1
ç
÷
z1z 2z 3
z1z 2z 3
è
ø è
ø
z
q
q
q
220
Solution of Advanced Problems in Mathematics for JEE
(d) The triangle formed by joining z 1 , z 3 and z 2 is isosceles and right angled at z 3 .
B(iz)
A(z)
3.
O
C(i2z)
Method I : Multiplying a complex number by i rotates a vector for z in the anticlockwise
direction by an angle of 90°.
\
Ð AOB = ÐBOC = 90°
As shown in figure, the DABC is a right angled isosceles triangle.
Method II : Let z , iz , i 2 z are vertices A , B and C of the triangle ABC.
\
| AB| =|BC | also | AB|2 + |BC |2 =| AC|2
Since,
| AB| =|BC| also | AB|2 + |BC|2 =| AC|2
\
the DABC is a right angled isosceles triangle.
7. ( z + i ) 4 = 1 + i
æ p 2m p ö
z = -i + 2 1/ 8 cos ç +
÷
4 ø
è8
æ 2 1/ 8 × 2 ö
÷
Square side length = ç
ç
÷
2
è
ø
æ1
3 ö÷
8. z = 4 çç - i
2 ÷ø
è2
æ 2mp 60° ö
Roots = 4 1/ 4 cos ç
÷
4 ø
è 4
m = 0 , 1, 2 , 3
m =1
2
9. a + bw + cw = a
a + bw 2 + cw = a
æ 1
æ 1
3 i ö÷
3 i ö÷
|a| = 1 Þ a + b ç - +
+ cç - =1
ç 2
ç 2
2 ÷ø
2 ÷ø
è
è
10. Check option for z = w
w 62 + w + 1 = 0
w2 + w + 1 = 0
w155 + w + 1 = 0
w2 + w + 1 = 0
221
Complex Numbers
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
Sol.
f( z ) + f( z ) = f( z ) + f( z )
Þ
Þ
(az + b) + ( a z + b ) = a z + b + a z + b
(a - a )( z - z ) = 0
Im (a ) = 0
(Im( z ) ¹ 0)
f( z ) + f( z ) = 0
Þ
a( z + z ) + (b + b ) = 0
Þ
Re(b) = 0
(Q a = a )
(Re( z ) = 0)
2
| f( z )| > ( z + 1)
Þ
2 2
a z
2
+ b 2 > z 2 + 2z + 1
Þ (a 2 - 1) z 2 - 2 z + (b 2 - 1) > 0 " z Î R
Paragraph for Question Nos. 3 to 5
Sol.
|a - b| = 2 7
2
Þ
|(a + b) - 4ab| = 28
Þ
|z 12 - 4( z 2 + m )| = 28
Þ
|m - ( 4 + 5 i )| = 7
greatest (|m|) = 16 + 25 + 7
least |m| = 7 - 16 + 25
Paragraph for Question Nos. 6 to 7
Sol. C 1 :|z - z 1|2 + |z - z 2|2 = 10
Þ C 1 : ( x - 5) 2 + y 2 = 1
C 2 :|z - z 1|2 + |z - z 2|2 = 16
Þ C 2 : ( x - 5) 2 + y 2 = 4
Paragraph for Question Nos. 8 to 9
Z 2 - Z1
Z - Z 1 i q/ 2
Z 3 - Z1
Z - Z 1 i q/ 2
= 4
×e
,
= 4
e
|Z 2 - Z 1| |Z 4 - Z 1|
|Z 3 - Z 1| |Z 4 - Z 1|
Þ
Þ
B(Z3)
( Z 2 - Z 1 )( Z 3 - Z 1 ) ( Z 4 - Z 1 ) 2 q i
=
e
|Z 2 - Z 1||Z 3 - Z 1| |Z 4 - Z 1|2
( Z 2 - Z 1 )( Z 3 - Z 1 )
(Z 4 - Z1 )
Similarly, others.
2
=
AB × AC
( IA) 2
I(Z4)
q/2
Sol.
q/2
B(Z1)
(Z1+Z2)
B(Z2)
222
Solution of Advanced Problems in Mathematics for JEE
Exercise-4 : Matching Type Problems
1. Let BC = n , CA = n + 1, AB = n + 2
æ z - z3 ö
æ z - z1 ö
÷÷ = 2 arg çç 3
÷÷ = ÐC = 2 ÐA
(A) arg çç 1
z
z
è 2
è z 2 - z1 ø
3 ø
sin C sin A
sin 2 A sin A
\
=
Þ
=
c
a
n+2
n
n+2
(n + 2) 2 + (n + 1) 2 - n 2 n + 2
Þ
=
2n
2(n + 2)(n + 1)
2n
Þ
cos A =
Þ
n(n 2 + n + 5) = (n 2 + 3n + 2)(n + 2) Þ n 2 - 3n - 4 = 0 Þ n = 4
\
biggest side = n + 2 = 6
®
®
®
®
Þ
n 2 + (n + 1) 2 = (n + 2) 2 Þ n = 3
1
Area = × 3 × 4 = 6 = D
2
(B) ( c - a ) × ( b - c ) = 0 Þ ÐC = 90° Þ a 2 + b 2 = c 2
\
®
\
(C)
®
®
®
®
®
| a ´ b + b ´ c + c ´ a| = 2 D = 12
a1 b2 - a 2 b1
4
= tan A =
a1 a 2 + b1 b2
3
\
\
3
5
(n + 2) 2 + (n + 1) 2 - n 2 3
=
2(n + 2)(n + 1)
5
5(n 2 + 6n + 5) = 6(n 2 + 3n + 2)
Þ
Þ
\ cos A =
n 2 - 12n - 13 = 0 Þ n = 13
1
1
S - c = ( a + b - c) = (13 + 14 - 15) = 6
2
2
(D) Altitudes are in H.P. Û sides are in A.P.
Also, b > a + c , a > b + c , c > a + b Þ least value of a = 2
\ least value of b = 3
3. (A) {0 , 1, w + 1} m = {0 , 1, - w 2 } m
0 , 1, - 1, - w 2 , - w, w
(B) 2w,( x 2 - x + 10) = 0
roots are 2 + 3w, 2 + 3w 2
Last number is 3.
(C) Central angle = 60°
Equilateral D
(D) Put z = 1
z 1 = 1, z 2 = w, z 3 = w 2
223
Complex Numbers
Exercise-5 : Subjective Type Problems
Im(z)
1.
O
O
p/4
p/4
Re(z)
2 £ |z| £ 4
1
Probability =
4
2.
Im(z)
5
4
3
2
1
O
3. z + z = 2 |z - 1|
arg ( z 1 - z 2 ) =
Re(z)
1 2 3 4 5
Þ y2 = 2x - 1
p
Þ y1 - y 2 = x1 - x 2
4
y 12 - y 22 = 2 ( x 1 - x 2 ) = 2 ( y 1 - y 2 ) Þ y 1 + y 2 = 2 ( y 1 ¹ y 2 )
❑❑❑
224
Solution of Advanced Problems in Mathematics for JEE
12
MATRICES
Exercise-1 : Single Choice Problems
écos q ù
é sin q ù
1. A = ê
ú [cos q sin q] + ê - cos q ú [sin q - cos q]
sin
q
ë
û
ë
û
é cos 2 q
cos q sin q ù é sin 2 q
- sin q cos q ù
=ê
ú+ê
ú =I
2
sin q û ë - sin q cos q
cos 2 q û
ësin q cos q
0 0 -1
2. | A| = 0 -1
-1 0
0 =1
0
æ 0 0 -1ö æ 0 0 -1ö æ 1 0 0 ö
ç
֍
÷ ç
÷
A = ç 0 -1 0 ÷ ç 0 -1 0 ÷ = ç 0 1 0 ÷
ç -1 0 0 ÷ ç -1 0 0 ÷ ç 0 0 1 ÷
è
øè
ø è
ø
é3 0 0 ù
3. A = ê0 3 0 ú
ê
ú
êë0 0 3 úû
2
det (adj (adj ( A))) =| A|4 = 27 4
ì27 4 ü 1
í
ý=
î 5 þ 5
4. A -1 B -1 = B -1 A -1 Þ C = ( A -1 + B -1 ) 5 = ( I ) 5
5. A 4 = I Þ A( A 3 ) = I
7. (adj A) A =| A|I
| A| = xyz - 8 x - 4 y - 3 z + 28 = 2l - l = l
8. ( x - 2) + ( x 2 - x + 3) + ( x - 7) = 0
x 2 + x - 6 = 0 Þ ( x + 3)( x - 2) = 0
225
Matrices
1 é0 3 ù
é -1 3 ù
9. A = ê
Þ A -1 = ê
ú
9 ë3 1úû
ë 3 0û
- tan q ù é cos 2 q
é 1
- sin q cos q ù é a -bù
10. ê
ê
ú =ê
ú
ú
1 û ësin q cos q
cos 2 q û ë b a û
ëtan q
a = cos 2q , b = sin 2q
Þ
P2 =I -P
11.
or
P 3 = P - P 2 = 2P - I
or
P 4 = 2I - 3P
or
P 5 = -3 I + 5P
or
P 6 = 5I - 8 P
12. |adj (adj ( A))| =| A|( n -1)
2
| A| = x + y + z = 12
x ³ 1, y ³ 1, z ³ 1
Þ
11
Þ
C 2 = 55
é a bù
é d -c ù
13. Let A = ê
; adj( A) = ê
ú
ú
ë c dû
ë -b a û
T
é d -bù
é a bù
=ê
; adj (adj ( A)) = ê
ú
ú
ë -c a û
ë c dû
14. M = A 2m × A -1
M=
A 2m +1
a 2 + b2
If A 2 = ( a 2 + b 2 ) × I Þ A 2m = ( a 2 + b 2 ) m × I
A 2m +1 = ( a 2 + b 2 ) m × A
15. A 2 + 5 A + 6 I = I
( A + 2 I )( A + 3 I ) = I
Þ A + 2 I and A + 3 I are inverse of each other.
é3 -5 ù é12 -5 ù é 1 0 ù
16. AB = ê
úê
ú =ê
ú =I
ë7 -12 û ë 7 -3 û ë0 1û
é 3 -2 ù
17. adj( A) = ê
ú
ë -2 2 û
18. AA 1 = I
écos q 2 sin q ù é cos q sin q ù é 1 0 ù
ê sin q cos q ú ê2 sin q cos q ú = ê0 1ú
ë
ûë
û ë
û
Þ
écos 2 q + 4 sin 2 q
3 sin q cos q ù é 1 0 ù
ê
ú =ê
ú Þ sin q = 0
sin 2 q + cos 2 q û ë0 1û
ë 3 sin q cos q
226
Solution of Advanced Problems in Mathematics for JEE
æ 1 2ö
æ cos q sin q ö
20. If A = ç
÷, P = ç
÷ , Q = P T AP , we have
è 0 1ø
è - sin q cos q ø
PQ 2014 P T =
P( P T AP )( P T AP ) ¼( P T AP )P T
2014 times
= ( PP T ) A( PP T ) A( PP T ) ¼( PP T ) A( PP T )
Matrix multiplication is associative.
æ cos q sin q ö æ cos q - sin q ö
PP T = ç
֍
÷
è - sin q cos q ø è sin q cos q ø
æ 1 0ö
=ç
÷ = I2
è 0 1ø
Hence,
PQ 2014 P T = A 2014
æ1
A =ç
è0
æ1
A3 =ç
è0
2ö
æ 1 2ö æ 1 2ö æ 1 4ö
÷ Þ A2 =ç
֍
÷ =ç
÷
1ø
è 0 1ø è 0 1ø è 0 1 ø
4öæ 1 2ö æ 1 6ö
֍
÷ =ç
÷
1 ø è 0 1ø è 0 1ø
æ 1 6ö æ 1 2ö æ 1 8ö
A4 =ç
֍
÷ =ç
÷
è 0 1ø è 0 1ø è 0 1ø
æ 1 2n ö
æ 1 4028 ö
An =ç
÷ and A 2014 = ç
÷
1 ø
è0 1 ø
è0
Þ
M
æMö
adj ç ÷ =
2
2
è ø
1
1
22. | A -1| =
=
| A| 5
21.
2
æ1
ö
= ç |M|÷
8
è
ø
2
|( AB ) T | =| AB| =| A × ( adj A)| =| A||
× adj ( A)|= 5 ´ 5 2 = 5 3
1
1
\|| A -1|( AB ) T |=| ( AB ) T |= | AB|= 1
5
53
Exercise-2 : One or More than One Answer is/are Correct
3.
Also,
and
A a Ab = A a +b
é1 0 0ù
A 0 = ê0 1 0 ú = I
ê
ú
êë0 0 1úû
A a A -a = A a -a = A 0 = I
227
Matrices
A a-1 = A -a
we get
However, A a-1 = - A a and A a2 = -I do not hold.
4. A( A 2 - I ) - 2( A 2 - I ) = 0
( A 2 - I )( A - 2 I ) = 0
Exercise-3 : Matching Type Problems
1. (A) Possible non-negative value of | A| = 2 , 4 , 8
(B) Sum is 0.
(C) |adj(adj(adj A)))| =| A|
least absolute value of | A| = 2
Þ | A| = ± 2
(D) least | A| = -8
16
|4 A -1| =
= -2
| A|
2. (A) Since A is idempotent, A 2 = A 3 = A 4 =¼¼ = A. Now,
( A + I ) n = I + n C 1 A + n C 2 A 2 +¼¼ + n C n A n
= I + n C 1 A + n C 2 A +¼¼ + n C n A
= I + ( n C 1 + n C 2 +¼¼ + n C n ) A
= I + (2 n - 1) A
Þ
2 n - 1 = 127 Þ n = 7
(B) We have,
( I - A)( I + A + A 2 +¼¼ + A 7 )
= I + A + A 2 +¼¼ + A 7 + ( - A - A 2 - A 3 - A 4 ¼¼ - A 8 )
= I - A8
=I
(if A 8 = 0)
(C) Here matrix A is skew-symmetric and since | A| =| A T | = ( -1) n| A|, so | A|(1 - ( -1) n ) = 0.
As n is odd, hence| A| = 0. Hence A is singular.
(D) If A is symmetric, A -1 is also symmetric for matrix of any order.
æ
ö
÷ 1
n ç
1
ç 1 ÷ = 1 dx
5. (A)
n r =1 ç r ÷
x
ç
÷ 0
è nø
å
ò
228
Solution of Advanced Problems in Mathematics for JEE
(B) D = 4 cos t cos 2t
(C) 3 x 2 + 2 px + g < 0
æ 5ö
f ç- ÷ =0
è 3ø
f( -1) = 0
(D) (2 x - 2) 2 + 1 + ||b - 1|-3| =|sin y|
b -1= ±3
|sin y| = 1
Exercise-4 : Subjective Type Problems
1. ( AB ) 2 = AB × AB = A 3 B 2
( AB ) 3 = ( AB ) 2 × AB = A 3 B 2 × AB = A 7 B 3
( AB ) 4 = ( AB ) 3 × AB = A 7 B 3 × AB = A 15 B 4
Þ ( AB ) 10 = A 1023 B 10
æ 3
ö
32 33
1
2. l = lim 18 ç
+
+
+¼¼ ÷ = 18 ´
=9
2
4
6
ç
÷
n® ¥
1ö
æ
3
3
è3
ø
3ç1 - ÷
3ø
è
æ 2
ö
22
1
m = lim 12 ç
+
+¼¼ ÷ = 12 ´
= 12
2
4
ç
÷
n® ¥
1ö
æ
2
è2
ø
2ç1 - ÷
2ø
è
a1 a 2 a 3
4. b1 b2 b3 = a1 b2 c 3 - a1 c 2 b3 …six elements
c1
c2
c3
All cannot be simultaneously 1.
5. First element of matrix A10 = 286 (10 th of sequence 1, 2, 6, 15, …)
Trace of A10 = 286 + 297 + 308 + 319 +¼ + 385 = 3055
❑❑❑
229
Permutation and Combinations
13
PERMUTATION AND COMBINATIONS
Exercise-1 : Single Choice Problems
1.
7
7
= 81;
9 9
8
7
= 72;
9
= 72
8 9
8!
8!
æ
ö 2
æ
ö
2. ç
÷ ´ C1 ´ 3! + ç
÷ ´ 3 ! = 8400
è 3 !3 !2 !2 ! ø
è 3 !2 !2 !2 ! ø
æ 3!
ö
3. Number of ways = 6 ´ ç ´ 3 !÷ = 108
è 2!
ø
5!
= 240
2!
4.
4
C1 ´
5.
6
C 2 ´ 1 ´ 4 ! = 360
x 2 - 5x + 3 = 0
6.
Þ
5
b
a + b = 5, ab = 3
Sum of roots =
7.
a
a b a 2 + b 2 19
+ =
=
b a
ab
3
C 4 ´ 8 C 6 + 5 C 5 ´ 8 C 5 = 196
8. (1 + 2 + 3 +¼¼ + 22) 21 C 10
9. x =
2009 ´ 2008 ´ 2007 + 1
1
= 2008 +
2008 ´ 2007 ´ 2007
2009 ´ 2007
Þ
10.
[ x] = 2008
N = p1n p 2 p 3 ¼¼ p m +1
No. of factors = (n + 1) 2 m
11. Number of ways = (11)! ´ 2 12
230
Solution of Advanced Problems in Mathematics for JEE
12.
5 5
m = 5 ´ 5 ´ 8C 2 ´ 2 !
4 5
n = 4 ´ 5 ´ 8C 2 ´ 2 !
13. Three different digits (not including zero)
9
C 3 ´ 2!
Two digits (not including zero)
9
C2 ´2
Three digits (including zero)
9
C2 ´1
14. Let no. of elements in A = n
No. of elements in B = m
2 n - 2 m = 1920 = 2 7 ´ 15
n = 11, m = 7
n( A È B ) = n( A) + n( B ) - n( A Ç B ) = 15
C ¼¼¼ = 4 ! = 24
D¼¼¼ = 4 ! = 24
M ¼¼¼ = 4 ! = 24
S C …… = 3! = 6
Þ
15.
S D …… = 3! = 6
S M C DW=1
S M C WD=1
16. P = All A’s together =
5!
;
3!
n( P Ç Q ) = 3 ! ;
Q = All B’s together =
n( P È Q ) =
6!
4!
5! 6 !
+
- 3 ! = 50 - 6 = 44
3! 4!
17. 5 6 ´ 6 7 ´ 7 8 ´ 8 9 ´ 9 10 ´ 10 11 ´¼¼´ 30 31
No. of zero’s = no. of 5’s
= 6 + 11 + 16 + 21 + (2 ´ 26) + 31 = 137
18. ( x - y)( x + y) = 10 ´ 337
Þ
x - y = 10 and x + y = 337
347
(not possible)
x=
2
19. Total number of different things = n + 2
231
Permutation and Combinations
20. Let the numbers are 10 - d , 10 , 10 + d.
d Î {-9 , - 8 , - 7 , ¼¼ , 7 , 8 , 9}
22. m = 2 ´ 5 ! ´ 5 !
n = 4 ! ´ 5!
23. Total ways = 4 ´ 4 ! = 96
25.
4
C 2 ´ 5 2 ´ (21) 2 = 66150
26. Total all letters are different.
Þ
10 5 - 10 C 5 ´ 5 ! = 69760
29. M = 1440
M = 25 ×32 ×5
No. of divisions = 6 ´ 3 ´ 2 = 36
P = Product of divisors = (1440) 18
P = 2 90 × 3 36 × 518
Hence, x = 30
30. Case-1 : All digits same = 9
Case-2 : Excluding zero :
(i) No’s having 3 digits same : 9 C 2 ´ 2 C 1 ´
4!
= 288
3!
(ii) No’s having 2 digits same, 2 other same : 9 C 2 ´
4!
= 216
2 !2 !
Case-3 : Including zero :
(i) No’s having 3 zero’s : 9
3!
= 27
2!
3!
(iii) No’s having 1 zero = 9 C 1 ´ = 27
2!
(ii) No’s having 2 zero’s : 9 C 1 ´
Hence, total no’s = 576
31. Case-I : When two T’s contain exactly one vowel between them,
5 ! ´ ( 5 C 1 ´ 5 C 4 ´ 4 !) = 15 ´ 5 ! ´ 5 !
Case-II : When two T’s also contain consonant between them,
4 ! ´ ( 5 C 2 ) ´ ( 7 C 5 ´ 5 !) = 42 ´ 5 ! ´ 5 !
32. 6 6 6 6 6 0 ® 6
6!
4 !2 !
6!
666642®
4!
666633®
232
Solution of Advanced Problems in Mathematics for JEE
666444®
6!
3 !3 !
33. Five 4 runs + one 0 run =
6!
5!
Four 4 runs + two 2 runs =
6!
4 !2 !
Three 4 runs + two 3 runs + one 2 runs =
Two 4 runs + four 3 runs =
Þ
6!
3 !2 !
6!
2 !4 !
N = 96
34.
7
35.
x 1 + x 2 + x 3 + x 4 + x 5 = 101
Let x 1 = 2 k1 + 1, x 2 = 2 k 2 + 1, x 3 = 2 k 3 + 1, x 4 = 2 k 4 + 1, x 5 = 2 k 5 + 1
C 2 = 21
Þ
k1 + k 2 + k 3 + k 4 + k 5 = 48 ;
48+ 5-1
C 5-1
36. Total ways = (largest number is 4)
6 4 - ( 4 4 - 3 4 ) = 1121
37.
6
C 3 ´ 4!
38. If two points are selected from one side of main diagonal = 6 C 2 .
Then other two points are selected on other side of main diagonal = 1.
Total ways = 6 C 2 ´ 1 = 15
39. (9 - x 1 ) + (9 - x 2 ) + (9 - x 3 ) + (9 - x 4 ) + (9 - x 5 ) = 43
Þ
x1 + x 2 + x 3 + x 4 + x 5 = 2
Number of ways = 2+ 5-1 C 5-1 = 6 C 4 = 15
Exercise-2 : One or More than One Answer is/are Correct
1. Case-I : All five letters are different.
= 5!
Case-II : Two letters are same and remaining are different.
5!
3
C 1 ´ 4 C 3 ´ = 720
2!
Case-III : Two alike, two other alike and remaining different.
5!
3
C 2 ´ 3C 1 ´
= 270
2 !2 !
Total number of words = 1110
233
Permutation and Combinations
100
2.
å 100 C k ( x - 2)100-k × 3 k = ( x + 1)100
k =0
Coeff. of x 50 = 100 C 50
3.
Total - Row 1 - Row 2
|2
8
{|2 for N }
C 5|6 -|6 -|6
|2
4. = (four odd) + (4 even) + (3 even + 1 odd) + (2 even + 2 odd)
= 5C 4 ´ 4 ! + 4C 4 ´ 4 ! + 4C 3 ´ 5C 1 ´ 4 ! + 4C 2 ´ 5C 2 ´ 4 ´ 4
= 1584
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1. 0
2. Digit 6 always come at last three place digit 5 always come at last four place and digit 4 always
come at last five place.
3
C 1 ´ 3 C 1 ´ 3 C 1 ´ 3 ! = 162
Exercise-4 : Matching Type Problems
1. (A)
6! 7
´ C 2 = 7560
2!
(B) 5 !´ 6 C 2 = 1800
4!
= 720
2 !2 !
2. (A) Total ways - (No repeating letter is at odd position)
11!
11!
-0 =
2 !2 !2 !
(2 !) 3
(C) 7560 - 1800 = 5760
(B)
7!
4!
´ 8C 4 ´
= 210 ´ 7 !
2 !2 !
2!
(C) MM
TT
HEICS
7 ! ´ 8 C 2 ´ 1 = 28 ´ 7 !
æ 4 !ö æ 7 ! ö 4 ! 7 !
(D) ç ÷ ´ ç
÷=
è 2 ! ø è 2 !2 ! ø (2 !) 3
(D) 4 ! ´ 5 C 4 ×
234
Solution of Advanced Problems in Mathematics for JEE
Exercise-5 : Subjective Type Problems
1.
9
C 4 ´ 5 C 4 = 630
2.
9
C2 ´
4.
10
7!
9!
=
2 !2 ! 8
C 3 - 8 C 3 = 64
5. Case-I: If Ravi is include.
7
C 5 ´ 9 C 8 = 189
Case-II: If Ravi is not include.
7
C 6 ´ [ 8 C 7 + 9 C 8 ] = 119
Total number of ways = 308
6.
6
C 4 - 4C 2 = 9
7. 5 ! - (1 + 5 C 2 ´ 1) = 109
8. Let other two sides are a and b.
\
a + b > 11
a3
b2
0 < a £ 11, 0 < b £ 11
9.
c3
a2
c2
b1
a1
c1
( a1 , a 2 , a 3 ),( b1 , b2 ) and ( c1 , c 2 , c 3 ) are alike things so these can be arranged is
8!
4 ×5 ×6 ×7 ×8
=
= 560
2 !3 !3 !
2 ´6
10.
n
C 2 - n = 14 Þ n = 7
11. x 1 + x 2 + x 3 +¼¼ + x 7 + x 8 = 93
x 1 ³ 0 , x 2 ³ 6 , x 3 ³ 6 , ¼¼ x 7 ³ 6 , x 8 ³ 0
x 1 + x ¢2 + x ¢3 +¼¼ + x 7¢ + x 8 = 57
No. of ways = 64 C 7
12.
4
C 4 ( 2 C 1 ) 4 = 16
13. Let
x 1 objects of one type
x 2 objects of second type
x 3 objects of third type
x 1 + x 2 + x 3 = 3n
235
Permutation and Combinations
0 £ x 1 £ 2n, 0 £ x 2 £ 2n , 0 £ x 3 £ 2n
Number of ways = 3n + 2 C 2 - 3 ´ n +1 C 2 = 3n 2 + 3n + 1
14. x + y + z + w = 15
x ³ 0, y ³ 6, z ³ 2, w ³ 1
x + y¢ + z ¢ + w¢ = 6
Number of ways = 9 C 3 = 84
❑❑❑
236
Solution of Advanced Problems in Mathematics for JEE
14
BINOMIAL THEOREM
Exercise-1 : Single Choice Problems
153
1. Let x = 2 2
a = x2 + 2 x + 1
N = x 16 - 1
N = ( x 4 - 1)( x 4 + 1)( x 8 + 1)
N = ( x 4 - 1)( x 2 + 2 x + 1)( x 2 - 2 x + 1)( x 8 + 1)
Let y = 2 204
b = y2 - y + 1
3.
4
N = y 6 - 1 = ( y 3 - 1)( y 3 + 1)
= ( y 3 - 1)( y + 1)( y 2 - y + 1)
C 2a 2 = - 6C 3a 3
Þ
a =-
3
10
5. a n = (2 + 3 ) n
Let a ¢n = (2 - 3 ) n Þ a n + a ¢n = integer
[a n ] + {a n } + a ¢n = integer Þ {a n } = 1 - a ¢n
Þ
lim (a n - [a n ]) = lim [1 - (2 - 3 ) n ] = 1 - 0 = 0
n ®¥
n ®¥
20
20
20
20
N = C 7 - C 8 + C 9 - C 10 +¼¼ - 20 C 20
So,
6.
(Q0 > {a n }, a ¢n < 1)
= ( 20 C 7 + 20 C 9 + 20 C 11 +¼¼ + 20 C 19 ) - ( 20 C 8 + 20 C 10 +¼¼ + 20 C 20 )
= ( 20 C 0 + 20 C 2 + 20 C 4 + 20 C 6 ) - ( 20 C 1 + 20 C 3 + 20 C 5 )
= (1 + 190 + 4845 + 38760) - (20 + 1140 + 15504)
= 43796 - 16664 = 27132 = 3 ´ 4 ´ 7 ´ 19 ´ 17
1
é
ù
7. log 2 ê1 + (2 12 - 2)ú = log 2 2 11 = 11
[QS n C r = 2 n ]
2
ë
û
æ 1 ö
8. T r +1 = n C r × x n -r × y r = 12 C r × x 12-r × ç
÷
è x3 ø
r
237
Binomial Theorem
12 - 4 r = 0 Þ r = 3
12 ´ 11 ´ 10
T 4 = 12 C 3 =
= 220
3 ×2 ×1
3
4 5
1
1
+ +
+¼¼ = 4 ! 5! 6 !
3 ! ( k + 3)!
9.
52
52
r
é1
1
ù
1
1
å ( r + 1)! = å êë r ! - ( r + 1)!úû = 3 ! - 53 !
r =3
r =3
k = 50
Þ
n
10.
f( x) =
å[( r + 1) 2 n C r - r 2 n C r -1 ]
r =1
f(n) = (n + 1) 2 - 1
f(30) = 960
12.
n
n
2
n
C 1 × a + C 2 × a + C 3 × a 3 +¼¼ n C n × a n = (1 + a ) n - 1
2p i
æ
ö
ç where a = e n = a 2 ÷
ç
a1 ÷
è
ø
13. 2 30 × 3 20 = 2 10 × (6) 20 = 1024 (7 - 1) 20 = 1024 (7 K + 1) = 7 k ¢ + 1024 = 7 k ¢ + 1022 + 2
14.
26
Þ
C 0 + 26 C 1 + 26 C 2 +¼¼ + 26 C 26 = 2 26
2( 26 C 0 + 26 C 1 +¼¼ + 26 C 13 ) = 2 26 + 26 C 13
15. (1 + x + x 2 ) n = a 0 + a1 x + a 2 x 2 + a 3 x 3 +¼+ a 2n x 2n
differentiate w.r.t. x
n(1 + x + x 2 ) n -1 (1 + 2 x) = a1 + 2 a 2 x + 3 a 3 x 2 +¼ + 2n × a 2n x 2n -1
n × 3 n = a1 + 2 a 2 + 3 a 3 +¼ + 2n a 2n
Put x = 1
0 = a1 + 2 a 2 w + 3 a 3 w 2 +¼ + 2n a 2n w 2n -1
Put x = w
Put x = w
2
2
4
0 = a1 + 2 a 2 w + 3 a 3 w +¼ + 2n a 2n w
(1) + (2) + (3)
n × 3 n -1 = a1 + 4 a 4 + 7 a7 + 10 a10 +¼
16.
…(1)
n
C r + n C r -1 = n +1 C r
3
C 0 + 3 C 1 + 4 C 2 + 5 C 3 +¼ + 99 C 97 = 100 C 97
17. Last digit of 9 ! = 0
Last digit of 3 9966 = 9
Hence last digit 9.
18. x = T7 = n C 6 (3 1/ 3 ) n -6 × ( 4 -1/ 3 ) 6
4n -2
…(2)
…(3)
238
Solution of Advanced Problems in Mathematics for JEE
y = T n -5 = n C n -6 (3 1/ 3 ) 6 × ( 4 -1/ 3 ) n -6
y = 12 x
n
C n -6 (3 1/ 3 ) 6 ( 4 -1/ 3 ) n -6 = 12 × n C 6 (3 1/ 3 ) n -6 ( 4 -1/ 3 ) 6
12 = (12 1/ 3 ) 12-n Þ n = 9
Þ
æ2ö
20. t r +1 = 15 C r ( x 2 ) 15-r × ç ÷
è xø
r
Coeff. of x 15 = 15 C 5 × 2 5
Coeff. of x 0 = 15 C 10 × 2 10
21. (1 + x) 2 (1 + y) 3 (1 + z ) 4 (1 + w) 5
General term = 2 C a 3 C b 4 C d 5 C e x a + b + d + e
2
å
22.
C a ´ 3 C b ´ 4 C d ´ 5 C e = 14 C 12 or 14 C 12 =
a + b + d + e =12
n
n
1 n
r × nC r + 2
× Cr;
r =0
r =0 r + 1
å
å
n ×2
Þ
n -1
n
n
2
14 ´ 13
= 91
2
n
å n -1 C r -1 + n + 1 å n +1 C r +1
r =0
r =0
2
+
× (2 n +1 - 1)
n+1
Exercise-2 : One or More than One Answer is/are Correct
1. N = 20 C 7 - 20 C 8 + 20 C 9 - 20 C 10 +¼¼ - 20 C 20
= ( 20 C 7 + 20 C 9 + 20 C 11 +¼¼ + 20 C 19 ) - ( 20 C 8 + 20 C 10 +¼¼ + 20 C 20 )
= ( 20 C 20 + 20 C 2 + 20 C 4 + 20 C 6 ) - ( 20 C 1 + 20 C 3 + 20 C 5 )
2. For B and D put x = 1, - 1
For A differentiate with respect to x then put x = 0
1
For C replace x with
x
4
3.
å( -1) r 16C r = 16C 0 - 16C 1 + 16C 2 - 16C 3 + 16C 4 = 1365
r =0
4. 2 ´
1 n
1 n
´ C1 = 1 +
´ C 2 Þ n = 8, 1
2
22
r
æ 1ö
T r +1 = C r ç ÷ x
è2ø
8
16- 3r
4
Þ r = 0, 4, 8
5. LHS = (1 + 2 x 2 + x 4 )(1 + C 1 x + C 2 x 2 + C 3 x 3 +¼¼ )
239
Binomial Theorem
RHS = a 0 + a1 x + a 2 x 2 + a 3 x 3 +¼¼
Comparing the coefficients of x , x 2 , x 3 ,¼¼
Now,
2 a 2 = a1 + a 3
2( n C 2 + 2) = n C 1 + ( n C 3 + 2 n C 1 )
2
or
n 3 - 9n 2 + 26n - 24 = 0
\
(n - 2)(n 2 - 7n + 12) = 0
or
(n - 2)(n - 3)(n - 4) = 0
n = 2, 3, 4
\
n
6.
n(n - 1)
n(n - 1)(n - 2)
+ 4 = 3n +
2
6
n
n
(Q8 + 52 = 36 + 24)
æ n
öæ n
è i =0
ø è j=0
öæ n
ö
ø è k =0
ø
å å å n C i × n C j × n C k = çç å n C i ÷÷ çç å n C j ÷÷ çç å n C k ÷÷ = 2 3n
i = 0 j= 0 k = 0
7. (
100
C6 +
100
C7 ) + 3(
100
C7 +
100
C 8) + 3(
100
C8 +
100
C 9 ) + ( 100 C 9 + 100 C 10 )
= 101 C 7 + 3 101 C 8 + 3 101 C 9 + 101 C 10
( 101 C 7 + 101 C 8 ) + 2 ( 101 C 8 + 101 C 9 ) + ( 101 C 9 + 101 C 10 ) = 102 C 8 + 2 × 102 C 9 + 102 C 10
= ( 102 C 8 + 102 C 9 ) + ( 102 C 9 + 102 C 10 ) = 103 C 9 + 103 C 10 = 104 C 10
15
8.
15
C 2r
>
C 2r +1
1
2r + 1 1
6 r - 13
13
15
Þ
> Þ
<0 Þ
< r<
2
15 - 2 r 2
2 r - 15
6
2
9. f( x) = 1 + x 111 + x 222 +¼ + x 999
if f( x) is divided by x + 1, then remainder f( -1) = 0
if f( x) is divided by x - 1, then remainder f(1) = 10
f( x) = (1 + x 222 + x 444 + x 666 + x 888 ) + x 111 (1 + x 222 + x 444 + x 666 + x 888 )
= (1 + x 111 )(1 + x 222 + x 444 + x 666 + x 888 )
Exercise-3 : Matching Type Problems
n
2. (B) P =
å nCr =2n
r =0
m
Q=
å m C r (15) r = (1 + 15) m = 16 m
r =0
Þ
n = 4m
(C) 1 + 6 + 120 + 56 K
Reminder = 15
240
Solution of Advanced Problems in Mathematics for JEE
3. (A)
a 2 + b 2 + ab ( a - b)( a 2 + b 2 + ab) a 3 - b 3
=
=
a+b
( a - b)( a + b)
a 2 - b2
4+ 3
1
+¼ = (( 169 ) 3 - 13 ) = 1098
2
3 +1
5+ 3
7+ 5
8
8
(B) (2 cos 2 q - 3 sin q) = ( -2 sin 2 q - 3 sin q + 2)
5
5
3
Greatest value = 5 at sin q = (Q 4 £ q £ 6)
4
+
8 + 15
+
12 + 35
(C) Let ( 3 + 1) 6 = I + f
and ( 3 - 1) 6 = f ¢ Þ ( 3 + 1) 6 + ( 3 - 1) 6 = 416 = I + 1
I = 415 = 1 ´ 5 ´ 83
Þ
(D) (1 + x)(1 + x 2 ) ¼(1 + x 128 ) =
1 - x 256 1 - x n +1
=
1- x
1- x
n + 1 = 256
Þ
Exercise-4 : Subjective Type Problems
2. Coefficient of x 60 = -6 + 5 + 8 - 6 = 1
7. (1 + x) 3n = 3n C 0 + 3n C 1 x + 3n C 2 x 2 +¼¼ + 3n C 3n x 3n
Put x = 1
2 3n = 3n C 0 + 3n C 1 + 3n C 2 +¼¼ + 3n C 3n
Put x = w
( - w 2 ) 3n = 3n C 0 + 3n C 1 w + 3n C 2 w 2 +¼¼ + 3n C 3n
Put x = w 2
( - w) 3n = 3n C 0 + 3n C 1 w 2 + 3n C 2 w 4 +¼¼ + 3n C 3n
2 3n + ( - w 2 ) 3n + ( - w) 3n = 3 [ 3n C 0 + 3n C 3 +¼¼ + 3n C 3n ]
5
10.
å 20 C 2K -1 = 2 18 Þ 2 108 = 2 3 (2 5 ) 21 = 8 (33 - 1) 21
K =1
Remainder = -8 or 3
11. f(n) = n C 0 a n -1 - n C 1 a n -2 +¼¼
Þ
f ( n) =
( a - 1) n
a
f(2007) + f(2008) = 3 7 K
Þ
3 9 + ( a - 1) 3 9
= 37 K Þ K = 9
a
13. (360 + 1) 44 - 1 = 44 C 0 × (360) 44 + 44 C 1 × (360) 43 +¼ + 44 C 43 × (360) 1
= 360 [ 44 C 0 × (360) 43 + 44 C 1 × (360) 42 +¼ + 44 C 43 ]
241
Binomial Theorem
14. (3|x -2| + (3|x -2|-9 ) 1/ 5 ) 7
T 6 = 7 C 5 × (3|x -2| ) 2 × 3|x -2|-9 = 567
Þ
3 3|x -2|- 9 = 27 Þ | x - 2| = 4 Þ x = 6 , - 2
10
15. 1 +
10
å 3 r × 10C r + å r × 10C r
r =1
r =1
1 + ((1 + 3)
10
-
10
C 0 ) + 10 × 2 9 = 4 10 + 5 × 2 10 = 2 10 ( 4 5 + 5)
a = 1, b = 5
if a , b lies between the roots of f( x) = 0
f(1) < 0 Ç f(5) < 0
-k 2 < 0 Ç 16 - k 2 < 0
16. S n = n C 0 n C 1 + n C 1 n C 2 +¼ + n C n -1 × n C n = 2n C n -1
S n +1 = 2n + 2 C n
S n +1 2n + 2 C n 15
=
=
2n
Sn
4
C n -1
Þ
(2n + 2)(2n + 1) 15
=
n(n + 2)
4
Þ
n 2 - 6n + 8 = 0
❑❑❑
242
Solution of Advanced Problems in Mathematics for JEE
15
PROBABILITY
Exercise-1 : Single Choice Problems
2.
[where x = P( A)]
f( x) = 3 x + 4 1 - x
f( x) max. = 5 at x =
3.
9
25
P( A È B ) = 1 - P ( A È B ) =
P( A Ç B ) =
5
6
1
3
, P ( A) =
4
4
P ( B ) = P ( A È B ) + P ( A Ç B ) - P ( A) =
Þ
1
3
n
31
æ 1ö
4. 1 - ç ÷ =
Þ n =5
32
è2ø
5. Required probability =
6. Required probability =
3!´ 2
1
=
9!
140
3 !3 !3 !
3 !(3n - 3)!
[(n - 1)!]3
(3n)!
(n !) 3
7. If product of two numbers equal to third number, then possibilities are (2 , 3 , 6), (2 , 4 , 8),
(2 , 5, 10).
3
1
Probability =
=
10
C 3 40
8. P =
3 2 2 1 4
´ ´ ´ =
3 3 3 3 27
243
Probability
9. Total word = n =
7!
2 !2 !
T ITANIC
I TTANIC
A TTINIC
6! 6!
6!
m 5
+
+
Þ P= =
2 ! 2 ! 2 !2 !
n 7
n!
3
6
10. Probability =
=
=
n
32
64
n
Favourable word = m =
(n - 1)!
Þ
n
n -1
=
1×2 ×3
Þ n=4
43
11. Total case = n = 9 ´ 10 3
Favourable case = m = (9 ´ 10 3 ) - 6 4
P =1-
64
9 ´ 10
3
=
107
125
12. Total case = n = 6!
Favourable case = m = (3 ! ´ 2 !) + (2 ! ´ 2 !) = 16
16 1
Probability =
=
6 ! 45
13. E 1 “No card is king from removed cards”
E 2 “1 card is king from removed cards”
E 3 “2 card is king from removed cards”
E 4 “3 card is king from removed cards”
E 5 “4 card is king from removed cards”
F = 3 cards are drawn from pack those are kings.
S
P( F ) =
48
æ F ö
i =1
=
3
C 2 × 10 C 4
13
C
26
=
14.
4
C
å P( E i ) × P ççè E i ÷÷ø = 52 C 26 × 26 C3 +
C6
´
3
4
52
C 26 × 26 C 3
48
C 25 × 4 C 1
52
C 26
3
×
C3
26
( 48 C 26 + 48 C 25 ) =
C3
+0+0+0
4 ´ 49 C 26
52
C 26 × 26 C 3
1
(13)(17)(25)
1 15
=
7 286
15. Let f be function from {1, 2 , ¼ , 10} to itself total functions possible is 10 10 . The number of
one-one onto functions possible is 10!.
Hence, the probability of selected function to be one-one onto is
10 !
10
10
=
9!
10 9
.
244
Solution of Advanced Problems in Mathematics for JEE
16. Let the friends come to the restaurant at 5 h x min and 5h y min, respectively, where
x , y Î [0 , 60].
Hence, the sample space consists of all points ( x , y) lying in 60 ´ 60 square as shown above
and for favourable cases,| x - y| £ 15, that is -15 £ x - y £ 15 which is shown by shaded region
in the graph shown below :
y
45
60
45
15
0
15
60
x
Hence, the probability that they will meet is given by :
1
2 ´ ´ 45 ´ 45
2
7
æ3ö
2
1=1-ç ÷ =
60 ´ 60
4
16
è ø
17. Total ways = 91 C 3
Favourable ways = (Common ratio is 2) + (Common ratio is 3) = 16 + 2 = 18
Exercise-2 : One or More than One Answer is/are Correct
4
1. Probability =
C 3 ´ 11 C 5
15
C8
´
1
8
=
7 195
1öæ
1öæ
1öæ
1ö
1 ö
æ
æ
2. Probability = ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ¼¼ ç 1 ÷
2 øè
4 øè
6 øè
8ø
2012 ø
è
è
1 3 5 7
2011
2012 !
= ´ ´ ´ ´¼¼´
=
2 4 6 8
2012 2 2012 (1006 !) 2
2 1
= or i = 1, 2 , 3.
4 2
1
Also for i ¹ j , P( E i Ç E j ) = = P( E j ) P( E i ). Therefore, E i and E j are independent for i ¹ j .
4
1
Also,
P( E 1 Ç E 2 Ç E 3 ) = ¹ P( E 1 ) P( E 2 ) P( E 3 )
4
3. We have P( E i ) =
E 1 , E 2 , E 3 are not independent.
3
4. Max. ( P( A Ç B )) = P( A) =
5
\
245
Probability
Min. ( P( A Ç B )) = P( A) + P( B ) - 1 =
4
15
P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B ) =
19
- P( A Ç B )
15
3
- P( A Ç B )
5
æ A ö P( A Ç B ) P( B ) - P( A Ç B )
Pç ÷=
=
P( B )
P( B )
èBø
P ( A Ç B ) = P ( A) - P ( A Ç B ) =
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1
2 1 3 1 4 1 4 2
´1+
´ +
´ +
´ =
=
10
10 2 10 3 10 4 10 5
3 1
´
æ B 3 ö P( B 3 Ç E 2 )
1
10
3
÷÷ =
2. P çç
=
=
2
1
3
1
4
1
E
P
(
E
)
3
2
è 2ø
´ +
´ +
´
10 2 10 3 10 4
1. P( E 1 ) =
Paragraph for Question Nos. 3 to 5
3. Mr. A’ s 3 digit number is always greater than Mr. B’s 3 digit numbers then A should always pick
digit 9.
8
Probability =
8
4. Probability =
9
C3 ´1
8
=
C3 ´ C3
1
9
C3
=
8
C 3 ´ 8C 2
9
C3 ´ C3
=
1
3
1
84
5. P( E ) = A picks 9 or A does not pick 9 and his number is greater than B
8
C3
1 2 1æ
1 ö÷ 37
= + × ç1 ×
=
8
8
ç
3 3 2è
C3
C 3 ÷ø 56
Paragraph for Question Nos. 6 to 7
6. Let a n = number of ways of outcomes of n tosses when no 2 consecutive heads occur
Also,
\
a n = a n -2 + a n -1
a1 = 2
a2 = 3
a 3 = 5, a 4 = 8
a10 = 144
(H or T)
(TT or HT or TH)
……
246
Solution of Advanced Problems in Mathematics for JEE
\ Probability =
144
2 10
7. [HT HT HTH] T, T, T
Number of ways of arranging =
Probability =
4!
=4
3!
4
2 10
Paragraph for Question Nos. 8 to 10
8. 6n > 2 n , n Î N
\ n = 1, 2 , 3 , 4
4 æ Number of solutions of x + y > 4 , 1 £ x , y £ 6 ö
9.
´ç
÷
6 è
36
ø
æ Number of solutions of x + y + z > 8 , 1 £ x , y , z £ 6 ö
´ç
÷
63
è
ø
4 30 160 100
= ´
´
=
6 36 216 243
4 30 æ
160 ö 4 30 56
35
10. Probability = ´
´ ç1 ´
=
÷= ´
6 36 è
216 ø 6 36 216 243
Paragraph for Question Nos. 11 to 12
11. Let p1 be the probability of being an answer correct from section 1. Then p1 = 1 5 . Let p 2 be
the probability of being an answer correct from section 2. Then p 2 = 1 15 .
1 1
1
Hence, the required probability is ´
=
5 15 75
12. Scoring 10 marks from four questions can be done in 3 + 3 + 3 + 1 = 10 ways so as to answer 3
questions from section 2 and 1 question from section 1 correctly.
10
Hence, the required probability is
3
C 3 10 C 1 1 æ 1 ö
ç ÷ .
20
5 è 15 ø
C4
Exercise-5 : Subjective Type Problems
æ 2 1 2 1 1 2 ö æ 2 1 2 ö 2 416
1. ç + × + × × ÷ ç + × ÷ =
è 3 3 3 3 3 3 ø è 3 3 3 ø 3 729
6
5. Probability =
C 5 + 7 C 4 + 8 C 3 + 9 C 2 + 10 C 1 + 1
2
10
=
9
64
247
Probability
3
6. p =
7
C1
C1
=
3
7
7. Total ways =
6!
´ 3 ! = 90
2 !2 !2 !3 !
Favourable cases = 90 - [3 ! + 3 C 1 ´ 3 C 1 ´ 2 ´ 2] = 48
Þ
p=
48 8
=
90 15
9. E 1 ® be the event of both getting the correct answer
E 2 ® both getting wrong answers.
E ® both obtaining same answer.
1 1
1
1öæ
1 ö 77
æ
P( E 1 ) =
=
, P( E 2 ) = ç 1 - ÷ ç 1 ÷=
8 12 96
8
12
è
øè
ø 96
æ E ö
æ E ö
1
÷÷ = 1; P çç
÷÷ =
P çç
E
E
1001
è 1ø
è 2ø
1
1×
E
13
æ
ö
96
Pç 1 ÷=
=
1
1
77
E
14
è
ø 1×
+
×
96 1001 96
10. Total ways = 9 C 7 ´ 7 !
Favourable ways Þ 9 C 7 ´ 7 ! - ( 7 C 3 ´ 3 !) ´ ( 6 C 4 ´ 4 !)
P( E ) = 1 -
( 7 C 3 ´ 3 !) ´ ( 6 C 4 ´ 4 !)
9
C7 ´ 7!
=1-
15 7
=
36 12
3
1 ì1 1
1 1 ü 1 ìïæ 1 1 1 ö üï 27
í ´ + 2 ´ ´ ý + ´ íç ´ ´ ÷ ý =
2 î2 2
4 4 þ 4 ïè 4 4 2 ø ï 128
î
þ
1
1
12. 1st 2 nd
´
a
4 6
1 1
2 nd 1st
´
b
4 6
1 1
1st 1st
c
´
4 36
1 1
2 nd 2 nd
´
d
4 36
c+d
2
=
a+ b+ c+ d 5
❑❑❑
11.
248
Solution of Advanced Problems in Mathematics for JEE
16
LOGARITHMS
Exercise-1 : Single Choice Problems
1. log 10 x = A
x>0
log 10 ( x - 2) = B, x - 2 > 0 Þ x > 2
Þ
A 2 - 3 AB + 2 B 2 < 0
Þ
Þ
Case-I :
and
Case-II :
( A - 2 B )( A - B ) < 0
(log x - 2 log( x - 2))(log x - log( x - 2)) < 0
log x - 2 log ( x - 2) < 0
…(1)
log x - log( x - 2) > 0
log x - 2 log ( x - 2) > 0
and
log x - log( x - 2) < 0
From (1) & (2),
x Î ( 4 , ¥)
…(2)
(log e x) 2 = (log e x) - (log e x) 2 + 1
2.
2
2(log e x) - log e x - 1 = 0
(2 log e x + 1)(log e x - 1) = 0
log e x = -
3.
S = (a
1
(not possible)
2
log e x = 1
log 3 7 log 3 7
log 7 11 log 7 11
)
+ (b
)
= 27 log 3 7 + 49 log 7 11 + 11
+ ( c log11 25 ) log11 25
log11 25
= 469
0
4.
1ö
æ
a 2 - 3 a + 3 > ç x + ÷ and a 2 - 3 a + 3 > 0
xø
è
a 2 - 3a + 2 > 0
( a - 1)( a - 2) > 0 Þ a Î ( -¥, 1) È (2 , ¥)
5
5. P =
= log 120 x 5 ; (120) P = x 5 = 32 Þ x = 2
log x 120
(log e x > 0)
249
Logarithms
6.
x=
z 1/ 3
z 1/ 6
, y=
2
5
z 1/ 3 z 1/ 6
1
×
= z 3/ 2 Þ z =
2
5
10
If xy = z 3/ 2 ;
7. log x (log 3 (log x y)) = 0 Þ y = x 3 , log y 27 = 1 Þ y = 27
-3
5
+4=
2
2
4 log 3 2
3
3-a
9. a =
Þ log 3 2 =
; log 6 16 =
1 + log 3 2
1 + 2 log 3 2
2a
8. log 10-2 10 3 + log 10-1 10 -4 =
10. log 2 (log 2 (log 3 x)) = 0 Þ x = 9
log 2 (log 3 (log 2 y)) = 0 Þ y = 8
x y 7
x y 2
11. Let log 3 a = x , log 3 b = y; + = and + =
3 2 2
2 3 3
12. a = log 2 5; b = log 5 8; c = log 8 11; d = log 11 14
2 abcd = 2 log 2 14 = 14
14.
log 8 17 log 2 2 17
=
log 9 23
log 3 23
15. Case-I : 2 x - 3 > 1
Case-II : 0 < 2 x - 3 < 1
3x - 4 > 1
5
x> Þ x>2
3
16. p £ log 10 N < p + 1
0 < 3x - 4 < 1
5
3
5
x< Þ < x<
3
2
3
Þ P = 10 p + 1 - 10 p
- q £ log 10 1 N < - q + 1 Þ Q = 10 q - 10 q - 1
17. n + 1 = number of digits = 1 + characteristic
18. log 10 (0.15) 20 = 20 (log 10 15 - 2) = -16.478
19. log 2 (log 4 (log 10 10 16 )) = log 2 (log 4 16) = 1
20. 2 log x - log(2 x - 75) = 2
x2
= 100 Þ x 2 - 200 x + 7500 = 0
2 x - 75
21. x
log x a×log a y ×log y z
= x log x z = z
22. x x x = x 3 x / 2
x ¹ 0, 1
If
x x=
3
9
x Þ x=
2
4
x = 1, then it also satisfy.
250
Solution of Advanced Problems in Mathematics for JEE
(log 3 x) 2 = 2 log 3 x
23.
or
log 3 x = 0
log 3 x = 2
or
x =1
x =9
log 10 x + log 10 y = 2 Þ xy = 100
x - y = 15
Þ
24.
x = 20,
Þ
…(1)
…(2)
y =5
1/ 2
25.
Þ
7
æ x + 1 æç 2 x - 3 ö÷ ö
ç
3è
xø÷
3
2
=
2
ç
÷
ç
÷
è
ø
2
1 14
x+ x- =
3
x 3
Þ 5 x 2 - 14 x - 3 = 0
2
26.
25
( 2 x - x 2 + 1)
+9
( 2 x - x 2 + 1)
2
2
3 2x - x +1 52x - x +1
= 34
×
3
5
2
Let 3 2 x - x + 1 = a and 5 2 x - x + 1 = b
34
a 2 + b2 =
ab
15
15a 2 - 34 ab + 15b 2 = 0 Þ (3 a - 5b)(5a - 3 b) = 0
a 5
Case-1 : if =
b 3
2 x - x 2+ 1
Þ
æ3ö
ç ÷
è 5ø
Þ
2x - x 2 + 1 = -1 Þ x 2 - 2x - 2 = 0
=
5
3
Sum of two values of x = 2
a 3
Case-2 : if =
b 5
æ3ö
ç ÷
è 5ø
2 x - x 2+ 1
=
3
5
Þ
2 x - x 2 + 1 = 1 Þ x = 0 and 2
Sum of all values of x is 4.
a x = by = cz = dw
27.
Þ
b = a x/ y , c = a x/z , d = a x/w
log a ( bcd) = log a
æx x xö
çç + + ÷÷
æ1 1 1ö
x x x
y z wø
aè
= + + = xç + + ÷
y
z
w
çy
è
z
w ÷ø
251
Logarithms
28. x =
4
4
( 5 + 1)( 5 + 1)( 8 5 + 1)( 16 5 + 1)
Multiply and divide by (1 - 16 5) then
x = -1 + 16 5
( x + 1) 48 = 5 3 = 125
1
log x log 18 ( 2 + 8 ) =
3
1
log x log ( 3 2 )2 3 2 =
3
1
æ 1ö 1
log x ç ÷ =
Þ x=
8
è2ø 3
29.
30.
f ( n) =
1
log 2 n
3
if log 8 n is integer
otherwise
=0
2011
å f(n) = log 8 2 3 + log 8 2 6 + log 8 2 9 = 1 + 2 + 3 = 6
n =1
log 0. 3 ( x - 1) < log ( 0. 3)2 ( x - 1) Þ ( x - 1) 2 > x - 1
32.
Þ ( x - 1)( x - 2) > 0
Also, for log to be defined ( x - 1) > 0
x Î (2 , ¥)
2
7 2 x - 5 x - 6 = ( 49) 3 log 2 2 = 7 3
33.
Þ
2 x 2 - 5x - 6 = 6
2 x 2 - 5 x - 12 = 0
Þ
(2 x + 3)( x - 4) = 0
34. (log 2 x) 4 + 16(log 2 x) 2 log 2
Þ
t 4 + 16t 2 ( 4 - t )
Þ
t 2 (t 2 + 64 - 16t )
Þ
t 2 ( t - 8) 2
16
x
(where log 2 x = t )
Since 1 £ x £ 256 Þ 0 £ t £ 8
Þ
Maximum of (t - 8) 2 t 2 lies at t = 4.
Hence, maximum ( 4 - 8) 2 × 4 2 = 256
252
Solution of Advanced Problems in Mathematics for JEE
37. Q
l>0
1 ± (1 - 4 log 16 l)
\
log 16 x =
Q
The given equation will have exactly one solution, if
1
1 - 4 log 16 l = 0 or log 16 l = = 4 -1
4
\
l = (16) 4
2
-1
= (2 4 ) 1/ 4 = 2 , - 2 , 2i , - 2i , where i = -1
But l is real and positive.
\ l =2
Number of real values = 1
38. Let x be the rational number, then according to question,
x = 50 ´ log 10 x
By trial x = 100
39. Q
x = log 5 (1000) = log 5 (5 3 ´ 8) = 3 + log 5 8
and y = log 7 (2058) = log 7 (7 3 ´ 6) = 3 + log 7 6
x - y = log 5 8 - log 7 6 > 0
Þ
æQ log 5 8 > 1, log 7 6 < 1ö
ç \ log 8 - log 6 > 0 ÷
è
5
7
ø
x> y
2 ö
4 ö
æ 2 ö
æ
æ
÷ + 5 log ç 5
÷ + 3 log ç 3
÷
40. 7 log ç
ç5´3÷
ç 23 ´3 ÷
ç 24 ´5÷
è
ø
è
ø
è
ø
\
4
= 7 {4 log 2 - log 5 - log 3} + 5 {2 log 5 - 3 log 2 - log 4} + 3 {4 log 3 - 4 log 2 - log 5}
= log 2
41. log 10 {tan 1° tan 2° tan 3°¼ tan 45°¼ tan 87° tan 88° tan 89° }
= log 10 {tan 1° tan 2° tan 3°¼ tan 45°¼ cot 3° cot 2° cot 1° }
= log 10 1 = 0
1 1 1
+ +
æ 1 1 1ö
æ7ö
42. log 7 log 7 7 2 4 8 = log 7 ç + + ÷ = log 7 ç ÷
è2 4 8ø
è8ø
= 1 - log 7 8 = 1 - 3 log 7 2
43. ( 4)
3
2
+ (9) log 2 2 = (10) log x 83
Þ
( 4) 1/ 2 + 9 2 = (10) log x 83
Þ
(83) 1 = (83) log x 10
1 = log x 10 Þ x = 10
æxö
æyö
æzö
log10 çç ÷÷
log10 ç ÷
log10 ç ÷
log10 x
log10 y
log10 z
zø
xø
è
è
èyø
(10
)
(10
)
(10
)
\
44.
log 2 3
253
Logarithms
45. Q
log x 2 log 2 x 2 = log 4 x 2
\ x > 0 , 2 x > 0 and 4 x > 0 and x ¹ 1, 2 x ¹ 1, 4 x ¹ 1
1 1
Þ x > 0 and x ¹ 1, ,
2 4
1
1
1
Then,
×
=
log 2 x log 2 2 x log 2 4 x
Þ
log 2 x × log 2 2 x = log 2 4 x
Þ
log 2 x × (1 + log 2 x) = (2 + log 2 x)
Þ
(log 2 x) 2 = 2
Þ
log 2 x = ± 2
\
x =2± 2
\
x = {2 - 2 , 2 2 }
46. Q 2 log 10 x - log x (0.01) = 2 log 10 x - log x (10 -2 )
= 2(log 10 x + log x 10)
æ log e x log e 10 ö
÷÷ ³ 2 × 2
= 2 çç
+
è log e 10 log e x ø
=4
47. Let
log 2 x = a
a 2 - 2a + 1 Þ a = 1
if
log 2 x = 1 Þ x = 2
48. log e ( e 2 x ln x ) = log e x 3
2 + (ln x) 2 = 3 ln x
Let ln x = a
a 2 - 3a + 2 = 0
Þ ( a - 2)( a - 1) = 0
Þ x1 = e 2 , x 2 = e
49. M = antilog 32 0.6 = (32) 0.6 = 2 3 = 8
N = 49 1 × 49 - log 7 2 + 5 - log 5 4
49 1 25
=
+ =
4
4 2
50. log 2 (log 2 (log 3 x)) = 0
Þ x =9
log 3 (log 3 (log 2 y)) = 0 Þ y = 8
51. |log 1/ 2 10 + |log 4 625 - log 2 5| =|log 1/ 2 10 + log 2 5| = 1
(Q x > 0 and x ¹ 1)
(Q AM ³ GM)
254
Solution of Advanced Problems in Mathematics for JEE
æ 1 ö
ç
÷
log 5 2 è 2 a ø
1
52. log 3 2 =
=
=
1 2 ab - 1
log 5 3
b2a
55. ( x - 3) 2 = 9 Þ x = 6
éæ 16 ö 7 æ 25 ö 5 æ 81 ö 3 ù
57. log a êç ÷ × ç ÷ × ç ÷ ú = 8
êëè 15 ø è 24 ø è 80 ø úû
log a 2 = 8
Þ
Þ a = 2 1/ 8
58. log 23 (2 7 ) - log 32 (3 -1/ 2 ) =
7 1 31
+ =
3 4 12
æ log 5 16 ö
÷
-ç
2
æ 1 ö æ 1 ö çè 2 log 5 9 ÷ø æ 1 ö æ 1 ö
÷÷ × çç
÷÷
÷÷
59. çç
= ç ÷ çç
è 27 ø è 27 ø
è 27 ø è 27 ø
- log 3 2
1
æ 1 ö - log 3 27 2 2
= ç ÷ ×2
=
27
è 27 ø
( x - 1)( x + 2)
( x - 1)( x + 2)
60. log 2
= log 2 4
Þ
=4
3x - 1
3x - 1
Þ
61. log 100 10 =
x 2 - 11x + 2 = 0
1
2
log 2 (log 4 2) = log 2 1 2 = -1
log 4 [log 2 (256) 2 ]2 = log 4 16 2 = 4
log 4 8 = log 22 2 3 =
3
2
62. l = log 5 (log 5 3) Þ 5 l = log 5 3
3 k+ 5
-l
= 3 k × 3 5-l = 3 k × 3 log 3 5 = 5 × 3 k
63. log 10 b 4 = 2 p × log 10 a 2
log 10 b
= log a b = p
log 10 a
64. 2 x = 3 y = 6 -z = k (let)
x = log 2 k , y = log 3 k , z = -log 6 k
1 1 1
+ + = log k 2 + log k 3 - log k 6 = 0
x y z
65. ( 2 - 1) 3 = 5 2 - 7
255
Logarithms
1 1
- log a b
17
1
3
3
2
66. 1 + log a b = Þ log a b = - Þ
=
4
4
1 + log a b
6
68. Let log y x = t
5t 2 - 26t + 5 = 0 Þ (5t - 1)(t - 5) = 0
Either x = y 5 or y = x 5
69. 1 -
1
1
=
Þ (log 3 x - 1) 2 = log 3 x Þ (log 3 x) 2 - 3 log 3 x + 1 = 0
log 3 x log 3 x - 1
1
1
x y z =4
log 2 y + log 2 z = 2 Þ
2
2
1
1
log 3 y + log 3 x + log 3 z = 2 Þ x × y × z = 9
2
2
1
1
log 4 z + log 4 x + log 4 y = 2 Þ x × y × z = 16 Þ xyz = 24
2
2
æ 1 ö log 71 / 49
- log1 / 55
71. ç
+7
÷ ×2
è 49 ø
1 1
´ +7
49 4
1
1
72. log 2 (3 - x) - log 2
+ log 2 (5 - x) = + log 2 ( x + 7)
2
2
70. log 2 x +
Þ
log 2 (3 - x)(5 - x) = log 2 ( x + 7)
x2 - 9x + 8 = 0 Þ x = 8
1
73. log 5 x = log x 5 Þ x = 5,
5
Þ
2
74. | x - 1|log 3 x -2 log x 9 = ( x - 7) 7
either x = 2 or
log 3 x 2 - 2 log x 9 = 7
(log 3 x - 4)(2 log 3 x + 1) = 0
75. 9
x -1
+ 7 = 4 (3
Let 3
x
x -1
+ 1)
=t
t2
æt
ö
+ 7 = 4 ç + 1÷
9
3
è
ø
Þ t 2 - 12t + 27 = 0
(t - 3)(t - 9) = 0
76. If a > 1
Þ
log a 10 > log a 3 > log a e > log a 2
log 10 a < log 3 a < log e a < log 2 a
256
Solution of Advanced Problems in Mathematics for JEE
4
78.
å
log 4 2 r =
r =1
4
r
å2 =5
r =1
79. log 3 2 + log 3 5 = log 3 10
log 3 9 < log 3 10 < log 3 27
80. sin 3q = 3 sin q - 4 sin 3 q
kæ 3
1 ö 3æ
1ö 4 æ
1ö
ça + 3 ÷ = ça + ÷ - ça + ÷
2è
2
a
8
a
è
ø
è
ø
a ø
=
3
-1 æ 3
1 ö
ça + 3 ÷
2 è
a ø
k = -1
Þ
81. 2 x - 3 > 0 Ç x 2 - 5 x - 6 > 0 Ç 2 x - 3 ¹ 1
3
x > Ç ( x - 6)( x + 1) > 0 Ç x ¹ 2
2
Þ (6 , ¥)
Exercise-2 : One or More than One Answer is/are Correct
6 (log x) 2 + log x - 1 = 0
1.
(3 log x - 1)(2 log x + 1) = 0
x = 10 1/ 3
or
x = 10 -1/ 2
3. 3(log 10 2) x 2 - (1 - log 10 2) x = 2 log 10 2 - x
log 10 2(3 x 2 + x - 2) = 0
log 10 2( x + 1)(3 x - 2) = 0
2
Roots of this eq. are x = -1,
3
Sum of coeff. = 2 log 10 2 (irrational)
Discriminant = b 2 - 4 ac = 25 (log 10 2) 2 (irrational)
4. A = min.( x 2 - 2 x + 7) " x Î R
2
B = min.( x - 2 x + 7) " x Î [2 , ¥)
Þ A =6
Þ B =7
257
Logarithms
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
1. If
a 1 = 4 , then 3 4 £ N < 3 5
If
a 2 = 2 , then 5 2 £ N < 5 3
Þ
81 £ N < 125
3. If
a 1 = 5, then 3 5 £ N < 3 6
If
a 2 = 3 , then 5 3 £ N < 5 4
If
a 3 = 2 , then 7 2 £ N < 7 3
Þ
243 £ N < 343
Paragraph for Question Nos. 4 to 5
2
2
Sol. | x - y | = 221
Paragraph for Question Nos. 6 to 7
Sol. (1 + 4 log p 2 (2 p)) 2 + (1 + log 2 p) 2 = (1 + log 2 4 p) 2
Let log 2 p = t
2
æ
æ 1 + t öö
2
2
ç1 + 2ç
÷ ÷ + (1 + t ) = (3 + t ) Þ t = log 2 p = 2
t
è
øø
è
Exercise-4 : Matching Type Problems
1. (A)
a = 3 (( 7 + 1) - ( 7 - 1)) = 6
b = 1296 = 36
(B)
a = ( 3 + 1) - ( 3 - 1) = 2
b = (3 + 2 ) - (3 - 2 ) = 2 2
(C)
a = ( 2 + 1), b = ( 2 - 1)
(D)
a =2 + 3,
b=2 - 3
Exercise-5 : Subjective Type Problems
1. N = 6 log10 40 × 6 2 log10 5 = 6 log10 1000 = 6 3 = 216
2.
log b ( a log 2 b ) = log a ( b log 2 b )
258
Solution of Advanced Problems in Mathematics for JEE
Þ
log b a = log a b
log a ( c - ( b - a) 2 ) = 3
Þ
a = b or a =
Þ
c = a3
1
(not possible)
b
Þ Minimum value of c = 8 at a = 2
3. log b 729 = 6 log b 3
if this is an integer, then b = 3 , 3 2 , 3 3 , 3 6
4. Case-1 : If
5
3
>1 Þ x>2
2
æ8
ö
Þ 3 x 2 - 2 x - 16 > 0 Þ x Î ç , ¥ ÷
è3
ø
5
5
3
0< x + <1 Þ - < x<2
2
2
3ö
æ
( x - 5) 2 > (2 x - 3) 2 Þ x Î ç - 2 , - ÷
2ø
è
( x - 5) 2 < (2 x - 3) 2
then
Case-2 : If
then
x+
there is no negative integral value of x.
1ö
æ
ç log100 x + ÷
6 (log a x )(log10 a )(log a 5)
2ø
5.
a
- 3 (log10 x -1) = 9 è
5
6 × 5(log10 x -1) - 3 (log10 x -1) = 3 (log10 x +1)
3 log10 x
+ 3 × 3 log10 x
3
10 log10 x
(log10 x -1)
6 ×5
=
×3
3
6 × 5(log10 x -1) =
æ 5ö
ç ÷
è3ø
log10 x - 2
=1
Þ
log 10 x - 2 = 0
Þ
x = 100
Integer part of log 3 100 is 4.
æ a + b ö log 5 a + log 5 b
6.
log 5 ç
÷=
2
è 3 ø
2
Þ
æ a + bö
log 5 ç
÷ = log 5 ( ab)
è 3 ø
Þ
( a + b) 2 = 9 ab Þ
a 4 + b 4 + 2 a 2 b 2 = 49 a 2 b 2
Þ
a 4 + b4
a 2 b2
= 47
a 2 - 7 ab + b 2 = 0
259
Logarithms
8. log 10 1 + x + 3 log 10 1 - x = 2 + log 10 1 - x + log 10 1 + x
log 10 1 - x = 1
Þ
1 - x = 10 Þ
x = -99 (not possible)
x 2 = 1 + 6 log 4 y
9.
y 2 - 2 x y - 2 2x +1 = 0
y = 2 x +1 and y = -2 x
Þ
if
y = -2 x
if
y = 2 x +1
Þ
(not possible, because y > 0)
log 2 y = x + 1
x 2 = 1 + 3 log 2 y
Þ
x 2 = 1 + 3 ( x + 1)
x2 - 3x - 4 = 0
Þ
( x - 4)( x + 1) = 0
x1 = 4
Þ
y 1 = 2 5 = 32
x2 = -1
Þ
y 2 = 2° = 1
log 2| x 1 x 2 y 1 y 2| = log 2 128 = 7
10. log 7 log 7 7 7 7 = log 7 log 7 (7 7 / 8 ) = log 7 (7 8) = 1 - 3 log 7 2
Þ
a =3
log 15 log 15 15 15 15 15 = log 15 log 15 (1515/16 )
æ 15 ö
= log 15 ç ÷ = 1 - 4 log 15 2
è 16 ø
Þ
Then
11.
b=4
a + b=7
log 1+ x (1 - y) + log 1- y (1 + x) = 2
1+ x =1- y
x = -y
\
log 1- y (1 + 2 y) + log 1- y (1 - 2 y) = 2
1
æ
ö
ç t + = 2 Þ t = 1÷
t
è
ø
260
Solution of Advanced Problems in Mathematics for JEE
log 1- y (1 - 4 y 2 ) = 2
1- 4y2 =1+ y2 -2y
5y2 - 2 y = 0
y = 0, y =
2
5
But y = 0 rejected.
12.
log b n = 2
log n (2 b) = log n 2 + log n b = 2
1
log n 2 + = 2
2
3
log n 2 =
Þ n = 2 2/ 3
2
if
log b n = 2
Þ b = n 1/ 2 = 2 1/ 3
n × b = 2 2/ 3 × 2 1/ 3 = 2
13.
log y x +
1
=2
log y x
log y x = 1 Þ
Þ
x=y
2
x + y = 12
Þ
x 2 + x - 12 = 0
Þ
( x + 4)( x - 3) = 0
x = - 4 or x = 3
Þ
but x > 0,
then x = 3
xy = 9
yx = xy
14.
if
x =2y
Þ
y 2 y = (2 y) y
2 y log y = y log(2 y)
Þ
if
then
y ¹0
then log y 2 = log(2 y)
y2 =2y Þ
y =2
x 2 + y 2 = 5 y 2 = 20
15. (log 2 4 + log 2 ( 4 x + 1)) log 2 ( 4 x + 1) = 3
Let
log 2 ( 4 x + 1) = t
261
Logarithms
t 2 + 2t - 3 = 0 Þ
t = -3 or 1
log 2 ( 4 x + 1) = 1 Þ 4 x = 1 Þ x = 0
17. x 2 + 4 x + 3 = 0 ( x > 0)
18. log 31 / 4 (log 3 5 x) = 4
Þ
log 5 x = 1
(log 3 5 x )
Þ log 3
Þ
x =5
❑❑❑
=1
262
Solution of Advanced Problems in Mathematics for JEE
17
STRAIGHT LINES
Exercise-1 : Single Choice Problems
1. Let ratio be l : 1 Þ
6l - 3
1
=0, l =
l+1
2
3.
y
C(2 +2p, p+2)
(0
–2,2
3 3
,1)
x
(p, 0)
p p
B 2 ,– 3
O
if ( a , sin a) lie inside the triangle, then a Î(0 , p)
711
9 ´ 79
4.
x=
=
13 + 11m 13 + 11m
if x is an integer, then m = 6
2
6.
æ yö
æ yö
7 ç ÷ + 2c ç ÷ - 1 = 0
x
è ø
è xø
m 1 + m 2 = 4m 1 m 2
Þ c =2
10.
Y
(0,8)
(0,3)
(8,0)
(–4,0)
O
X
263
Straight Lines
1 2
a = 72
2
13.
(a, 2a)
a = ± 12
(a, a)
Centroid º (16 , 16) or ( -16 , - 16)
14.
g( x) = ax + b
g(1) = 2
a + b=2
Þ
g(3) = 0
2a = - 2
Þ
a = -1
b=3
g( x) = - x + 3
-1
cot [cos (|sin x| + |cos x|) - sin -1 (|sin x|) + |cos x|]
|sin x| + |cos x|Î [1, 2 ]
Þ
cot [cos -1 1 - sin -1 1] = 0 = g(3)
15. Points A and B are mirror images about y = x.
Point P will lie on the ^ bisector of line joining A and B Þ P lie on y = x.
m1
m2
16. 4m 3 - 3 am 2 - 8 a 2 m + 8 = 0
m3
m 1 m 2 m 3 = -2
m3 =2
Þ
2x
17.
7x+y=8
(1,1)
–5
(Qm 1 m 2 = - 1)
y=
6
x–7y+6=0
æ y - mx ö
18. 2 x 2 + 3 y 2 - 5 x ç
÷ =0
è C ø
Coefficient of x 2 + coefficient of y 2 = 0
5+
5m
=0 Þ m + C =0
C
Then the equation of family of line is y = m( x - 1)
y=mx+C
(2a, a)
264
Solution of Advanced Problems in Mathematics for JEE
20. Equation of line BC is y = 7 x - 10
Equation of line CD is y = x + 2
(2 - 0)(10 - 0) 10
Area of rhombus =
=
(7 - 1)
3
D
C
(2, 4)
y = 7x
A
(0, 0)
(1,2)
B
y=x
A(3,3)
y=x
C(–6,0)
21.
y = –x
3
3
B
(–3,–3)
3
22. y = ( x - 9) + 6
4
23. Acute angle bisector is
7x - y
æ x - yö
÷÷
= - çç
50
è 2 ø
7x – y = 0
y = 2x
3 x + 4 y - 12
3 x + 4 y - 12
24. Either x = y or x =
or y =
Þ (1, 1)
5
5
Þ
æ 1 10 ö
25. Co-ordinate of point A ç - ,
÷
è 7 7 ø
1 1 1 1
Ar ( DABC ) = ´ ´ =
2 2 7 28
x–y=0
y
B
A
C
2)
(0,
(0,3/2)
O
26.
A(1,2)
G(4,1)
B
C
(x1,y1)
(4,b)
x+y=5
x=4
x +4+1
Co-ordinate of centroid G( 4 , 1) Þ 1
=4
3
Þ
x 1 = 7 and y 1 = - 2
x
265
Straight Lines
y=
27.
|x+
1
|x–
y=
|
O
–1
1
1|
2
The image of y =| x - 1| w.r.t. y-axis is y =| x + 1| Þ y = ± ( x + 1)
Required solution = ( y - ( x + 1))( y + ( x + 1)) = 0
Q(4,5)
P(1,4)
R(m,m)
28.
P¢(4,1)
X
O
Image of (1, 4) about the line y = x is ( 4 , 1) Þ P ¢( 4 , 1) Q ( 4 , 5) and R(m , m ) are collinear.
Þ
m =4
AD AB 10 2
29.
=
=
=
CD BC
5 1
B(5, 1)
A(–1, –7)
D
C(1, 4)
2
3
æ yö
æ yö
30. 4 c ç ÷ - ç ÷ + 6 = 0 has one root is - Þ c = -3
x
x
4
è ø
è ø
x y( a + c) 1
33.
+
+ =0
a
2 ac
c
Þ
a( y + 2) + c(2 x + y) = 0
Passes through a fixed point (1, - 2)
2
1æ y ö
2æ yö 1
ç ÷ + ç ÷ + =0
bè x ø
hè x ø a
34.
Þ
3m = -
m
2m
2b
b
ab 9
and 2m 2 =
Þ
=
h
a
h2 8
266
Solution of Advanced Problems in Mathematics for JEE
y
x
+
=1
2h 2k
if it passes through fixed point ( x 1 , y 1 )
x1
y
+ 1 =1
2h 2k
4
36. OA : OB = l : 1 Þ l =
3
35. Equation of line is
y
(0,2k)
P(h,k)
0
B(9/5, 9/10)
(2h,0)
x
4x + 2y – 9 = 0
1
O (0, 0)
l
2x + y + 6 = 0
A(–12/5, –6/5)
x = 2y
æ 7ö
37. G ç 1, ÷
è 3ø
A(1,1)
(–1,2)
(–3,3)B
(3,2)
C(5,3)
38. Diagonals are perpendicular.
39. Let point on the line x + y = 4 is ( a , 4 - a).
4( a) + 3 ( 4 - a) - 10
= 1 Þ a 2 + 4 a - 21 = 0
5
Þ
a1
a2
a1 + a 2 = - 4 Þ b1 + b2 = 12
40. Equation of altitude on BC
x + 4 y = 13
Equation of altitude on AB
Þ
7 x - 7 y + 19 = 0
æ 3 22 ö
Hç ,
÷
è7 7 ø
41. Equation of line is (3 x + 4 y + 5) + l ( 4 x + 6 y - 6) = 0
- ( 3 + 4l ) 7
1
Þ
´ = -1
Þ
l=
4 + 6l
5
2
5 -1 7 -5
42.
Þ x = 11
=
8 -2 x -8
A(–3,4)
x+y=1
B
2x+3y=6
4x–y+4= 0
C(–3/7, 16/7)
267
Straight Lines
S
R(7,5)
43.
(3,2)
P
(–1,–1)
Q(8,0)
S( -2 , 4)
a
1
1 a
44. Area = a + 1 a + 1 1 = 1
2 a+2
a
1
Þ
45.
( x - y) 2 = 1
Þ x - y = 1 and x - y + 1 = 0
46. AB subtend an acute angle at point C , then
a 2 + ( a + 1) 2 > 4
æ
- 7 - 1 ö÷ æç 7 - 1 ö÷
a Î çç -¥,
÷ È ç 2 , ¥÷
2
è
ø è
ø
Þ
48.
h = cos q
y
A
k = 2 sin q
h2 +
Þ
2
k
=1
4
1
q
P(h, k)
2
4x2 + y2 = 4
q
O(0,0)
50. Let the point of reflection is ( h, k).
B
x
h - a k - 0 -2( a + at 2 )
=
=
Þ x = -a
1
-t
1+ t2
51.
y
R(6, 7)
P
(1, 3)
x
O
R¢(6, –7)
52. Let ( x , y) and ( X , Y ) be the old and the new coordinates, respectively. Since the axes are
rotated in the anticlockwise direction, q = + 60° . Therefore,
é x ù = écos 60° - sin 60° ù é X ù
ëê y ûú ëê sin 60° cos 60° ûú ëê Y ûú
Þ
é x ù = é 1 2 - 3 2ù é X ù
ú
êë y úû ê 3 2
1 2 û êë Y úû
ë
268
Solution of Advanced Problems in Mathematics for JEE
éX
3 ù
Yú
ê
é xù = ê 2
2 ú
êë y úû ê 3
Y
X+ ú
êë 2
2 úû
Þ
Þ
x=
X
3
3
Y
Y and y =
X+
2
2
2
2
2
Þ
2
æX
ö
æ
ö
ç - 3 Y ÷ - ç 3 X + Y ÷ = a2
ç2
÷
ç
2 ø
2 ÷ø
è
è 2
Þ
( X 2 + 3Y 2 - 2 3 XY ) - (3 X 2 + Y 2 + 2 3 XY ) = 4 a 2
Þ
-2 X 2 + 2Y 2 - 4 3 XY = 4 a 2
Þ
Y 2 - X 2 - 2 3 XY = 2 a 2
which is the required equation.
53. The following figure depicts the condition. By observation from the figure, DABC is clearly an
obtuse angled and isosceles triangle.
y
4
C
4/3
B
O
4/3
x
4
A
Alternate solution : The following figure depicts the condition.
y
C
(–2, 2)
B(1, 1)
x
O
A
(2, –2)
x+3y–4=0
y = –x
3x+y–4=0
269
Straight Lines
From the figure, we get
A : 3 x + y = 4 and y = - x Þ x = 2 ; y = -2
B : (1, 1) by solving the equations.
C : x + 3 y - 4 = 0 and y = - x Þ x = -2 ; y = 2
Thus,
AB = BC = 1 + 9 = 10
AC = 4 2 + 4 2 = 4 2
10 + 10 - 16(2)
cos B =
<0
2( 10 )( 10 )
Therefore, the given triangle is isosceles and obtuse angled triangle.
y 2 - y1 y 3 - y 2
56.
=
Þ Points are collinear.
x 2 - x1
x3 - x2
57. 3 h = a cos t + b sin t + 1
3k = a sin t - b cos t
Þ (3 h - 1) 2 + (3 k) 2 = ( a cos t + b sin t ) 2 + ( a sin t - b cos t ) 2 = a 2 + b 2
y
x
58. Equation of line +
= 1.
a -1 - a
Lines passes from (4, 3).
62. The given triangle is equilateral. Therefore, the orthocentre of the triangle is same as centroid
of the triangle. Thus, the orthocentre, that is, the centroid is given by
æ 5 + 0 + ( 5 2) 0 + 0 + ( 5 3 2) ö æ 5 5 ö
ç
÷ ºç ,
÷
,
ç
÷ çè 2 2 3 ÷ø
3
3
è
ø
63. Using homogenization,
æ y - mx ö
æ y - mx ö
3x2 - y2 - 2x ç
÷ + 4yç
÷ =0
è C ø
è C ø
P
(0, 0) O
Coefficient of x 2 + Coefficient of y 2 = 0
2m ö æ
4ö
æ
ç3 +
÷ + ç -1 + ÷ = 0
C ø è
Cø
è
Q
C = -m - 2
64.
R(3, 3 3)
60º
P(–1, 0)
90º
60º
O(0, 0)
3x + y = 0
y = mx + C
270
Solution of Advanced Problems in Mathematics for JEE
Exercise-2 : One or More than One Answer is/are Correct
1. Let line be
x y
+ =1
a b
a + b = 9 and ab = 20
Þ
a = 5, b = 4 or
a = 4, b = 5
æ 4ö
2. Centroid is ç 4 , ÷.
è 3ø
3.
2
t2
3
3 -5
t -6 = 0 Þ t 2 + t - 6 = 0
-2 -1
2
æ yö
æ yö
bç ÷ + 6 ç ÷ + a = 0
x
è ø
è xø
4.
if
bm 2 + 6m + a = 0
m = 1 is root of the equation
Þ
if
a + b = -6
m = -1 is root of the equation
Þ
a + b=6
6. Co-ordinate of other two points
(0, 2 3)
(1 ± 2 cos q , 3 ± 2 sin q)
2
(1, 3)
æ
ö
æ
ö
ç 1 ± 2 ç 3 ÷ , 3 ± 2 æç 1 ö÷ ÷
ç 2 ÷
ç
è 2 ø ÷ø
è
ø
è
2
(1 + 3 , 3 + 1) and (1 - 3 , 3 - 1)
O
8. Image of A(3 , - 1) about angle bisector x - 4 y + 10 = 0 is A ¢( a , b).
a - 3 b + 1 - 2(3 + 4 + 10)
=
=
1
-4
17
6x+10y=59
A ¢(1, 7)
x + 10 ö
æ
Let point B ç x 1 , 1
÷ on the line x - 4 y + 10 = 0
4
è
ø
(2,0)
A(3,–1)
x–4y+10=0
Þ
If mid-point of AB lie on the line 6 x + 10 y = 59
æ x +3ö
æ x + 10 - 4 ö
÷ + 10 ç 1
÷ = 59
6ç 1
2
8
è
ø
è
ø
Þ
B(10 , 5)
x
B
C
A¢
271
Straight Lines
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 3 to 4
3. x + y = 2 and x - 3 y = 6
Meet at (3 , - 1)
4. Image of A(2 , - 4) about x + y = 2 lie on BC.
x2 - 2 y2 + 4
æ -4 ö
=
= -2 ç
÷
1
1
è 2 ø
A
I
C
B
Þ
x 2 = 6, y 2 = 0
Image of A(2 , - 4) about x - 3 y = 6 lie on BC.
x3 - 2 y3 + 4
8
=
= -2
1
-3
10
2
4
Þ
x3 = , y3 =
5
5
Equ. of line BC , x + 7 y = 6
æ4 2ö
Þ
B ç , ÷ and C(6 , 0)
è3 3ø
Exercise-4 : Matching Type Problems
n +1
2. (A)
å( 1 C r -1 + 2C r -1 + 3C r -1 +¼ + n C r -1 )
r =1
n +1
=
å
1
r =1
1
n +1
C r -1 +
2
å
r =1
3
2
n +1
C r -1 +
= 2 + 2 + 2 +¼ + 2
å
3
n +1
C r -1 +¼ +
r =1
n
= 2(2
å n C r -1
r =1
n
- 1)
(B) Family of line ( x + y + 2) + l(2 x - y + 4) = 0 always passes from ( -2 , 0).
If almost one tangent can be drawn from ( -2 , 0) then
S 1 = 4 - 8 g - 36 + 4 g 2 £ 0
g2 - 2 g - 8 £ 0
(C) 2 sin 7 x × cos 2 x = cos 2 x
Þ
1
2
p 5p 13 p 17 p 25p 29 p 37 p 41p
7x = ,
,
,
,
,
,
,
6 6
6
6
6
6
6
6
cos 2 x = 0 or sin 7 x =
x=
p 3p
,
4 4
(D) a + b = tan 65° tan 70° - tan 65° - tan 70°
272
Solution of Advanced Problems in Mathematics for JEE
tan 135° =
tan 65° + tan 70°
= -1
1 - tan 65° + tan 70°
Þ tan 65° tan 70° - tan 65° - tan 70° = 1
3. (A) cos 40° - 2 cos 40° sin 10° = cos 40° - (sin 50° - sin 30° )
1 1
1
(B) 3 2l
4 = 0 Þ (3 - 2l)(1 + 3l) = 0
1 1 -3l
k
2 - 2k 1
(C) -k + 1
2k
1 =0
-4 - k 6 - 2 k 1
Þ 2k 2 + k - 1 = 0
Þ k = -1,
k
¥
(D)
3
4
1
2
5
1
æ 1ö
æ 1ö
æ 1ö
æ 1ö
ç ÷ = ç ÷ + ç ÷ + ç ÷ +¼¼ =
4
è2ø
è2ø
è2ø
k =3 è 2 ø
å
Exercise-5 : Subjective Type Problems
1. D = 132
2.
ax + by + c = 3 x - 4 y + c
Þ
a = 3 , b = -4
Distance of 3 x - 4 y + c from A(3 , 1) is 1.
|9 - 4 + c|
Þ
=1
5
|c + 5| = 5
Also, 3 x - 4 y + c = 0 and 3 x - 4 y + 5 = 0 lie on same side of A
Þ
c + 5> 0
Þ
3.
c + 5 =5 Þ c =0
xy( x + y - 2) = 0
a + a4 -2 £ 0
Þ
( a > 0)
a =1
y
(0, 2)
O
(2, 0)
x
x+y=2
R
5. PQ = 3 2
QR £ PQ
6. x 2 ( y 2 - x 2 ) = 0
has 3 different lines x = 0, y = x and y = - x.
2,3
P(
)
(–1,6)
Q
273
Straight Lines
8. 2 < a <
10
3
y
(–2,10/3)
(–6,6)
y=x
(0,2)
(–2,2)
(–2,0)
x
(3,0)
O
y = –x
9. Describe a circle whose diameter is AB.
\ centre = (1, 0)
Radius = 2
Let ‘ m ’ the slope of the line passing through ( 4 , 1).
( y - 1) = m ( x - 4) intersect the circle
^ distance from centre < radius of circle.
-3m + 1
m2 + 1
Þ
(4,1)
A
(–1,0)
<2
9m 2 - 6m + 1 < 4m 2 + 4
æ 6 - 96 6 + 96 ö ì 1 ü
÷ - í , 1ý
m Îç
,
ç 10
÷ î5 þ
10
è
ø
12 6
l1 + l 2 =
=
10 5
5 (l 1 + l 2 ) = 6
10.
2
2
=
b
3
Þ b=
3
2
– 2, 0
b
2, 0
b
60°
60°
(0,–2)
❑❑❑
B(3,0)
274
Solution of Advanced Problems in Mathematics for JEE
18
CIRCLE
Exercise-1 : Single Choice Problems
3
= ( h - 1) 2 + ( k - 1) 2
2
Locus of point P( h, k) is
3
( x - 1) 2 + ( y - 1) 2 =
4
1. CP =
2.
d 2 - ( r1 - r2 ) 2 = 15 ;
4.
y
C
(1,1)
d 2 - ( r1 + r2 ) 2 = 5 Þ r1 r2 = 50
x
0
T
PT = 16 + 16 - 8 - 8 - 7 = 3
Þ
TT ¢ = 2 BT = 2 × 3 cos 45° = 3 2
1
(–4,4)
(0,6)
19
8. Let centroid be ( h, k).
cos a + sin a + 1
Þ
h=
,
3
k=
sin a - cos a + 2
3
Þ
3 h - 1 = cos a + sin a , 3 k - 2 = sin a - cos a
Þ
(3 h - 1) 2 + (3 k - 2) 2 = 2
Þ
1ö
2ö
2
æ
æ
çx - ÷ +çy - ÷ =
3ø
3ø
9
è
è
2
2
3
P
(4,4)
5. It will be circle with diametric ends as (1, 1) and ( 4 , 2) i . e. , point of
intersection.
6.
P(h,
k)
1
3
O
B(1,1)
T¢
275
Circle
P
9.
(1,1)
Q(–3,0)
Length of tangent = PQ = 4 2 + 12 - 5 = 12
10.
D(12,17)
a
a cosq
q
A
90
O
q
°–
a sinq
a
C
a
a cosq
90°– q
q
a cosq B L a sinq M
OA = a sin q = 12 ,
DL = a sin q + a cos q = 17
a cos q = 5
C = ( a cos q + a sin q , a cos q) = (17 , 5)
12. Centroid divide the line joining orthocentre and circumcentre in 2 : 1.
y
B(0,1)
G
2
H
(h,k)
Þ
1
h = 1 + cos q ,
2
A(1,0)
O(0,0)
1+cosq , 1+sinq
3
3
O
x
(cosq, sinq)C
k = 1 + sin q
2
( x - 1) + ( y - 1) = 1
13. Co-ordinate of centre is C(1, 1).
( x - 1) 2 + ( y - 1) 2 = 1
x2 + y2 - 2x - 2 y + 1 = 0
(1,1)
x+y=2
(1,2)
(3/2,3/2)
y=x
(2,1)
276
Solution of Advanced Problems in Mathematics for JEE
14.
a = 2 R cos
p
6
A
a = 4 3 cm
Þ
Area of DABC =
3 2
a = 12 3 cm 2
4
15. Image of centre C 1 (5, 0) about the line y = x + 3 is
x 2 - 5 y 2 - 0 - 2 ( 5 + 3)
=
=
1
-1
12 + 12
Þ
C 2 ( -3 , 8)
Equation of reflected circle is
p/6
B
p/6
C
x2 + y2 – 10x = 0
C1
(5,0)
y = x+3
(x2, y2)
( x + 3) 2 + ( y - 8) 2 = 25
x=4
19.
C
x=0
(0,0)
20. Let the equation of line is 3 x + 4 y = C
C
= 3 Þ C = 15 (in first quadrant)
5
21. C 1 (5, 0), C 2 (3 , - 1), C 3 (3 2 , 2) do not lie on straight line.
22. Let equation of diameter is 3 x + 5 y = C
Þ
C =7
23. Equation of circle is
( x + 1)( x - 2) + ( y - 2)( y - 3) + l ( x - 3 y + 7) = 0
If its radius is 5.
Þ
l = ±1
25. Equation of circle is ( x - 1)( x - 3) + ( y - 4)( y - 2) = 0
y
(1,4)
A
(2,3)
O
26. Equation of tangent at O(0 , 0).
Þ
x(0) + y(0) + g( x + 0) + f( y + 0) = 0
gx + fy = 0
B(3,4)
(3,2)
C
x
277
Circle
27. Equation of normal at O(0 , 0)
(0,0)
O
y = -x
y=x
æ
æ 1 ö
æ 1 öö
Centre çç 0 ± çç ÷÷ , 0 ± çç
÷÷ ÷÷
2ø
è
è 2 øø
è
Þ
æ 1
1 ö æ 1
1 ö
Either çç ,
÷÷ or çç
,÷÷
2
2ø è 2
2ø
è
28. Here,
(Condition for external touch)
C 1 C 2 = r1 + r2
30. The triangle is right angled and the radical centre will be the orthocentre of the triangle.
32. Equation of common chord is 6 x + 14 y + (l + m ) = 0
If it passes through (1, - 4). Then, l + m = 50
x2 + y2 - 6x + 8 y = 0
33.
Distance of line from centre
9 - 16 - 25 32
=
5
5
(3,–4)
32
7
Shortest distance =
-5 =
5
5
p
p
34. Ð AOB =
Þ Ð ACB =
2
4
B
y
O
A
C(1,0)
x
y=7x+5
37. Equation of required circle :
S : ( x - 1) 2 + ( y - 1) 2 + l( x - y) = 0
S¢: x 2 + y 2 + 2 y - 3 = 0
Common chord of S = 0 and S ¢ = 0 is S - S ¢ = 0
( l - 2) x - ( l + 4) y + 5 = 0
Centre of S ¢ : (0 , - 1) lies on common chord Þ l = -9
S : ( x - 1) 2 + ( y - 1) 2 - 9 ( x - y) = 0
r=
Þ
2
9
2
40. Point lie inside the circle k + ( k + 2) 2 < 4 Þ 2 k 2 + 4 k < 0; -2 < k < 0
41. The length of the normal is
278
Solution of Advanced Problems in Mathematics for JEE
æ dy ö
y 1+ ç ÷
è dx ø
2
The length of radius vector of a point ( x , y) on the curve is | xi + yi|, that is
x 2 + y 2 , it is
given that
x 2 + y 2 =| y| 1 + ( y ¢) 2
Squaring on both sides of this equation, we get
x 2 + y 2 = y 2 [1 + ( y ¢) 2 ]
Þ
æ dy ö
x 2 + y 2 = y 2 + y 2ç ÷
è dx ø
2
2
Þ
Þ
Now,
æ dy ö
x =ç y
÷
è dx ø
dy
dy
y
= x or y =
= -x
dx
dx
dy
y
=x
dx
2
Þ
y dy = x dx
Integrating on both sides, we get
y2 x2
=
+c
2
2
Þ
x 2 - y 2 = 2 c or x 2 - y 2 = constant
This answer does not exist in the given options. So, consider the other alternative.
y dy = - x dx
Integrating on both sides, we get
y2
x2
=+c
2
2
Þ
x 2 + y 2 = constant
and this constant is > 0 in practical sense.
44. Length of chord = 2 3 2 - 5 = 4
(–1,–1)
5
||
(1,0)
3
||
279
Circle
47. Family of circles touching the line y = x at the point ( 4 , 4) is
( x - 4) 2 + ( y - 4) 2 + l( y - x) = 0
We need to find the member of this family which has length of chord = 6 2 on x + y = 0. For
different l’ s , we get different circles.
x 2 + y 2 - 8 x - 8 y + 32 + ly - lx = 0
x 2 + y 2 + x( -8 - l) + y( -8 + l) + 32 = 0
Þ
…(1)
y
C
R
2
2
3
4
E
D
A(4,4)
x
–x
y
=
=
y
x
O
Now,
OA = DC = 4 2
DE = 3 2 =
Therefore,
6 2
(given)
2
R 2 = (3 2 ) 2 + ( 4 2 ) 2
l2
= 50 Þ l2 = 100 Þ l = ± 10
2
Substituting l = -10 in eq. (1), we get
Þ
x 2 + y 2 + 2 x - 18 y + 32 = 0
[Substituting l = 10; in eq. (1); we get x 2 + y 2 - 18 x + 2 y + 32 = 0 , which does not exist in
the given options]
Note : From eq. (1), we get
R 2 = (Radius) 2 = g 2 + f 2 - c =
( l + 8) 2 ( l - 8) 2
l2
+
- 32 =
4
4
2
280
Solution of Advanced Problems in Mathematics for JEE
48. Slope of line normal to circle and perpendicular to line
1
m = - = tan q
2
Co-ordinate of point lie on normal at a dist. of 3 from centre
æ
æ -2 ö
æ 1 öö
ç 1 ± 3 çç
÷÷ , - 2 ± 3 çç
÷÷ ÷÷
ç
è 5ø
è 5 øø
è
C
(1, –2)
y = 2x + 11
Exercise-2 : One or More than One Answer is/are Correct
(3,3)
2.
(–3,3)
æpö
æp
ö
3. x 2 + y 2 - x ç ÷ + y ç - 2 sin -1 a ÷ = 0
2
2
è ø
è
ø
2
Þ
æpö
æp
ö
Length of chord = 2 ç ÷ + ç - sin -1 a ÷
è4ø
è4
ø
2
7. ( x + 2) 2 + ( y - 3) 2 is nothing but square of distance between ( x , y) and ( -2 , 3) where ( x , y) is
point lies on the circle.
Centre = ( - 4 , 5),
r = 16 + 25 + 40 = 9
A
Clearly, ( -2 , 3) is lies inside the circle.
\
b = PB 2 = (9 - 2 2 ) 2
\
a + b = 178, a - b = 72 2
8. Let point of intersection P( h, k)
Equation of chord of contact is
hx + ky = a 2
If it is tangent to x 2 + y 2 - 2 ax = 0
Þ
C
a = PA 2 = (9 + 2 2 ) 2
PC = 2 2
ha - a 2
h2 + k 2
=a
P
B
281
Circle
9. Equation of tangent to circle
(5,5)
3
9
y - 3 = ( x - 2) ± 13 1 +
2
4
Þ
(2,3)
2 y = 3 x + 13 , 2 y = 3 x - 13
x2 - 2 y2 - 3
æ 13 ö
Þ ( -1, 5)
=
= -ç ÷
3
2
è 13 ø
x3 - 2 y3 - 3
æ -13 ö
=
= -ç
÷
3
-2
è 13 ø
(–1,1)
(0,6)
Þ (5, 1)
10. 2 x 2 = 98 Þ x 2 = 49 Þ x = ± 7
(–6, 5)
(8, 5)
y=5
7
7
98
(–6, –9)
y = –9
(1, –2)
x
(8, –9)
x
x = –6
x=8
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
Sol. P1 ( -2 , - 1)
1ö
æ 16
P2 ç ,- ÷
7ø
è 7
æ1
ö
P 3 ç , - 1÷
è2
ø
P1
C1(0,–2)
C2 –3,– 1
2
5
5 /2
P3
P2
3 5
C3(2,2)
Paragraph for Question Nos. 4 to 6
2
2
4. S : x + y + x (2l - 9) + y (3l - 12) + 53 - 27l = 0
C : x2 + y2 - 4x - 6 y - 3 = 0
282
Solution of Advanced Problems in Mathematics for JEE
Equation of line : S - C = 0
or
x(2l - 5) + y(3l - 6) + 56 - 27l = 0
or
5 x + 6 y - 56 = 0 or 2 x + 3 y - 27 = 0
23
Þ
x = 2, y =
3
5. Centre of C lies on common chord of S and C.
Þ (2 , 3) lies on x(2l - 5) + y(3l - 6) + 56 - 27l = 0
Þ
S : x 2 + y 2 - 5x - 6 y - 1 = 0
6. Difference of squares of lengths of tangents from A and B is 3, which is equal to| AP 2 - BP 2|.
Paragraph for Question Nos. 7 to 8
7. Max. dist. between any two arbitrary points = 2
y
y=x
(1,1)
x
(1,–1)
y
x
8.
O
Paragraph for Question Nos. 9 to 10
Sol. Let P( h, k)
L1 = h 2 + k 2 - 4
L2 = h2 + k 2 - 4h
L3 = h2 + k 2 - 4k
If
L41 = L22 L23 + 16
Þ ( h 2 + k 2 - 4) 2 = ( h 2 + k 2 - 4 h)( h 2 + k 2 - 4 k) + 16
Þ ( h + k)( h 2 + k 2 - 2 h - 2 k) = 0
C1 : x + y = 0
C2 : x2 + y2 - 2x - 2 y = 0
283
Circle
Exercise-5 : Subjective Type Problems
1. Equation of chord of contact w.r.t. P
A
hx + ky = 1
Equation of common chord is
P(h,k)
( l - 3) x + ( 2l + 2) y + 3 = 0
l - 3 2l + 2
Þ
=
= -3
h
k
Þ Equation of locus is 6 x - 3 y - 8 = 0
2. By using system of circles any circle passing through (1, 1) and ( -2 , 1) is
B
( x - 1)( x + 2) + ( y - 1) 2 + l( y - 1) = 0
…(1)
x2 + y2 -1=0
…(2)
Given circles
Now radical axis of (1) and (2) is
( x - 2 y) + l( y - 1) = 0
Q Radical centre of given circles is (0 , 0).
So, eq. (3) is passing through (0 , 0).
…(3)
\
l =0
Put l = 0 in eq. (1) we get required circle.
y
3
3.
= tan q =
x
4
y)
(x,
4
q
3
0
5
4. x 2 + y 2 - 26 x + 12 y + 210
( x - 13) 2 + ( y + 6) 2 + 5
(13,–6)
(5,0)
5.
S º x 2 + y 2 + 2 gx + 2 fy + c = 0
Þ
(1, 1) satisfy circle.
Þ
Þ
2 g + 2 f = -c - 2
2g + 2 f + c = -2
c =0
…(1)
284
Solution of Advanced Problems in Mathematics for JEE
and
g + f = -1
\ Length of tangent = 8 + 4 g + 4 f + c = 2
6. Length of common external tangent
9
90
…(2)
°–
B
r1
r 2 + r22 - d 2
= 1
2 r1 r2
–B
0°
B
5
r2
–C
90°
cos(90° - B + 90° - C + 30° ) = cos 60°
A
r1
C1
90°–C
…(1)
30 °
d 2 - ( r1 - r2 ) 2 = 5
C2
r2
C
(2, 2)
9.
y = –1
x=5
x=2
From diagram common points are 3.
10. (C 1 C 2 ) 2 = r12 + r22
18 = 2 r 2 Þ r 2 = 9
y
(2,4)
(2,3/2)
O
11.
x
(2,–1)
12. PQ = PA = PB
A
2
2
2
2
( h - 1) + ( k - 2) = 6 - ( h - 3) - ( k - 4)
Þ
Q C(3,4)
(1,2)
h2 + k 2 - 4h - 6k - 3 = 0
13. c = 3 , a 2 + b 2 = 36
Length of chord AB = 2 r 2 - p 2
æ
ö
2c
÷
=2 c -ç
ç
2
2 ÷
è a +b ø
P(h,k)
2
2
=2 2
❑❑❑
B
285
Parabola
19
PARABOLA
Exercise-1 : Single Choice Problems
1.
P ¢Q ¢ = PQ cos(90° - q)
4
=
(t 2 < 1)
2
t +1
P¢
y
P(1,2)
q
( P ¢Q ¢) min = 2 2
Q¢
x
O
Q(1,–2)
2. Equation of circle with SP as diameter
9ö
æ
( x - 4) ç x - ÷ + y( y - 6) = 0
4ø
è
25
æ 25 ö
Centre ç
, 3 ÷ and radius =
8
8
è
ø
Equation of normal at P( 4 , 6) is
4 x + 3 y = 34
æ 25
ö
+ 9 - 34 ÷
ç
æ 25 ö
÷
Length of chord = 2 ç ÷ - ç 8
5
ç
÷
è 8 ø
ç
÷
è
ø
2
2
=
15
4
3. The diagonals are the focal chord.
2
AS = 1 + t = c (say)
1
1
1 1ö
æ 1
+
=1
+
= ÷
çQ
c æ 25
ö
è AS CS a ø
- c÷
ç
è 4
ø
Þ
5
c = ,5
4
B
A
S
D
C
286
Solution of Advanced Problems in Mathematics for JEE
æ1 ö
æ1
ö
A ç , 1÷ , B( 4 , 4), C( 4 , - 4) and D ç , - 1÷
4
4
è
ø
è
ø
1
15
Area of trapzium = (2 + 8) ´
2
4
2
4. For normal chord t 2 = - t 1 t1
Also chord substends an angle of 90° at the vertex
Þ t 22 = 8
t 1t 2 = - 4
\
9. ( y - x + 2) + l( y + x - 2) = 0
The family of lines passes through (2 , 0).
The chord is x = 2 and end points are (2 , ± 4).
2
10.
t 2 = - t1 t1
2 2
t 1,
P(
t 2 + t 22
2t + 2t 2
and k = 1
h= 1
2
2
Put the value of t 2 and eliminate t 1 we get
h -2 =
11. The
parabola
4
k2
is
+
k2
2
Þ a = 2, b = 4, c = 2
( y - 1) 2 = 4( x - 1).
The
t 1)
Q(t22, 2t 2)
coordinates
of
P(1 + t 12 , 1 + 2t 1 )
and
Q (1 + t 22 , 1 + 2t 2 ).
Here S(2 , 1) is the focus. The coordinates of T are G.M. of abscissa and A.M. of ordinates of P
and Q.
ST 2 = 16
Þ
\ SP × SQ = ST 2
12. Let P(t 1 ) and Q (t 2 ) are point of y 2 = 8 x
2t 12 + 2t 22 = 17 and (2t 12 )(2t 22 ) = 11
ST 2 = SP × SQ = 2(1 + t 12 ) 2(1 + t 22 ) = 34 + 4 + 11
ay = x 2
13.
Þ
x1 =
aB
2A
ST = 49
-a
-1
A
Þ
=
=B
æ dy ö 2 x 1
ç ÷
dx
è ø
and
y1 =
1 a
B 2
(slope of normal)
put ( x 1 , y 1 ) in ay = x 2
287
Parabola
dy
y
=
dx 2 x
2 dy 1
= dx
y
x
14.
P
(x,y)
(0,y/2)
Þ
Þ
15.
2 log y = log x + log c
y 2 = cx
put (3 , 1)
(–x,0)
( x - a ) 2 = - ( y - (a + 4))
4
x+
y=
The curve passes through (2 , 0)
(2 - a ) 2 = - (0 - (a + 4))
(0,4)
2
a - 5a = 0 Þ a = 0 or a = 5
( x - 5) 2 = - ( y - 9) put y = 0 Þ x = 2 , 8
A (–2,0)
B(2,0)
16. y = (tan 60° ) x is the focal chord.
C
P
Coordinates of P and Q are intersection of y = 3 x with parabola
æ 4 -4ö
P( 4 , 4 3 ), Q ç - ,
÷
3ø
è 3
(–2,0)
Find ^ bisector of PQ.
60°
(0,0)
Q
17. The director circle of the parabola is its directrix ( x + 11 = 0). Now apply condition of tangency.
18. The following figure depicts the condition. Chord of contact of a point A( x A , y A ) with respect
to y 2 = 4 x is y A y = 2( x + x A ). Since this chord passes through the point (3, 1), we have
y A = 2 ( x A + 3)
y
B
(xA, yA)
A
(3,1)
x
O
C
AB and AC are tangents to the parabola.
BC is chord of contact of point A with respect to the parabola y 2 = 4 ax.
288
Solution of Advanced Problems in Mathematics for JEE
Given that point A lies on x 2 + y 2 = 25, we have
x 2A + y 2A = 25
Þ
x 2A + 4 ( x A + 3) 2 = 25
Þ
x 2A + 4 ( x 2A + 9 + 6 x A ) = 25
Þ
5 x 2A + 24 x A + 36 - 25 = 0
Þ
5 x 2A + 24 x A + 11 = 0
Exercise-2 : One or More than One Answer is/are Correct
1. Equation of normal at P( at 2 , 2 at ) is
P(at2, 2at)
y = - tx + 2 at + at 3
G( 4 a + at 2 , - 2 at )
Þ
Locus of point G( h, K ) is
(at2, –2at) Q
y 2 = 4 a( x - 4 a)
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
Sol. Tangent and normal are angle bisectors of focal radius and perpendicular to directrix.
Circle ‘ C ’ circumscribing DABP is
( x - 5)( x + 3) + ( y - 4)( y + 4) = 0
Length of latus rectum = 4(2 2 ) = 8 2
9 x+
y=
x+
5
y=
P
x–y–1=0
B(–3,–4)
2)
(3 ,
(5,4
)
A
G(h,K)
289
Parabola
Exercise-5 : Subjective Type Problems
1.
b = 2a 2 + 4a - 2
…(1)
2
…(2)
- b = 2a - 4a - 2
(1) & (2) Þ 4a 2 - 4 = 0 Þ a = ± 1
\
A
(a,b)
–1
B
(–a,–b)
Put a = 1, b = 2 + 4 - 2 = 4
A (1, 4), B ( -1, - 4)
AB 2 = l 2 = ( 4 + 64 ) 2 = 68
æ a + b æ a + bö2 ö
æ a + b - a 2 - b2 ö
÷
2. R = ç
, -ç
,
÷ ÷ , M = çç
÷
ç 2
2
è 2 ø ÷ø
è 2
ø
è
- b2 + a 2
( x - b)
b-a
y + b 2 = - ( b + a)( x - b)
a
PQ = y + b 2 =
P
y = - ( b + a)( x - b) - b 2
b
D 1 = [[- ( a + b)( x - b) - b 2 ] + x 2 ] dx
ò
a
= - ( a + b)
1
Area of DPQR = D 2 =
2
( x - b) 2
x3
- b2 x +
2
3
a
-a 2
b
-b 2
2
a+b
æ a + bö
-ç
÷
2
è 2 ø
R 1 ® R 1 - R 2 , R 2 ® R 2 - R 3 , we get D 2 =
3.
Þ
Þ
Similarly,
1
1
1
( a - b) 3
8
m AB ´ m BC = -1
-2
-2
´
= -1
(t 1 + t 2 ) (t 2 + t 3 )
(t 1 + t 2 )(t 2 + t 3 ) = - 4
m AD ´ m CD = -1
Þ
(t 1 + t 4 )(t 3 + t 4 ) = - 4
Þ
(t 1 + t 2 )(t 2 + t 3 ) = (t 1 + t 4 )(t 3 + t 4 )
Solving this
t2 + t4
= -1
t1 + t 3
❑❑❑
b
R
b
=
a
( a - b) 3
6
M
Q
290
Solution of Advanced Problems in Mathematics for JEE
20
ELLIPSE
Exercise-1 : Single Choice Problems
1. Length of perpendicular from C(0 , 0) to the tangent at P(2 3 cos q , 2 2 cos q) is
CF =
Equation of normal at P is
-1
cos 2 q sin 2 q
+
12
8
2 3x 2 2 y
= 12 - 8 which meets the major axis at
cos q
sin q
æ 2
ö
Gç
cos q , 0 ÷
3
è
ø
CF ´ PG = 8
(–4,3)
2.
(0,3)
y=3
(4,3)
(2,0)
The minimum length of intercept will be possible when
y = 3 or y = -3 Þ AB = 8
x+y=7
3.
291
Ellipse
dy
x
== -1
dx
2y
Put x = 2 y in the equation of ellipse
The point lies in I quad Þ (2 , 1)
4. Equation of tangent at P is
Þ
y
b2
P ae, a
ex + y = a
2
10
e= , a=
3
3
x
2
b
Q ae, – a
10 5
b=
9
and
Length of latus rectum =
2 b 2 100
=
a
27
5. Area bounded by circle & ellipse = pa 2 - pab = pa( a - b)
6.
S 1 F1 + S 2 F 2
³ ( S 1 F1 )( S 2 F 2 ) = 16
2
Q Product of perpendiculars from two foci of an ellipse upon any tangent is equal to the
square of semi-minor axis.
f( k 2 + 2 k + 5) > f( k + 11)
7.
Þ
k 2 + 2 k + 5 < k + 11 Þ k Î ( -3 , 2)
8. Since sides of the square are tangent and perpendicular to each other,
so the vertices lie on director circle
æ 10 ö
x 2 + y 2 = 16 + b 2 = ç ÷
è 2 ø
Þ
9. T = S 1
2
b=3
A p( 4)(3)
=
= 12
p
p
æ xq + py ö
2
2
Þ px + qy + ç
÷ - 1 = p + q + pq - 1
2
è
ø
10
5 2
5 2
Þ p 2 + q 2 = - pq Þ p = 0 , q = 0
10. The combined equation of pair of tangents drawn from a point ( x 1 , y 1 ) to the ellipse
y2
x2
Sº
+
- 1 = 0 is T 2 = SS 1 . Therefore,
2
2
a
b
2
ö
æ x2
ö æ x2
yy
y2
y2
æ xx 1
ö
ç 2 + 21 - 1÷ = ç 2 + 2 - 1÷ ç 12 + 12 - 1÷
ç
÷
ç
÷
è a
ø
b
b
b
èa
øè a
ø
292
Solution of Advanced Problems in Mathematics for JEE
2
æ x2
ö æ 42
ö
æ 4x
ö
+ 2 y - 1÷ = ç
+ y 2 - 1÷ ç
+ 2 2 - 1÷
ç
ç 9
֍ 9
÷
è 9
ø
è
øè
ø
3 x 2 + 7 y 2 - 16 xy + 8 x + 36 y - 52 = 0
Þ
2 h 2 - ab
a+b
where, a = 3 , b = 7 and h = -8. Therefore,
tan a =
Þ
2 64 - 21
43
=
10
5
Note : a is acute angle between the pair of tangents. Therefore,
tan a =
( a + b - c) 2 = a 2 + b 2 + c 2 + 2 ab - 2 ac - 2 bc
Alternate solution : Any line passing through the point (4, 2) is given by
y - 2 = m ( x - 4)
y = mx - 4m + 2
For this line to be tangent to the given ellipse, put this y into the equation of the ellipse and
make
D =0
That is,
x2
+ (mx - 4m + 2) 2 = 1
9
(1 + 9m 2 ) x 2 + x(36m - 72m 2 ) + 16 (9)m 2 - 16 (9)m + 27 = 0
Now,
D = 0 Þ B 2 - 4 AC = 0
Þ (36m - 72m 2 ) 2 - 4 (1 + 9m 2 )(16 × 9m 2 - 16 × 9m + 27) = 0
Þ (36m ) 2 (1 - 2m ) 2 - 36 (1 + 9m 2 )(16m 2 - 16m + 3) = 0
Þ
m 2 (1 + 4m 2 - 4m ) - 36 (16m 2 - 16m + 3 + 9 × 16m 4 - 9 × 16m + 27m 2 ) = 0
Þ
7m 2 - 16m + 3 = 0
Now,
tan a =
m1 - m 2
=
1 + m 1m 2
2
Þ
( m 1 + m 2 ) 2 - 4m 1 m 2
1 + m 1m 2
3
æ 16 ö
ç ÷ -4×
2
7 æç 16 2 - 4 × 3 × 7 ö÷
è 7 ø
tan a =
=
÷
3
10 ç
7
1+
è
ø
7
293
Ellipse
43
æ 1 ö
= ç ÷ 4 ( 43) =
5
è 10 ø
where a is the acute angle between the tangents.
Exercise-2 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1. SS ¢ = 2 ae
3ö
÷
( x 1 - x 2 ) + ( y 1 - y 2 ) = 4(2) ç
÷
è 2 ø
2
2
2æ
ç
x2
2
( x 1 + x 2 ) 2 + ( y 1 + y 2 ) 2 - 4 ( x 1 x 2 + y 1 y 2 ) = 12
S¢
C
x1
(2 h) 2 + (2 k) 2 - 4 (1 + 1) = 12
(h,k) y
2
S
y1
(Q x 1 x 2 and y 1 y 2 are ^ distance of the foci from their tangents
= b 2 = 12 )
Þ
h2 + k 2 = 5
2. (2 x - h)( h) = 1
Þ
x=
1 + h2
2h
(2 y - k)( k) = 1
Þ
y=
1 + k2
2k
S¢(2x–h, 2y–k)
C
(x,y)
S(h,k)
❑❑❑
294
Solution of Advanced Problems in Mathematics for JEE
21
HYPERBOLA
Exercise-1 : Single Choice Problems
1. The normal is y - 4 =
1
( x - 1). Put the value of y in xy = 4 we get co-ordinates.
4
3. c 2 = a 2 m 2 - b 2 Þ c 2 = l2 m 2 - (l3 + l2 + l) 2
c2 ³ 0
Þ m 2 ³ (l2 + l + 1) 2
3
9
l2 + l + 1 has minimum value Þ m 2 ³
4
16
4.
3
x and the double ordinate be
2
ö
æ
ö
3
h 2 - 4 ÷÷ and P ¢ çç h, h 2 - 4 ÷÷
2
ø
è
ø
The asymptotes are y = ±
æ
3
P çç h,
2
è
Þ
( PQ )( PQ ¢) = 3
5. 2 ae = 5 and 2 a = 3
5
Þ
e=
3
1
1
5
Þ
+
= 1 Þ e¢ =
2
2
4
¢
e
(e )
6. The equation of normal at (2 sec q , tan q) is 2 x cos q + y cot q = 5
1
Equal intercepts
Þ
sin q =
2
25
2
2
Also touches ellipse
Þ a +b =
Q c 2 = a 2m 2 + b 2
3
7. Let locus of point be ( h, k).
Equation of chord of contact is hx + ky = 4
æ 4 - hx ö
For tangent, x ç
÷ = 1 has two equal roots.
è k ø
(1,2)
(5,5)
295
Hyperbola
hk = 4 Þ
Þ
xy = 4
2
2
8.
æ x cos a + y sin a ö
x2 y
-ç
÷÷ = 0
16 18 çè
p
ø
Þ
Coeff. of x 2 + coeff. of y 2 = 0 Þ
P = ± 12
æ dö
The chord x cos a + y sin a ± 12 = 0 is tangent to the circle x 2 + y 2 = ç ÷
è2ø
2
Þ
d
=6
4
9. Let the rectangular hyperbola be x 2 - y 2 = a 2 and the point be ( a sec q , a tan q).
æ 2 a ö æ a cos q ö æ 2 a sin q ö
a1 a 2 + b1 b2 = ( a cos q) ç
÷ + ç÷ç
÷
è cos q ø è sin q ø è cos q ø
Exercise-2 : One or More than One Answer is/are Correct
æ 1ö
3. Let ç t , ÷ be any point on xy = 1
è tø
xy = 1
Þ
Þ
Þ
Þ
Þ
xy ¢ + y = 0
-y
y¢ =
x
1
y¢ = t2
-b
=t2
a
a and b are of opp. sign.
❑❑❑
296
Solution of Advanced Problems in Mathematics for JEE
22
COMPOUND ANGLES
Exercise-1 : Single Choice Problems
2.
a sin x + b(2 cos c cos x) = a
a - a sin x
cos c =
2 b cos x
1
= (a sec x - a tan x) differentiate w.r.t. x
2b
Þ
3.
a sec x tan x - a sec 2 x = 0
a
sin x =
a
tan x ×
3 tan x - tan 3 x
< -1
1 - 3 tan 2 x
æ 3t - t 3 ö
÷ + 1< 0
tç
ç 1 - 3t 2 ÷
è
ø
1-t4
1 - 3t
5.
<0
Þ
(t - 1)(t + 1)
(3t 2 - 1)
<0
æ 1
ö
t Î çç
, 1÷÷
è 3 ø
Þ
4.
2
(Let tan x = t )
8
8
r =1
r =1
å tan( r A) tan {( r + 1) A} = å éêë
tan( r + 1) A - tan( rA) - tan A ù tan 9 A - 9 tan A
= -10
úû =
tan A
tan A
f( x) = 2 cosec 2 x + sec x + cosec x
1 + sin x + cos x
=
sin x cos x
f ¢( x) =
sin 3 x + sin 2 x - cos 3 x - cos 2 x
2
2
sin x cos x
=0
Þ
x=
p
4
297
Compound Angles
2
f( x) min =
6.
at x =
2 -1
p
4
cosec q + cosec (60° – q) - cosec (60°+q)
where q = 10°
1
1
10. (2 sin x cos x + 2 cos 2 x) = (sin 2 x + cos 2 x + 1)
2
2
tan A tan B
11.
=
=k
( k > 0), if 2 sin A = 3 sin B
3
5
2 tan A
Þ
2
3 tan B
=
1 + tan A
Þ
2
1 + tan B
2 3k
1 + 3k
2
=
3 ´ 5k
1 + 5k
2
Þ k=
1
5
12. Gives equations can be written as
2 cos a + 9 cos d = -6 cos b - 7 cos g
…(1)
2 sin a - 9 sin d = 6 sin b - 7 sin g
…(2)
Square and add equation (1) and (2),
13.
Þ
4 + 36 + 36 [cos a cos d - sin d sin a ] = 36 + 49 + 84 [cos b cos g - sin b sin g]
Þ
36 [cos(a + d)] = 84 [cos(b + g)]
cos(a + d) 84 7 m
=
= = ;
cos(b + g) 36 3 n
1 + sin q + 1 - sin q
2
1 - sin q
3
14. A =
2 rp
=
2
= -2 sec q
cos q
2p
4p
6p
2p
Þ
a +b =
4p
8p
å cos 7 = cos 7 + cos 7 + cos 7 = cos 7 + cos 7 + cos 7 = B
r =1
15.
m + n = 10
x 1
=
z 3
y 1
tan a = =
z 2
tan a + tan b
tan(a + b) =
=1
1 - tan a tan b
tanb =
é p 2p ù
17. f( x) = -2 sin 2 x + sin x + 2 " x Î ê ,
ë 6 3 úû
Let sin x = t
é1 ù
f(t ) = -2 t 2 + t + 2 " t Î ê , 1ú
ë2 û
p
4
298
Solution of Advanced Problems in Mathematics for JEE
1
3
18. 1 + (cos 2 A - sin 2 B ) - cos A cos B = 1 + cos( A + B ) cos( A - B ) - [cos( A + B ) + cos( A - B )] =
2
4
19. (2 sin x - cosec x) 2 + (tan x - cot x) 2 = 0
\
sin 2 x =
1
Ç tan 2 x = 1
2
20. cos 2 A = sin A × tan A Þ cos 3 A = sin 2 A
æ 3 + 1ö
æ 3 + 1ö
æ 3 + 1ö
÷ sin x + ç
÷ cos x = ç
÷
21. f( x) = çç
÷
ç 2 ÷
ç 2 ÷ (sin x + cos x)
è 2 ø
è
ø
è
ø
22. A = B + C
tan A tan B tan C = tan A - tan B - tan C
23.
E = sin A + sin 2 B + sin 3C
3
4 3
E = + 2 × × -1
5
5 5
15 24
39 - 25 14
=
+
-1=
=
25 25
25
25
cos A cos C + cos A cos C
24.
(Q A + B + C = p)
= cot C
cos A sin C + cos A sin C
æa + gö
æa -gö
2 cos ç
÷ sin ç
÷
2
2 ø
sin a - sin g
æa + gö
è
ø
è
25.
=
= cot ç
÷ = cot b
cos g - cos a
æa + gö
æa -gö
è 2 ø
2 sin ç
÷ sin ç
÷
è 2 ø
è 2 ø
Þ
26. cos
x
x
x
x
x
× cos
cos
LL cos × cos =
256
128
64
4
2
sin x
æ x ö
256 sin ç
÷
è 256 ø
(sin 7a + sin 5a ) + 5(sin 5a + sin 3a ) + 12(sin 3a + sin a )
27.
sin 6a + 5 sin 4a + 12 sin 2a
2 sin 6a cos a + 5(2 sin 4a cos a ) + 12(2 sin 2a cos a )
=
= 2 cos a
sin 6a + 5 sin 4a + 12 sin 2a
28. tan 2 A + tan 2 B + tan 2 C = tan A tan B + tan B tan C + tan A tan C
Þ
Þ
tan A = tan B = tan C
p
A = B =C =
3
29. log|sin x| |cos x |+ log|cos x| |sin x|= 2
log|sin x| |cos x |= 1 Þ |cos x |=|sin x |
3
30. f( x) = sin x + cos x = 1 - 3 sin x cos x = 1 - sin 2 2 x
4
6
6
2
Þ
2
299
Compound Angles
31. y =
=
2 sin a
é(1 + sin a ) - cos a ù 2 sin a[(1 + sin a ) - cos a ]
´ê
=
1 + cos a + sin a ë(1 + sin a ) - cos a úû
(1 + sin a ) 2 - cos 2 a
1 + sin a - cos a
1 + sin a
tan 3 A
32.
1 + tan 2 A
+
cot 3 A
1 + cot 2 A
=
sin 3 A cos 3 A
+
cos A
sin A
=
sin 4 A + cos 4 A 1 - 2 sin 2 A cos 2 A
=
sin A cos A
sin A cos A
= sec A cosec A - 2 sin A cos A
33.
1 - sin q
1 + sin q
2
2
+
=
=
1 + sin q
1 - sin q
1 - sin 2 q |cos q|
34. y = (sin 2 q + cosec 2 q + 2) + (cos 2 q + sec 2 q + 2) = 7 + (tan 2 q + cot 2 q) ³ 9
35. log 3 sin x - log 3 cos x - log 3 (1 - tan x) - log 3 (1 + tan x) = -1
æ tan x ö
log 3 ç
÷ = -1 Þ
è 1 - tan 2 x ø
36. sin q + cosec q = 2
tan x
=
2
1 - tan x
1
Þ
3
tan 2 x =
2
3
1
æ
ö
sin q = cosec q = 1 ; ç x + ³ 2 ÷
x
è
ø
Þ
37. (tan q + cot q) (tan 2 q + cot 2 q - 1) = 52
(tan q + cot q) {(tan q + cot q) 2 - 3} = 52
Let
tan q + cot q = t
t 3 - 3t - 52 = 0
2
Þ
t =4
tan q + cot q = (tan q + cot q) 2 - 2 = 14
38.
2
-5 £ 3 sin x - 4 cos x £ 5
10 £ 3 sin x - 4 cos x + 15 £ 20
log 20 10 £ log 20 (3 sin x - 4 cos x + 15) £ log 20 20
x2 + y2 =9
39.
Let x = 3 cos q , y = 3 sin q
4 a 2 + 9 b 2 = 16
Let a = 2 cos f , b =
4
sin f
3
4 a 2 x 2 + 9 b 2 y 2 - 12 abxy = (2 ax - 3 by) 2
= (12 cos q cos f - 12 sin q sin f) 2 = 144 cos 2 (q + f)
300
40.
Solution of Advanced Problems in Mathematics for JEE
A 2 = sin 2 - sin 3
y=sinx
sin 3 > sin 2
A2 < 0
B 2 = cos 2 - cos 3
cos 3 > cos 2
O
1.57 3
2 3.14
B2 < 0
y=cosx
1
3
O
2 3.14
x
1.57
Both A and B are not real numbers.
41. (2 x + 2 - x - 2 cos x) (3 x + p + 3 - x -p + 2 cos x)(5 p- x + 5 x -p - 2 cos x) = 0
If
2 x + 2 -x
= cos x
2
Þ
x =0
If
3 x + p + 3 - x -p
= - cos x
2
Þ
x = -p
5 p- x + 5 x -p
= cos x
2
There are two real values of x.
If
(Not possible)
e sin x - e - sin x - 4 = 0
42.
e 2 sin x - 4 e sin x - 1 = 0
Þ
e sin x = 2 ± 5
If
e sin x = 2 + 5
Þ
sin x = ln (2 + 5)
[ln(2 + 5) > 1, Not possible]
If
e sin x = 2 - 5
(2 - 5 < 0) Not possible
There is no solution.
43.
4 sin 4 a + 4 sin 2 a × cos 2 a + 4 cos 2 ( p / 4 - a / 2)
= 4 sin 2 a + 2 [1 + cos( p / 2 - a )]
= 2 |sin a | + 2 + 2 sin a
= -2 sin a + 2 + 2 sin a = 2
(If p < a <
3p
then sin a < 0)
2
x
301
Compound Angles
p
p ö
æ
cos
ç sin
÷
p
p öç
æ
12
12
÷
44. ç cos
- sin
+
÷
p
p ÷
12
12 ø ç
è
sin
ç cos
÷
12
12 ø
è
p
p
cos
- sin
12
12 = 2 1 - sin p / 6 = 2 2
=
p
p
sin p / 6
sin
× cos
12
12
tan 100°+ tan 125°
45.
tan(100°+125° ) =
=1
1 - tan 100° tan 125°
tan 100°+ tan 125°+ tan 100° tan 125° = 1
Þ
46. If sin x + sin 2 x = 1 Þ sin x = 1 - sin 2 x Þ sin x = cos 2 x
cos 8 x + 2 cos 6 x + cos 4 x = sin 4 x + 2 sin 3 x + sin 2 x
= sin 2 x(sin 2 x + 2 sin x + 1)
= (1 - sin x) (2 + sin x)
= 2 - sin x - sin 2 x = 1
47. Let x = 5cos q, y = 5sin q
(Q 3 x + 4 y > 0)
0 < 3 x + 4 y £ 25
æqö
48. 5 cos 2q + 2 cos 2 ç ÷ + 1 = 0
è2ø
10 cos 2 q + cos q - 3 = 0
Þ
cos q =
1
3
, 2
5
4
where 0 < b < p and tanb > 0
5
3
then
cosb =
5
4
é3
ù
5 ê sin(a + b) - cos(a + b)ú coseca = 5
5
ë5
û
49. sinb =
é
pö
pö
p
pö
p
p öù
5p ö
æ
æ
æ
æ
æ
50. sin ç x + ÷ + cos ç x + ÷ = 2 êcos sin ç x - ÷ + sin × cos ç x + ÷ ú = 2 sin ç x +
÷
6
6
4
6
4
6
12 ø
è
ø
è
ø
è
ø
è
øû
è
ë
5p p
p
This attained maximum value when x +
=
Þ x=
12 2
12
51.
sin 2 x - cos 2 x = 2 a - 1
- 2 £ 2a - 1 £ 2
1- 2
1+ 2
£ a£
2
2
52.
(cos 12°× cos 24°× cos 48°× cos 84° ) (cos 36° cos 72° ) × cos 60°
( - cos 12°× cos 24°× cos 48°× cos 96° ) (cos 36° cos 72° ) × cos 60°
302
Solution of Advanced Problems in Mathematics for JEE
é sin(2 4 ´ 12° ) ù æ 5 + 1
5 - 1ö 1
1
÷´ =
´
êú ´ çç
÷
4
4
4
2
1
28
2 sin 12° û è
ø
ë
53. 2 cos 2 q + cos q + 1
7
1
y min = at cos q = 8
4
y max = 4 at cos q = 1
y max 32
=
y min
7
–1
3
–1/4
4
æ 3 tan x - tan x ö
÷ + 1 < 0 ; tan x - 1 < 0
54. tan x ç
ç 1 - 3 tan 2 x ÷
3 tan 2 x - 1
è
ø
Þ
(tan 2 x + 1)(tan x + 1)(tan x - 1)
( 3 tan x + 1)( 3 tan x - 1)
<0
p
p
< x<
6
4
p
p
55. a = 2 R sin = 2 r tan
n
n
Þ
R
p
n
r
a/2
56. (cos12°+ cos 132° ) + (cos 84°+ cos 156° )
12°+132°
12°-132°
84°+156°
84°-156°
= 2 cos
cos
+ 2 cos
cos
2
2
2
2
= 2 cos 72° cos 60°+2 cos 120° cos 36°
=2 ´
57.
5 -1 1
5 +1
1
æ 1ö
´ + 2 ´ç- ÷´
=4
2
2
4
2
è
ø
1 é 2 sin q cos q 2 sin 3q cos 3q 2 sin 9q cos 9q 2 sin 27q cos 27q ù
+
+
+
2 êë cos q cos 3q cos 9q cos 3q cos 9q cos 27q
cos 27q cos 81q úû
=
1 é sin(3q - q) sin(9q - 3q) sin(27q - 9q) sin(81q - 27q) ù
+
+
+
2 êë cos q cos 3q cos 3q cos 9q cos 9q cos 27q cos 27q cos 81q úû
1
1 é sin 80q ù
= [tan 81q - tan q] = ê
2
2 ë cos q cos 81q úû
–1
303
Compound Angles
æ 4 cos 20° + 1 ö 2 sin 40° + sin 20° 2 sin(60° - 20° ) + sin 20°
58. sin 20° ç
=
= 3
÷=
cos 20°
cos 20°
è cos 20° ø
59. Let us draw the graph of
æ xp ö
f( x) = sin ç
÷
è 2 ø
and
g( x) = cos( xp)
On the same xy-plane as shown in the following figure.
y
g(x)
1
0
1
2
3
2
3
2
5
2
4
7
2
x
f(x)
æ5 7ö
From this graphical representation, it is clear that y is strictly increasing in ç , ÷
è2 2ø
Because for all values of x,
5
7
< x<
2
2
æ xp ö
That is,
sin ç
÷<0
è 2 ø
and
cos( xp) < 0
dy
which imply that
>0
dx
which means that y is strictly increasing.
æ sin 8q ö
60. 8 sin q sin 3q ç
÷ = cos 6q
è 4 sin 2q ø
sin 3q sin 8q = cos 6q cos q
cos 5q - cos 11q = cos 7q + cos 5q
cos 7q + cos 11q = 0
2 cos 9q × cos 2q = 0
1
1
3
61. tan A = - Þ sin A =
; cos A = 3
10
10
63. (2 cos q) 2 = (1 - sin q) 2 Þ sin q = 1 or sin q =
-3
5
304
Solution of Advanced Problems in Mathematics for JEE
1
= 2 Þ sin q = 1
sin q
65. tan 2 q + cot 2 q = a Þ tan 3 q + cot 3 q = a + 2 ( a - 1) = 52
5
66. tan A = - tan C =
12
3
4
cos B = - cos D = Þ tan D =
5
3
64. sin q +
67.
tan 2 q - sin 2 q = tan 2 q sin 2 q =|tan q sin q|
68.
sin 10° + sin 20°
= tan 15° = 2 - 3
cos 10° + cos 20°
69. (sin 2 q) 3 + (cos 2 q) 3 = (sin 2 q + cos 2 q)(sin 4 q + cos 4 q - sin 2 q cos 2 q)
= 1 - 3 sin 2 q cos 2 q
70.
tan x + 1 sec 2 x + 2
(tan x + 1) 2 - (sec 2 x + 2)
Þ
tan x - 1 tan 2 x - 1
tan 2 x - 1
Þ
2 tan x - 2
2
Þ
tan x - 1
71.
2
tan x + 1
cot a - tan a
- [cos 450° + cos(2a - 180° )]
cot a + tan a
Þ (cos 2 a - sin 2 a ) + cos 2a = 2 cos 2a
æ 1 + tan a ö æ 1 + tan a ö
72. ç
÷ ×ç
÷+1
è 1 - tan a ø è 1 - tan a ø
æp
ö
æp
ö
æp
ö
1 + tan 2 ç + a ÷ = sec 2 ç + a ÷ = cosec 2 ç - a ÷
è4
ø
è4
ø
è4
ø
73.
tan a + sin a
= tan a
1 + cos a
74. (cos 2a + cos 5a ) - (cos 3a + cos 4a )
7a
3a
7a
a
2 cos
× cos
- 2 cos
× cos
2
2
2
2
7a é
3a
aù
a
7a
2 cos
cos
- cos ú = -4 sin sin a cos
2 êë
2
2û
2
2
75. cos 2 g =
Þ
1 - tan 2 g
1 + tan 2 g
=
æ 1 + sin a sin b ö
÷÷
1 - çç
è cos a cos b ø
æ 1 + sin a sin b ö
÷÷
1 + çç
è cos a cos b ø
(cos a cos b) 2 - (1 + sin a cos b) 2
2
(cos a cos b) + (1 + sin a sin b)
2
=
2
2
[1 + cos(a - b)][cos(a + b) - 1]
(cos a cos b) 2 + (1 + sin a sin b) 2
£0
305
Compound Angles
76. x =
2p
(II nd quadrant)
3
sin 50 x
1
æ 101x ö
× cos ç
÷=x
2
2
è
ø
sin
2
p
2p
10 p
1
77. cos 3 0° + cos 3 + cos 3
+ cos 3 p +¼ + cos 3
=3
3
3
8
1 - 2(cos 60° - cos 80° ) 2 cos 80°
78.
=
=1
2 sin 10°
2 sin 10°
cos x + cos 2 x + cos 3 x +¼ + cos 100 x =
79. ( x + 5) 2 + ( y - 12) 2 = 14 2
Let x = -5 + 14 cos q , y = 12 + 14 sin q
Þ
x 2 + y 2 = 365 + 336 sin q - 140 cos q
80. tan q = l has three distinct solution in [0 , 2p] Þ l = 0 and q = 0 , p , 2 p.
1 + tan a
1 - tan a
2
+
=
1 - tan a
1 + tan a
1 - tan 2 a
4
æ3
ö
82. 3 sin q + 4 cos q = 5 ç sin q + cos q ÷ = 5 sin(q + 53° )
5
è5
ø
81.
n
83. f(n) =
Õ cos r
r =1
f( 4) = cos 1 × cos 2 × cos 3 × cos 4 < 0
f(5) = cos 1 × cos 2 × cos 3 × cos 4 × cos 5 < 0
84.
( p 2 - q 2 ) 2 ( 4 tan A sin A) 2
=
= 16
pq
tan 2 A - sin 2 A
85. 0 < sin a < cos a < 1
æ pö
a Î ç 0, ÷
è 4ø
(sin a ) cos a < (sin a ) sin a
(cos a ) cos a < (cos a ) sin a
A
5A
86. 32 sin sin
= 16 (cos 2 A - cos 3 A)
2
2
= 16 [(2 cos 2 A - 1) - ( 4 cos 3 A - 3 cos A)]
87. cos a cos b - sin a sin b + sin a cos b - cos a sin b = 0
cos a(cos b - sin b) + sin a(cos b - sin b) = 0
(cos b - sin b)(cos a + sin a ) = 0
cos a = - sin a
tan a = -1
(Q cos b ¹ sin b)
306
Solution of Advanced Problems in Mathematics for JEE
88. 2 x = 3 y = 6 -z = k
x = log 2 k , y = log 3 k , z = - log 6 k
1 1 1
+ + =0
x y z
2
æ -21 ö
æ -27 ö
89. (sin a + sin b) 2 + (cos a + cos b) 2 = ç
÷ +ç
÷
65
è
ø
è 65 ø
1170
æ a -b ö
2 + 2 cos(a - b) =
= 4 cos 2 ç
÷
2
è 2 ø
(65)
2
90. m 2 = a 2 + b 2 + 2 ( a 2 cos 2 q + b 2 sin 2 q)( a 2 sin 2 q + b 2 cos 2 q)
= a 2 + b 2 + 2 a 2 b 2 + ( a 4 + b 4 - 2 a 2 b 2 ) sin 2 q cos 2 q
n
91. Q =
å
sin(3 r q) cos(3 r q)
r = 0 cos(3
r
q) cos(3 r +1 q)
=
1 n
1
tan(3 r +1 q) - tan(3 r q) = P
2 r =0
2
å
92. When 270° < q < 360° , we have
qö
æ
2 (1 + cos q) = ç 2 cos 2 ÷
2
è
ø
which is non-negative. Now, the above equation can be written as
q
2 (1 + cos q) = 2 cos
2
= -2 cos
q
2
q
q
æ
ö
çQcos < 0 when 135° < < 180° ÷
2
2
è
ø
Now, let us consider that 2 + 2 (1 + cos q)
which is not-negative. That is,
2 + 2 (1 + cos q) = 2 - 2 cos
q
2
= 2 1 - cos
= 2 sin
= 2 sin
q
q
= 2 2 sin 2
2
4
q
4
q
4
93. We know that - 2 £ sin x + cos x £ 2
3p
When x = , we have sin x + cos x = - 2
4
q
135° q
æ
ö
< < 90° ÷
çQsin > 0 when
4
2
4
è
ø
307
Compound Angles
when x = -
3p
, we have y = - 2 + 1 < 0
4
which implies that options (1) and (2) are incorrect.
p
Now, at x = , we have sin x + cos x = 2
4
That is, (sin 4 x + cos 4 x) 2 ¹ 2. Therefore, y ¹ 2 + 2 for any x Î R .
which implies that option (4) is incorrect.
Note : The maximum value of sin x + cos x is
(sin 4 x + cos 4 x) 2 is 2, for x =
2 , for x =
p
and the maximum value of
4
p
.
16
94. (cos x + cos y) 2 + (sin x + sin y) 2 = ( - cos z ) 2 + ( - sin z ) 2
2 + 2 cos( x - y) = 1
sin 50° sin 70° + sin 10° sin 70° - sin 10° sin 50°
1
1
1
95.
+
=
sin 10° sin 50° sin 70°
sin 10° sin 50° sin 70°
1
(cos 20° - cos 120° + cos 60° - cos 80° - cos 40° + cos 60° )
2
=
1
sin 30°
4
1æ3
ö
ç + cos 20° - 2 cos 60° cos 20° ÷
2è2
ø
=
=6
1
sin 30°
4
Exercise-2 : One or More than One Answer is/are Correct
1. cot 12° cot 24° cot 48° [cot 28° cot(60°-28° ) cot(60°+28° )] = (cot 12° cot 48° )(cot 24° cot 84° )
=
cot 36° cot 72°
´
=1
cot 72° cot 36°
cot 4 x - 2(1 + cot 2 x) + a 2 = 0
2.
Þ
cot 4 x - 2 cot 2 x + a 2 - 2 = 0
Þ
(cot 2 x - 1) 2 = 3 - a 2
to have atleast one solution
3 - a2 ³ 0
Þ
a2 - 3 £ 0
a Î [- 3 , 3 ]
308
Solution of Advanced Problems in Mathematics for JEE
Integral values –1, 0, 1
Sum = 0
\
3. (A)
tan 1 > tan -1 1 Þ tan 1 >
p
4
(B)
sin 1 > cos 1
sin 57.3° > cos 57.3°
(C)
(not possible)
tan 1 < sin 1
Because tan 57.3 > 1 > sin 57.3°
p
(D)
cos 1 <
4
æpö
Þ
cos (cos 1) > cos ç ÷
è4ø
4. (A) tan 1 > 1 and sin 1 < 1, then log sin 1 tan 1 < 0
(B) 1 + tan 3 < 1 and cos 1 < 1, then log cos1 (1 + tan 3) > 0
(C) cos q + sec q > 2 and log 10 5 < 1, then log log 10 5(cos q + sec q) < 0
(D) 2 sin 18° < 1 and tan 15° < 1, then log tan 15° 2 sin 18° > 0
æaö
æaö
2 tanç ÷
1 - tan 2 ç ÷
è2ø ,
è2ø
5. Put sin a =
cos a =
a
æ
ö
æaö
1 + tan 2 ç ÷
1 + tan 2 ç ÷
è2ø
è2ø
sin(2a + b) 3
6. Given
=
sin b
1
Option (C)
sin(2a + b) + sin b 3 + 1
=
sin(2a + b) - sin b 3 - 1
(Use C and D method)
tan(a + b) = 2 tan a
Option (B)
3 sin b = sin(2a + b)
2 sin b = sin(2a + b) - sin b
2 sin b = 2 cos(a + b) sin a
Option (D)
3 sin b = sin{a + (a + b)}
3 sin b = sin a cos(a + b) + cos a sin(a + b)
Subtract from (B) option
2 sin b = cos a sin(a + b)
cos b
cos(2a + b)
Option (A) cot b - 3 cot(2a + b) =
-3
sin b
sin(2a + b)
cos b
cos(2a + b) 2 sin(a + b) sin a
=
-3
=
= 4 tan a (from D)
sin b
3 sin b
sin b
Also
cot a + cot(a + b) =
3
cot a
2
(from C)
309
Compound Angles
Now multiply the two relations.
7.
Þ
8.
sin( x + 20° ) = sin( x + 40° ) + sin( x - 40° )
sin( x + 20° ) - sin( x - 40° ) = sin( x + 40° )
cos( x - 10° ) = sin( x + 40° ) = cos[90°-( x + 40° )]
x = 30° now check the option, only (a) and (b) satisfy
2S(cos x cos y) + 2S(sin x sin y) + 3 = 0
(S cos x) 2 + (S sin x) 2 = 0
Þ
S cos x = 0
and
S sin x = 0
cos 3 x + cos 3 y + cos 3 z = 4(cos 3 x + cos 3 y + cos 3 z ) - 3(cos x + cos y + cos z )
= 12 cos x cos y cos z
9. 0 < sin x < 1, 0 < cos x < 1
If
sin n x + cos n x = 1
n =2
n
n
n<2
n
n
n>2
sin x + cos x > 1
sin x + cos x < 1
10. If
Þ
Þ
If
x = sin(a - b) sin( g - d)
2 x = cos(a - b - g + d) - cos(a - b + g - d)
y = sin(b - g) sin(a - d)
2 y = cos(b - g - a + d) - cos(b - g + a - d)
z = sin( g - a ) sin(b - d)
2z = cos( g - a - b + d) - cos( g - a + b + d)
2 x + 2 y + 2z = 0
Þ x + y + z =0
x + y + z = 0 then x 3 + y 3 + z 3 = 3 xyz
X 2 + 4 XY + Y 2 = ( x cos q - y sin q) 2 + ( x sin q + y cos q) 2
11.
+ 4( x cos q - y sin q)( x sin q + y cos q)
2
2
2
2
= x + y + 4 { x sin q cos q - y sin q cos q + xy(cos 2 q - sin 2 q)}
= x 2 (1 + 4 sin q cos q) + y 2 (1 - 4 sin q cos q) + 4 xy(cos 2 q - sin 2 q)
Þ
cos 2 q - sin 2 q = 0
p
q=
4
( 0 £ q £ p / 2)
x 2 + 4 XY + Y 2 = 3 x 2 - y 2
Þ A = 3 and B = -1
12. (A)
2( a + d) = 2( b + c)
1
1 - tan 2 20°
=
tan 40°
2 tan 20°
(B)
tan 50° =
Þ
tan 20° + 2 tan 50° = tan 70°
310
Solution of Advanced Problems in Mathematics for JEE
2a + 2b = 2c
Þ
(D)
tan 20°-2 tan 10° = tan 20° tan 2 10° > 0
tan 20° > 2 tan 10°
b > a and d > c
1
1
1
13. (A) (2 sin 75° cos 75° ) = sin 150° =
2
2
4
Þ
Þ
7
(B) log 28
2 = 2 + log 2 (irrational)
(C) log 53 × log 65 = log 63 = 1 + log 23 (irrational)
3
(D) 8 - log 27 = 8 -1/ 3 =
1
2
14. a - b = sin x cos x(cos 2 x - sin 2 x) =
a + b = sin x cos x =
15.
1
1
sin 2 x cos 2 x = sin 4 x
2
4
1
sin 2 x
2
2 + 2 + 2 cos 4q = 2 + 2(1 + cos 4q)
= 2 + 4 cos 2 2q
= 2 + 2 |cos 2q|
If p < 2q < 3 p / 2 then
p
3p
<q<
2
4
2 + 2 |cos 2q| = 2 - 2 cos 2q = 2 |sin q|= 2 sin q
If
3p
3p
< 2q < 2 p then
<q< p
2
4
2 + 2 |cos 2q| = 2 + 2 cos 2q = 2 |cos q|= -2 cos q
16. 1 + tan a + tan 2 a = tan 3 a
1 + tan 2 a = tan a(tan 2 a - 1)
1
1
1
18. a >
Þa>
; 1£
£4
6
6
2
2
2
sin x + cos x
1 - 3 sin x cos x
1 - 3 sin x cos 2 x
19. log 10 sin x + log 10 cos x + 2 log 10 cot x + log 10 tan x = -1
Þ
log 10 (sin x × cos x × cot x) = k = log 10 cos 2 x = -1
20. tan A + tan B + tan C = tan A tan B tan C
3
6
3
6
+
+ tan C =
×
× tan C
tan C tan C
tan C tan C
Þ
tan 2 C = 9 Þ tan C = 3
311
Compound Angles
21.
(1 - cot x)
2
sin x
= (1 - cot x) × cosec 2 x
= (1 - cot x)(1 + cot 2 x)
22. f( x) =
23. y =
1é
2p ö
4p öù
æ
2æ
2 sin 2 x + 2 sin 2 ç x +
÷ + 2 sin ç x +
÷
ê
2ë
3 ø
3 ø úû
è
è
tan x 1 - 3 tan 2 x
=
tan 3 x 3 - tan 2 x
tan 2 x =
24.
1-3y
>0
3-y
2 sin( A - B ) = cos B (sin B - sin 3 B ) - sin B (cos B + cos 3 B )
= -sin B cos B
1
sin 2 B
= - sin 2 B Þ sin( A - B ) = 2
2 2
25. a >
1
6
6
1
Þa>
sin x + cos x
2
2
1 - 3 sin x cos x
; 1£
1
2
1 - 3 sin x cos 2 x
26. 1 + tan a + tan 2 a = tan 3 a
Þ
1 + tan 2 a = tan a(tan 2 a - 1)
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
1. q = 286.5° (IV quadrant) l < 0, m > 0
2. tan( -1042° ) = - tan(1080° - 38° ) = tan 38° < tan 45°
3. q = 4011
. ° (I quadrant) l > 0, m > 0
Paragraph for Question Nos. 4 to 6
a = sin a
2p ö
æ
b = sin ç a +
÷
3 ø
è
4p ö
æ
c = sin ç a +
÷
3 ø
è
p = cos a
2p ö
æ
q = cos ç a +
÷
3 ø
è
4p ö
æ
r = cos ç a +
÷
3 ø
è
£4
312
Solution of Advanced Problems in Mathematics for JEE
2p ö
4p ö
æ
æ
4. a + b + c = sin a + sin ç a +
÷ + sin ç a +
÷
3 ø
3 ø
è
è
æpö
= sin a + 2 sin(a + p) cos ç ÷ = 0
è3ø
2p ö
2p ö
4p ö
4p ö
æ
æ
æ
æ
5. ab + bc + ac = sin a sin ç a +
÷ + sin ç a +
÷ sin ç a +
÷ + sin a sin ç a +
÷
3 ø
3 ø
3 ø
3 ø
è
è
è
è
=
1é
2p
2p ö
2p
4p
4 p ö ù -3
æ
æ
cos
- cos ç 2a +
- cos(2a + 2 p) + cos
- cosç 2a +
÷ + cos
÷ =
2 êë
3
3
3
3
3 ø úû 4
è
ø
è
2p ö
4p ö
4p ö
2p ö
2p
3
æ
æ
æ
æ
6. qc - rb = cos ç a +
=
÷ sin ç a +
÷ - cos ç a +
÷ sin ç a +
÷ = sin
3 ø
3 ø
3 ø
3 ø
3
2
è
è
è
è
Paragraph for Question Nos. 7 to 8
7.
tan A = 7 + 4 3 = cot C
A
tan A + cot C = 2 7 + 4 3
= 2(2 + 3 ) = 4 + 2 3
= 3 +1
1
B
æ 2 + 6ö
æ AC ö
÷ = log 2 = 1
8. log AE ç
÷ = log 2 çç
÷
2
CD
è
ø
è 1+ 3 ø
Paragraph for Question Nos. 9 to 10
9. In a DABC , cot A + cot B + cot C ³ 3 Þ cot q ³ 3
10. cot q - cot A = cot B + cot C Þ sin( A - q) =
sin( B - q) =
sin 2 A sin q
sin B sin C
sin 2 B × sin q
sin 2 C sin q
and sin(C - q) =
sin A sin C
sin A sin B
Paragraph for Question Nos. 11 to 12
x
x
+ sin
2
2
11. f( x) =
x
x
cos
- sin
2
2
cos
p x
12. If < < p Þ f( x) =
2 2
x
x
x
+ sin
1 - tan
2
2 =
2
x
x
x
- cos - sin
1 + tan
2
2
2
- cos
F 6 + 2
E
2
1
D
C
313
Compound Angles
Exercise-4 : Matching Type Problems
1. (A) If A + B = 45° then (1 + tan A) (1 + tan B ) = 2
(B)
a 2 - 5a £ 6 sin x " x Î R
a 2 - 5a £ -6
a 2 - 5a + 6 £ 0
Þ ( a - 3) ( a - 2) £ 0
4
1ö
1
æ 4
ö
æ
ç a + ÷ - ç a + 4 + 2÷
a
è
ø
a
è
ø
(C)
2
4
=
1ö
1 ö
æ 2
æ
ça + ÷ -ça + 2 ÷
a
è
ø
a ø
è
2
2
1ö
1
æ
2
ça + ÷ + a + 2
a
è
ø
a
1ö
1
æ
2
ça + ÷ + a + 2
a
è
ø
a
2
1ö
1 ö
æ
æ
= ç a + ÷ - ç a2 +
÷ =2
aø
è
a2 ø
è
3
(D)
å( x - k) 2 = ( x - 1) 2 + ( x - 2) 2 + ( x - 3) 2 = 0 No real root
k =1
2. (A) y =
1 - tan 2 ( p / 4 - x)
1 + tan 2 ( p / 4 - x)
= cos( p / 2 - 2 x) = sin 2 x
æ 5 sin x - 12 cos x + 26 ö
(B) 0 £ log 3 ç
÷£1
13
è
ø
2
1ö
7
æ
(C) y = -2 sin 2 x + cos x + 3 = 2 cos 2 x + cos x + 1 = 2 ç cos x + ÷ +
4
8
è
ø
(D) y = 4 sin 2 q + 4 sin q cos q + cos 2 q = (2 sin q + cos q) 2
4. (A)
Þ
Þ
(B)
Þ
æ1
ö
cos 2 x = ç - sin x ÷
è5
ø
2
(5 sin x - 4) (5 sin x + 3) = 0
4
3
or sin x =
5
5
q
cot = 1 + cot q
2
q
2 cos 2 = cos q + sin q
2
3p p
,
2 2
Þ
sin q = 1
(C)
f( x) = - sin 4 x + 8 sin 2 x + 2
Þ
f( x) Î [2 , 9]
(D)
log 2
Þ
q=-
(2 x 2 + 5 x + 27)
(2 x - 1)
2
³0
1ö
æ
çx> ÷
2
è
ø
314
Solution of Advanced Problems in Mathematics for JEE
2 x 2 - 9 x - 26 £ 0
13
-2 £ x £
2
Þ
Þ
5. (A) f( x) = -2 sin 2 x + sin x - 6
at sin x = -1
y min = -9
47
1
at sin x =
y max = 8
4
(B)
f( x) = 2 cos 2 x + 6
y min = 6; y max = 8
1
(C) f( x) = [4 sin 2 x - 1 + cos 2 x + 3 (1 + cos 2 x)]
2
1
= [2 + 4 sin 2 x + 4 cos 2 x]
2
= 1 + 2 (sin 2 x + cos 2 x)
y max = 1 + 2 2 ;
y min = 1 - 2 2
æp
ö
(D) f( x) = 2 sin ç + sin x ÷
è4
ø
Exercise-5 : Subjective Type Problems
1.
sin 80° sin 65° sin 35°
sin 80° sin 65°
sin 80° sin 65°
1
=
=
=
2 sin 35° cos 15°+2 sin 35° cos 35° 2(cos 15°+ cos 35° ) 4 cos 25° cos 10° 4
2. If A + B = 45°
Þ
3.
(1 - cot A) (1 - cot B ) = 2
(1 - cot 23° ) (1 - cot 22° ) = 2
4x2 - 7x + 1 = 0
7
tan A + tan B =
4
1
tan A × tan B =
4
tan( A + B ) =
tan A + tan B
7
=
1 - tan A + tan B 3
4 sin 2 ( A + B ) - 7 sin( A + B ) cos( A + B ) + cos 2 ( A + B )
=
4 tan 2 ( A + B ) - 7 tan( A + B ) + 1
1 + tan 2 ( A + B )
=1
315
Compound Angles
4.
(18 - 2) ´ 180° (n - 2) ´ 180°
+
+ 60° = 360° Þ n = 9
18
n
5. 10 (1 - cos 2a ) 2 + 15 (1 + cos 2a ) 2 = 24
1
3
Þ tan 2 a =
5
2
3p
pö
5p
3p ö
æ 3p p ö æ
æ 5p 3 p ö æ
6. tan ç
- ÷ ç tan
- tan ÷ + tan ç
- tan
÷ ç tan
÷
8øè
8
8ø
8 øè
8
8 ø
è 8
è 8
Þ
(5 cos 2a + 1) 2 = 0
Þ
cos 2a = -
7p
3p ö
9p
7p ö
9p
p
æ 7 p 5p ö æ
æ 9p 7p ö æ
+ tan ç
- tan
- tan
- tan = 0
÷ ç tan
÷ + tan ç
÷ ç tan
÷ = tan
8
8
8
8
8
8
8
8
8
8
è
øè
ø
è
øè
ø
2
p
p
p
2
p
p
æ
ö
æ
ö
2p
p 4 cos
+ 2 cos 2 ÷ sin
4 ç 1 + 2 cos
ç
÷ sin
cos
+ 2 cos 2
7
7ø
7
7 ø
7
è
è
7
7
7.
=
=
p
2p
4p
3p
cos cos
sin
sin
7
7
7
7
2p ö
p
æ
4 ç 1 + 2 cos
÷ sin
7 ø
7
è
=
=4
pæ
pö
sin ç 3 - 4 sin 2 ÷
7è
7ø
a 2 sec 2 200° = c 2 tan 2 200° + d 2 + 2 cd tan 200°
8.
b 2 sec 2 200° = c 2 + d 2 tan 2 200° - 2 cd tan 200°
Þ
a 2 + b2 = c 2 + d 2
( a sec 200° - c tan 200° ) 2 = d 2
( b sec 200° + d tan 200° ) 2 = c 2
Þ
Þ
( c 2 + d 2 ) (sec 2 200° + tan 2 200° ) + (2 bd - 2 ac) sec 200° tan 200° = c 2 + d 2
( c 2 + d 2 ) (2 tan 2 200° ) = (2 ac - 2 bd) sec 200° tan 200°
2( c 2 + d 2 ) 2 sec 200°
2
-2
=
=
=
ac - bd
tan 200° sin 200° sin 20°
p
9p
7p
9p
10 p
8p
7p
9p
9. 2 cos
cos
+ cos
+ cos
= cos
+ cos
+ cos
+ cos
=0
17
17
17
17
17
17
17
17
cos q - cos 3q + cos 3q - cos 9q + cos 9q - cos 17q cos q - cos 17q
10.
=
= tan 9q
sin 3q - sin q + sin 9q - sin 3q + sin 17q - sin 9q
sin 17q - sin q
Þ
11. 8 abc = 8 sin 10° sin 50° sin 70° = 1
a + b sin 10°+ sin 50° 2 sin 30° cos 20°
=
=
=1
c
sin 70°
sin 70°
1 1 1
1
1
1
sin 50° sin 70°+ sin 10° sin 70°- sin 10° sin 50°
+ - =
+
=
=6
a b c sin 10° sin 50° sin 70°
sin 10° sin 50° sin 70°
316
12.
Solution of Advanced Problems in Mathematics for JEE
1é
2p ö
4p öù
æ
3æ
4 sin 3 q + 4 sin 3 ç q +
÷ + 4 sin ç q +
÷
ê
4ë
3 ø
3 ø úû
è
è
=
ìæ
üù
1é
2p ö
4p ö
æ
÷ - sin(3q + 2 p) + 3 sin ç q +
÷ - sin(3q + 4 p)ýú
ê3 sin q - sin 3q + 3 sin íç q +
4ë
3 ø
3 ø
è
îè
þû
=
ù
1é ì
2p ö
4p öü
3
æ
æ
÷ + sin ç q +
÷ ý - 3 sin 3q ú = - sin 3q
ê3 ísin q + sin ç q +
4ë î
3 ø
3 øþ
4
è
è
û
n
13.
n
sin(2 r - 2 r -1 )
å cos 2 r cos 2 r -1 = å(tan 2 r - tan 2 r -1 ) = tan 2 n - tan 1
r =1
r =1
14. x = sec q - tan q, y = cosec q + cot q
1 + cos q 1 - sin q (1 - sin q)(1 + cos q)
y - x - xy =
=1
sin q
cos q
sin q × cos q
15.
cos 18° - cos 72° = 2 sin 45° sin 27°
= 2 sin 27°
4
16. 3(sin 1 - cos 1) + 6(sin 1 + cos 1) 2 + 4(sin 6 1 + cos 6 1)
= 3(1 - 2 sin 1cos 1) 2 + 6(1 + 2 sin 1cos 1) + 4(1 - 3 sin 2 1cos 2 1)
= 3(1 + 4 sin 2 1cos 2 1 - 4 sin 1cos 1) + 10 + 12 sin 1cos 1 - 12 sin 2 1cos 2 1
= 13
33
2
17. 3 sin 2 x + 2 cos x +
2
2
3 sin 2 x + 2 cos x
Let 3 sin 2 x + 2 cos x = t,
= 28
t 2 - 28t + 27 = 0 Þ t = 1, 27
If t = 1 Þ sin 2 x + 2 cos 2 x = 0
2 cos x(sin x + cos x) = 0
Þ x=
p 3p 7p
,
,
2 4 4
If t = 27
Þ sin 2 x + 2 cos 2 x = 3
(Not possible)
(sin 2a - cos 2a ) 2 + 8 sin 4a = 1 + 7 sin 4a = 1
(at a =
p 3p 7p
,
, )
2 4 4
18. (sin q + cosec q) 2 + (cos q + sec q) 2
= 5 + cosec 2 q + sec 2 q
= 7 + tan 2 q + cot 2 q
³9
19. tan 20° + tan 40° + tan 80° - tan 60°
sin 20° cos 80° + sin 80° cos 20° sin 40° cos 60° – sin 60° cos 40°
=
+
cos 20° cos 80°
cos 40° cos 60°
317
Compound Angles
sin 100°
sin 20°
cos 20° cos 80° cos 40° cos 60°
sin 80°
2 sin 20°
=
cos 20° cos 80° cos 40°
sin 80° cos 40° - sin 40° cos 80°
sin 40°
=
=
cos 20° cos 40° cos 80°
cos 20° cos 40° cos 80°
8 sin 40° sin 20°
=
= 8 sin 40°
sin(8 ´ 20° )
=
1 + cos 10 x cos 6 x = 2 cos 2 8 x + sin 2 8 x
20.
2 + cos 16 x + cos 4 x = 2(1 + cos 16 x) + 1 - cos 16 x
cos 4 x = 1
np
x=
(n = 0 , ± 1, ± 2 , ± 3 .......)
2
Þ
If 360° < k < 540°
k = 450° (n = 5)
Þ
cos 20°+2 sin 2 55° = 1 + 2 sin k°
21.
= cos 20° + 1 - cos 110°
= 1 + cos 20° + sin 20°
= 1 + 2 sin( 45°+20° )
k = 65
cos 96°+ cos 6°
2 cos 51° cos 45°
23. tan 19 x =
== - cot 51° = tan 141°
cos 96°- cos 6°
2 sin 51° sin 45°
Þ
Þ 19 x = 180° n + 141
2 sin 40° + sin 20° 2 sin(60° - 20° ) + sin 20°
24.
=
cos 20° cos 30°
cos 20° cos 30°
3p
sin
2p
4p
6p
7 × cos 4 p - 1 = -3
25. cos
+ cos
+ cos
-1=
p
7
7
7
7
2
sin
7
k
11
26. (cos 2 A - cos 3 A) =
2
8
k
11
[2 cos 2 A - 1 - 4 cos 3 A + 3 cos A] =
2
8
Þ
k=4
27. 3 sin 2 x + 4 cos 2 x = 3 + cos 2 x
28. tan a + tan b = 12
tan a × tan b = -3
318
Solution of Advanced Problems in Mathematics for JEE
tan(a + b) =
tan a + tan b
=3
1 - tan a tan b
æ sin 18° cos 9°
ö
+ç
- cos 18° ÷
sin 9°
ø
2 sin 33° sin 57° è
cos 24° cos 33° sin 9°
+
=2
sin 57° cos 24° sin 9°
æ 1 + cos 2q ö æ 1 + cos 4q ö æ 1 + cos 8q ö
30. tan q ç
֍
֍
÷
è cos 2q ø è cos 4q ø è cos 8q ø
29.
cos 24° cos 33°
2
sin q æç 2 cos 2 q ö÷ æç 2 cos 2 2q ö÷ æç 2 cos 2 4q ö÷ 8 sin q cos q cos 2q cos 4q sin 8q
=
=
= tan 8q
cos q çè cos 2q ÷ø çè cos 4q ÷ø çè cos 8q ÷ø
cos 8q
cos 8q
31. y = sin 2 x + cos 2 x + tan 2 x + cot 2 x + cosec 2 x + sec 2 x + 6
y = 9 + 2(tan 2 x + cot 2 x) ³ 13
❑❑❑
319
Trigonometric Equations
23
TRIGONOMETRIC EQUATIONS
Exercise-1 : Single Choice Problems
1. tan 2 x - sec 2 y =
5a
- 3 = -2 - a 2
6
Þ 6 a 2 + 5a - 6 = 0
2. [tan( x + y) - cot( x + y)]2 + ( x + 1) 2 = 0
Þ
x = -1 and tan 2 ( x + y) = 1
x + y = np ±
3.
p
4
sin x + cos x = 1
pö
1
æ
sin ç x + ÷ =
4ø
è
2
x = np + ( -1) n
p p
4 4
4. sin 2 (sin x) - 3 sin(sin x) + 2 = 0
{sin(sin x) - 2}{sin(sin x) - 1} = 0
Equation has no solution.
5. tan 2 x = tan 6 x Þ sin 4 x = 0
4 x = p , 2 p , 3 p , ¼¼ , 11p
p 2p 3p
11p
x= ,
,
, ¼¼ ,
4 4 4
4
p 3 p 5p 7 p 9 p 11p
But ,
are rejected. So number of solutions = 5.
,
,
,
,
4 4 4 4 4
4
6. 3 sin 2 x - 6 sin x - sin x + 2 = 0
(3 sin x - 1)(sin x - 2) = 0
sin x ¹ 2 , then
1
sin x =
3
320
Solution of Advanced Problems in Mathematics for JEE
sin x =
1
has 6 solutions for x Î[0 , 5p]
3
7.
cos q + cos 2q = -1
Þ
2 cos 2 q + cos q = 0
Þ
cos q = 0 or cos q = -
Þ
8.
q=
1
1/2
p 2p 4p 3p
,
,
,
2 3 3 2
2p
p
0
1
2
3p/2
p/2
in [0 , 2p) max. (sin x , cos x) =
x
1
has two solutions.
2
9. (cot 2 x + 2 3 cot x + 3) + (cot 2 x + 1) + ( 4cosec x + 4) = 0
(cot x + 3 ) 2 + (cosec x + 2) 2 = 0
cot x = - 3
Þ
2
10.
and
cosec x = -2
2
sin x = sin 3 x
3 x = np ± x
np
np
np
x=
, x=
, hence general solution is
.
4
2
4
Þ
11. sin x > 0
Þ
8 sin 2 x cos 2 x = 1
Þ
2 sin 2 2 x = 1
Þ
cos 4 x = 0
x = (2n + 1)
12.
Þ
Þ
p
(n Î I )
8
cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x = 5
cos x = 1 Ç cos 2 x = 1 Ç cos 3 x = 1 Ç cos 4 x = 1 Ç cos 5 x = 1
2np
2np
2np
x = 2np Ç x = np Ç x =
Çx=
Çx=
3
4
5
x = 2np
13. (2 sin x - cosec x) 2 + (tan x - cot x) 2 = 0
Þ
sin 2 x =
1
p
Ç tan 2 x = 1 Þ x = np ±
2
4
321
Trigonometric Equations
cos 3 3 x + cos 3 5 x = (2 cos 4 x cos x) 3
14.
cos 3 3 x + cos 3 5 x = (cos 5 x + cos 3 x) 3
3 cos 5 x cos 3 x(cos 5 x + cos 3 x) = 0
cos 5 x cos 3 x × cos 4 x cos x = 0
Þ
Þ
15. sin 100 x = 1 + cos 100 x Þ sin 100 x = 1 and cos 300 x = 0
p
16. sin q £ 1 and sec 2 4q ³ 1 Þ sin q = sec 4q = 1; q =
2
17. ( 4 sin 2 x + cosec 2 x) + (tan 2 x + cot 2 x) = 6
(2 sin x - cosec x) 2 + (tan x - cot x) 2 = 0
2 sin x = cosec x
Þ
4
and
tan x = cot x
2
18. sin q - 2 sin q + 1 = 2
(sin 2 q - 1) 2 = 2 = cos 4 q
(not possible)
19. cos( P sin x) = sin( P cos x)
æp
ö
cos( P sin x) = cos ç - P cos x ÷
2
è
ø
p
2
p
P sin x - P cos x = 2np 2
P sin x + P cos x = 2np +
20.
| x| + | y| = 2
æ px 2 ö
÷ =1
sin ç
ç 3 ÷
è
ø
–2 £ x £ 2
–2 £ y £ 2
px 2 p
=
3
2
x =±
3
2
æ p ö
21. x Î ç - , p ÷
è 2 ø
cos 2 x > |sin x|
sin x ³ 0
sin x < 0
2
1 - 2 sin x - sin x > 0
2
2 sin 2 x - 2 sin x + sin x - 1 < 0
(2 sin x - 1)(sin x + 1) < 0
(2 sin x + 1)(sin x - 1) < 0
2 sin x + 2 sin x - sin x - 1 < 0
–1
0
1/2
–1/2
0
1
322
Solution of Advanced Problems in Mathematics for JEE
-
1
1
< sin x £
2
2
æ p p ö æ 5p ö
, p÷
ç- , ÷ Èç
è 6 6ø è 6
ø
22. sin 4 x + cos 4 x = sin x cos x
1 - 2 sin 2 x cos 1 x = sin x cos x
2y2 + y -1=0
(2 y - 1)( y + 1) = 0
1
y=
2
2 sin x cos x = 1
y = -1
sin x cos x ¹ -1
sin 2 x = 1
p 5p
2x = ,
2 2
p 5p
x= ,
4 4
23. sin
5x
x
= 1 Ç sin = -1
2
2
24. cos 2q = sin 2 q Þ sin 2 q =
1
1
Þ sin q = ±
3
3
25. b sin q = -c - a cos q
b 2 (1 - cos 2q) = c 2 + a 2 cos 2q - 2 ac cos q
Þ ( a 2 + b 2 ) cos 2q - 2 ac cos q + ( c 2 - b 2 ) = 0
cos a × cos b =
c 2 - b2
…(1)
a 2 + b2
a 2 (1 - sin 2q) = c 2 + b 2 sin 2q - 2 bc sin q
( a 2 + b 2 ) sin 2q - 2 bc sin q + ( c 2 - a 2 ) = 0
sin a × sin b =
cos(a + b) = cos a cos b - sin a sin b =
c2 - a2
a 2 + b2
a 2 - b2
a 2 + b2
…(2)
323
Trigonometric Equations
Exercise-2 : One or More than One Answer is/are Correct
1. 2 cos 2 q + 2 2 cos q - 3 = 0
( 2 cos q + 1) 2 = 4 Þ cos q =
3.
1
2
or
-3
2
(Not possible)
4 sin 3 x + 5 ³ 4 cos 2 x + 5 sin x
Þ (sin x - 1)( 4 sin x + 1) 2 £ 0 " x Î R
4. 4 cos x (2 - 3 sin 2 x) + cos 2 x + 1 = 0
cos x (3 cos x + 2)(2 cos x - 1) = 0
p
Least difference =
6
5.
cos x cos 6 x = - 1
Case-1 :
cos x = 1 and cos 6 x = - 1
Not possible
Case-2 :
cos 6 x = 1 and cos x = -1
x = (2n - 1) p , (n Î I )
Þ
2 k = sin 2 2 x - 2 sin 2 x - 2
7.
Let
sin 2 x = t
t Î [-1, 1]
2 k = t 2 - 2t - 2
Þ
é 3 1ù
k Î ê- , ú
ë 2 2û
pöæ
3p ö æ
5p ö æ
7p ö
1
æ
4
2
8. f(q) = ç cos q - cos ÷ ç cos q - cos
÷ ç cos q - cos
÷ ç cos q - cos
÷ = cos q - cos q +
8
8
8
8
8
è
øè
øè
øè
ø
9.
4 sin 2 x cos 2 x + 4 sin 4 x - 4 sin 2 x cos 2 x
2
2
2
4 cos x - 4 sin x cos x
Þ
tan x = ±
= tan 4 x =
1
9
1
3
10. tan q(1 - sin 2 q) + cot q(1 - cos 2 q) + 1 + sin 2q = 0 Þ sin 2q = æ1
ö
3
11. 2 çç sin q +
cos q ÷÷ = - ( x - 3) 2 - 2
2
è2
ø
1
2
324
Solution of Advanced Problems in Mathematics for JEE
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
2
2
1. h( x) = f ( x) + g ( x) = 2 + 2 sin 4 x
8 cos 4 x ³ 0
cos 4 x ³ 0
p
Longest interval =
4
Þ
Þ
2. 2 + 2 sin 4 x = 4
Þ
sin 4 x = 1
x = ( 4n + 1)
Þ
3.
p
8
sin 3 x + cos x = cos 3 x + sin x
sin 3 x - sin x = cos 3 x - cos x
2 sin x cos 2 x = -2 sin 2 x sin x
or tan 2 x = -1
sin x = 0
3p 7p
or
x = 0, p
x=
,
8 8
Þ
Þ
Þ
Þ
Exercise-4 : Matching Type Problems
æ1
ö
cos x = ç - sin x ÷
è5
ø
2
2
1. (A)
Þ
Þ
(B)
Þ
Þ
(5 sin x - 4)(5 sin x + 3) = 0
4
or
sin x =
5
q
cot = 1 + cot q
2
q
2 cos 2 = cos q + sin q
2
3p p
sin q = 1 Þ q = ,
2 2
f( x) = - sin 4 x + 8 sin 2 x + 2
(C)
Þ
(D) log 2
Þ
f( x) Î [2 , 9]
(2 x 2 + 5 x + 27)
(2 x - 1)
2
³0
2 x 2 - 9 x - 26 £ 0
1ö
æ
çx> ÷
2ø
è
-
3
5
325
Trigonometric Equations
-2 £ x £
Þ
13
2
2. (A) sin x = 1, cos y = 1 or sin x = -1, cos y = -1
(B) f ¢( x) = cos x + sin x - K
k³ 2
Þ
2
(C) | x - 1| £ 1 and |2 x 2 - 5| £ 1
Þ
x2 =2
(D) sin x + sin y = sin( x + y)
æ x + yö
æ x - yö
æ x + yö
æ x + yö
Þ 2 sin ç
÷ cos ç
÷ = 2 sin ç
÷ cos ç
÷
è 2 ø
è 2 ø
è 2 ø
è 2 ø
Þ
Þ
if
if
if
if
æ x + yö
æ x - yö
æ x + yö
sin ç
÷ = 0 or cos ç
÷ = cos ç
÷
è 2 ø
è 2 ø
è 2 ø
np
np
or x =
x + y = 2np
, y=
2
2
x = 0, y = ± 1
1
1
x = , y =2
2
1
1
x =- , y =
2
2
y = 0, x = ± 1
Exercise-5 : Subjective Type Problems
1. Let
sin x - 1 = a , cos x - 1 = b, sin x = c
3
Þ
a + b 3 + c 3 = ( a + b + c) 3
Þ
a + b = 0 or b + c = 0 or c + a = 0
sin x + cos x = 2 or sin x + cos x = 1 or sin x =
Total solution = 5
Þ
2. sin y - 2014 cos y = 1
p
Þ
y=
2
2 sin 6 x
3.
<0
sin x - 1
Þ
Þ
sin 6 x > 0
æ pö æp pö
x Î ç 0, ÷ È ç , ÷
è 6ø è3 2ø
1
2
326
Solution of Advanced Problems in Mathematics for JEE
Þ
1 + tan 2 x - 2 2 tan x £ 0
é p 3p ù
é p p ö æ p 3p ù
Þ xÎê , ÷ Èç ,
xÎê ,
ú
ë8 8 û
ë 8 6 ø è 8 8 úû
4. sin 4 x - 4 sin 2 x + (2 + k) = 0
Let
sin 2 x = t
t Î[0 , 1]
t 2 - 4t + (2 + k) = 0
f(0) f(1) £ 0
( k + 2)( k - 1) £ 0 Þ - 2 £ k £ 1
5.
1
–p/2
–p/3
–p/6
O
p/6
p/3
p/2
–1
6. 2 sin 2 x + sin 2 2 x = 2
2 sin 4 x - 3 sin 2 x + 1 = 0 Þ (2 sin 2 x - 1)(sin 2 x - 1) = 0
sin 2 x + cos 2 x = tan x
2 tan x + 1 - tan 2 x = tan x(1 + tan 2 x)
Þ
tan 3 x + tan 2 x - tan x - 1 = 0 Þ (1 + tan x)(tan 2 x - 1) = 0
2 cos 2 x + sin x £ 2
2 sin 2 x - sin x ³ 0
sin x (2 sin x - 1) ³ 0
7. (3 cot q + 1)(cot q + 3) = 0
1
cot q = - and cot q = -3
3
p
p
q = a, p + a
q = - a, p + - a
2
2
8. (8 cos 4q - 3)(cot q - tan q) 2 = 12
æ 4 cos 2 2q ö
÷ = 12
8 (2 cos 2 2q - 1) - 3 ç
ç sin 2 2q ÷
è
ø
16 cos 4 2q - 8 cos 2 2q - 3 = 0
Þ ( 4 cos 2 2q - 3)( 4 cos 2 2q + 1) = 0
Þ
cos 2q = ±
3
2
327
Trigonometric Equations
9. 2 sin 2 x + 4 sin 2 x cos 2 x = 2
2 sin 4 x - 3 sin 2 x + 1 = 0 Þ (sin 2 x - 1)(2 sin 2 x - 1) = 0
1
sin x = ±
,±1
2
sin 2 x + cos 2 x = tan x
2 tan x
1 + tan 2 x
Þ
+
1 - tan 2 x
1 + tan 2 x
= tan x
tan 3 x + tan 2 x - tan x - 1 = 0
(tan x + 1) 2 (tan x - 1) = 0
2 cos 2 x + sin x £ 2
2 sin 2 x - sin x ³ 0
sin x(2 sin x - 1) ³ 0
❑❑❑
328
Solution of Advanced Problems in Mathematics for JEE
24
Exercise-1 : Single Choice Problems
1.
cot A + cot B cos A sin B + cos B sin A
sin 2 C
c2
=
=
=
cos C
cot C
(sin A sin B ) cos C
a 2 + b2 - c 2
(sin A sin B )
ab ×
sin C
2 ab
=
2. ÐBIC =
p æp - Aö
+ç
÷
2 è 2 ø
=
p æB +Cö
+ç
÷
2 è 2 ø
2c 2
18
=
8
æ 17
ö 2
- 1÷ c
ç
è 9
ø
A
H
I
B
1
3.
[(2 R cos A) 2 + a 2 ][(2 R cos B ) 2 + b 2 ][(2 R cos C ) 2 + c 2 ]
64
1
[(2 R cos A) 2 + (2 R sin A) 2 ][(2 R cos B ) 2 + (2 R sin B ) 2 ][(2 R cos C ) 2 + (2 R sin C ) 2 ]
64
=R6
4. B = 60°
2 sin 2 B = 3 sin 2 C Þ sin C =
1
2
Þ C = 45°
A
C 1
tan =
2
2 3
s-b 1
2
a+c
(A.M. ³ G.M.)
= Þ b= sÞ
= bÞ b³ 2
s
3
3
2
a
6. cos A cos B cos C
= 2 R cos A cos B cos C
tan A = 2 R cos A × cos B × cos C × Õ(tan A)
cos A
5. tan
å
å
C
329
Solution of Triangles
= 2 R cos A × cos B × cos C ×
sin A sin B sin C
= 2 R sin A sin B sin C
cos A cos B cos C
BD
AD
=
sin q sin 60°
CD
AD
In DCAD,
=
sin(75°-q) sin 45°
BD
CD sin 45°
Þ
sin 60° =
sin q
sin(75°- q)
8. In DBAD,
Þ
A
q
75°–q
sin q
BD sin 60°
1
=
=
sin(75°-q) CD sin 45°
6
B
2 bc
A
cos
b+ c
2
2 ac
B
Length of angle bisector BE =
cos
a+c
2
2 ab
C
Length of angle bisector CF =
cos
a+b
2
3
3
H. M. =
=
b+ c a+ c a+ b 1 1 1
+
+
+ +
2 bc
2 ac
2 ab
a b c
9. Length of angle bisector AD =
10. 2b = a + c
2 sin B = sin A + sin C
B
Bö
B
1
æ
æ A+Cö
æ A -C ö
2 ç 2 sin cos ÷ = 2 sin ç
÷ cos ç
÷ Þ sin =
2
2
2
2
2
2
2
è
ø
è
ø
è
ø
æ B - C ö b + c sin B + sin C
11. 2 cos ç
=
÷=
a
sin A
è 2 ø
A 1
Þ
sin =
Þ Ð A = 60°
2 2
4 + c2 - 1 1 æ
3ö
3
12. cos A =
= çc + ÷³
4c
4è
cø 2
I3
A
60°
60°
F
13.
60°
B
60°
60°
60°
60°
D
60°
I1
F
60°
C
I2
60°
1
60°+q
D
45°
3
C
330
Solution of Advanced Problems in Mathematics for JEE
If DABC is an equilateral triangle then DDEF and D I 1 I 2 I 3 are also equilateral triangle
Side of DDEF = 1 unit Þ Ar( DDEF ) =
3
4
1
x cos p = 1
AD =
1
1
3
x+
x+
x
x
1
AD max =
2
2x ×
14.
15. r =
A
X
120°
1/X
B
C
3a
3a
3a
,R =
, r1 =
Þ r , R , r1 are in A.P.
6
3
2
16. sin( B + C ) sin( B - C ) = sin( A + B ) sin( A - B )
sin 2 B - sin 2 C = sin 2 A - sin 2 B
Þ
2 sin 2 B = sin 2 A + sin 2 C
Þ
2b2 = a 2 + c 2
17.
(Using sine rule)
tan A + tan B
= tan( p - C )
1 - tan A tan B
7
7
tan C = Þ sin C =
4
65
tan( A + B ) =
Þ
Using sine rule
R=
18.
c
65
=
2 sin C 14
cos A cos B cos C
+
+
a
b
c
b 2 + c 2 - a 2 a 2 + c 2 - b 2 a 2 + b 2 - c 2 a 2 + b 2 + c 2 ( a + b + c) 2 - 2( ab + bc + ac)
+
+
=
=
2 abc
2 abc
2 abc
2 abc
2 abc
19.
a + c b + c a 2 + b 2 + ac + bc c 2 ( a + b + c) c 2 (2 s) 2 R c
+
=
=
=
=
=
b
a
ab
abc
4 RD
r
r
20. a 2 (sin B - 1) = b 2 + c 2 - a 2 = 2 bc cos A Þ cos A < 0
a
a
21. 2 R ¢ =
=
= 2R
sin( p - A) sin A
Þ
C
R¢ = R
O
A
B
331
Solution of Triangles
22.
a = d1 d 2
d1
d
= a cos q , 2 = a sin q
2
2
Þ
1 = 4 sin q cos q
1
sin 2q = Þ 2q = 30°
2
q
a
23. Q
(m + n) cot q = m cot a - n cot b
\
(2 + 1) cot q = 2 cot a - cot b
1
æp
ö
Put cot q = ,cot b = cot ç - a ÷ = tan a
3
2
è
ø
2
We have 1 =
- tan a Þ tan 2 a + tan a - 2 = 0
tan a
\
tan a = 1
B
l
l
=
2 sin 60°
3
Diagonal of square = 2 R Þ a 2 = 2 R \ a = R 2 =
25. cos q =
a b
\ a = 45°
24. Circumradius of equilateral D , R =
2
A
2
2 + ( 6 ) - ( 3 + 1)
2
4 6
=
l 2
3
q
D
2:1
\ Area of square =
C
2l 2
3
3 -1
2 2
26. If a , b, c are in A.P.
Þ
2 sin B = sin A + sin C Þ sin
B 1
=
2 4
s
=
r
6 cos
B
2
B
1 - 2 sin
2
= 3 15
æ A-Bö
1 - tan 2 ç
÷
è 2 ø = 31 Þ tan æ A - B ö = 1
27. cos( A - B ) =
ç
÷
æ A - B ö 32
è 2 ø 3 7
1 + tan 2 ç
÷
è 2 ø
C
1
æ A-Bö a-b
tan ç
cot
Þ cos C =
÷=
2
8
è 2 ø a+b
cos C =
a 2 + b2 - c 2 1
=
Þ c =6
2 ab
8
28. ( b + c) cos( B + C ) + ( c + a) cos(C + A) + ( a + b) cos( A + B )
= -[( b cos A + a cos B ) + ( c cos A + a cos C ) + ( c cos B + b cos C )] = -[a + b + c] = -30
332
Solution of Advanced Problems in Mathematics for JEE
30. Ð A =
p
2p
4p
,Ð B =
,ÐC =
7
7
7
æ
b 2 ö÷ æç
c 2 ö÷ æç
a 2 ö÷
( a 2 - b 2 )( b 2 - c 2 )( c 2 - a 2 ) = a 2 b 2 c 2 ç 1 11ç
a 2 ÷ø çè
b 2 ÷ø çè
c 2 ÷ø
è
2p ö æ
4p ö æ
p ö
æ
sin 2
sin 2
sin 2
ç
֍
֍
÷
7
7
7
2 2 2ç
÷ç1 ÷ç1 ÷ = a 2 b2 c 2
= a b c 1p
2
p
4
p
ç
֍
֍
÷
sin 2
sin 2
sin 2
ç
֍
֍
÷
7 øè
7 øè
7 ø
è
a
36. h =
2 3
Ar( DABC ) = Ar ( DAPB ) + Ar( DBPC ) + Ar( DAPC )
3 2 1
a = a( h + h1 + h2 )
4
2
2
2
Þ
h1 + h2 =
a
3
2
6 +7 -x
Þ x = 43
2 ´6 ´7
2 ab
p
39. CD =
cos
a+b
3
ab
=
a+b
42. a + b + c = 48
37. cos 60° =
B
p/3
p/3
C
a = 20
b + c = 28
Þ a + b > c, a + c > b
Þ 20 + b > 28 - b, 20 + c > 28 - c
Þ b> 4 ,c > 4
43. In an equilateral triangle
44.
D
a
b
A
a=b=c
( a + b + c)( b + c - a)( c + a - b)( a + b - c)
=
4b2 c 2
2 s(2 s - 2 a)(2 s - 2 b)(2 s - 2 c)
2 2
4b c
A
æ 2( s - a) ö æ ( s - b)( s - c) ö
2 A
= 4ç
cos 2 = sin 2 A
֍
÷ = 4 sin
bc
bc
2
2
è
øè
ø
45. R = 4 r
A
B
Cö
æ
R = 4 ç 4 R sin sin sin ÷
2
2
2ø
è
q
1 = 16 sin 2 × cos q = 8 (1 - cos q) cos q
2
A
q
B
q
p-2q
C
333
Solution of Triangles
46. DDMN ~
= DAMN Þ DM = AM
C
M
D
N
A
47.
B
a b c
b c a = - ( a 3 + b 2 + c 3 - 3 abc)
c a b
= - ( a + b + c)( a 2 + b 2 + c 2 - ab - bc - ac)
48. A =
p
2p
4p
,B =
,C =
7
7
7
2 öæ
2 öæ
2 ö
æ
ç1 - b ÷ç1 - c ÷ç1 - a ÷ = l
ç
a 2 ÷ø çè
b 2 ÷ø çè
c 2 ÷ø
è
2p ö æ
4p ö æ
p ö
æ
sin 2
sin 2
sin 2
ç
֍
֍
÷
7 ÷ ç1 7 ÷ ç1 7 ÷ =l
ç1 p ÷ç
2p ÷ ç
4p ÷
ç
sin 2
sin 2
sin 2
ç
֍
֍
÷
7 øè
7 øè
7 ø
è
2p ö æ
4p ö æ
pö
æ
2 p
2 2p
2 4p
- sin 2
- sin 2
- sin 2 ÷
ç sin
÷ ç sin
÷ ç sin
7
7 ֍
7
7 ֍
7
7 ÷ =l
ç
ç
÷
ç
÷
ç
÷
2 p
2 2p
2 4p
sin
sin
sin
ç
֍
֍
÷
7
7
7
è
øè
øè
ø
Þ
49. r1 =
l = 1(sin 2 A - sin 2 B = sin( A - B ) × sin( A + B ))
D
D
D
, r2 =
, r3 =
s-a
s-b
s-c
r1 r2 r3
=
s3
( s - a)( s - b)( s - c)
r3
s-a s-b s-c
+
+
s
s
s ³ æ s - a öæ s - böæ s - c ö
ç
֍
֍
÷
3
è s øè s øè s ø
3
B
50. sin A =
5
4
5
sin B =
3
5
sin C = 1
90°
C
4
A
334
Solution of Advanced Problems in Mathematics for JEE
A
B
C
A
B
C
cos cos + 4 R cos sin cos
2
2
2
2
2
2
C
2 cos 2
2
A
B
A
Bö
æ
2 R ç sin cos + cos sin ÷
2
2
2
2ø
è
=
= 2R
C
cos
2
r + r2
51. 1
=
1 + cos C
4 R sin
sin 2 a + cos 2 a - (1 + sin a cos a ) 1
=
2 sin a cos a
2
55. Since we need to compute the radius of an escribed circle, we would be needing the length of
all the sides of the given triangle ABC.
From the question, we already know AB = AC = 5.
For finding the length of side BC , let us draw a line AD which is the bisector of angle BAC , as
shown in the figure below.
53. cos q =
ÐBAD = ÐDAC = 15°
Therefore,
sin 15° =
y
C
BD BD
3 -1
and sin 15° =
=
AB
5
2 2
Therefore,
BD = 5 sin 15° =
We also know that
BC = 2 BD
Therefore,
BC =
B
5 ( 3 - 1)
2 2
A
N
x
5 ( 3 - 1)
2
Now, we know that the required radius
æ A ö æ AB + BC + CA ö
æ Aö
r1 = s tan ç ÷ = ç
÷ tan ç ÷
2
2
è ø è
ø
è2ø
æ
ö
5 ( 3 - 1)
ç5+
+ 5÷
æ 10 2 + 5 3 - 5 ö
ç
÷
2
÷ (2 - 3 )
=ç
(tan 15° ) = çç
÷
÷
2
2 2
è
ø
çç
÷÷
è
ø
a
- C cos B
2
æ a 2 + c 2 - b2 ö
a
÷
= -Cç
ç
÷
2
2 ac
è
ø
56. ED = BE - BD =
=
b2 - c 2
2a
A
C
E D
B
335
Solution of Triangles
57. 2R(sin A cos B cos C + cos A sin B cos C + cos A cos B sin C )
= 2R(sin( A + B ) cos C + cos A cos B sin C )
abc D rs
= R(2 sin A sin B sin C ) =
= =
4R 2 R R
58. In DAFE ,
b cos A
= 2R 1
sin B
A
Þ
R 1 = R cos A
Similarly, R 2 = R cos B
and
R 3 = R cos C
R 1 + R 2 + R 3 = R (cos A + cos B + cos C ) £
3
R
2
59. Ar ( DABC ) = Ar ( DOAB ) + Ar ( DOBC ) + Ar ( DOAC )
1
8 = R 2 (sin a + sin b + sin g)
2
4p
Þ
sin a + sin b + sin g =
5
Exercise-2 : One or More than One Answer is/are Correct
1. x 2 - r( r1 r2 + r2 r3 + r1 r3 ) x + ( r1 r2 r3 - 1) = 0
x 2 - ( r1 r2 r3 ) x + ( r1 r2 r3 - 1) = 0
Roots are 1 and r1 r2 r3 - 1
Þ
A
60°
2.
b
c
90°
B=H
30°
a
C
s
tan 30° Þ s is irrational Þ D is irrational
3
r1 = s tan 30° = 3 r (rational)
3. R = 2 r , r = ( s - a) tan 30° =
4. D + E + F =
p
2
5. a = 4 , b = 8 , Ð C = 60°
cos C =
1 a 2 + b2 - c 2
=
Þ c=4 3
2
2 ab
E
F
B
D
C
20 ö
æ
2
çQ R =
÷
p ø
è
336
Solution of Advanced Problems in Mathematics for JEE
6. If
Þ
r r2
s-a s-c
= Þ
=
r1 r3
s
s-b
a 2 + b2 = c 2
Þ
Ð C = 90°
7. Ð BOC = 2 Ð A
Ð BIC = p 2 + A 2
Ð BHC = p - A
8.
3x2 - 4x + 3 < 0
Þ ( 3 x - 1)( x - 3 ) < 0
1
Þ
30° < A , B < 60°
3
< x< 3
Þ 60° < C < 120°
2
9. cos 2q = 2 cos q - 1
1
= 2 cos 2
2
p
1
2 cos 2 = 1 +
8
2
cos 2
p
-1
8
p
2 +1
=
8
2 2
cos C =
a 2 + b2 - c 2
then solve it
2 ab
11. (3 sin A + 4 cos B ) 2 + ( 4 sin B + 3 cos A) 2 = 37 ; 9 + 16 + 24 sin( A + B ) = 37
12.
b+ c 2
=
a
1
A
B
13. Ð A , ÐB , ÐC
2
cos 60° =
2
b+c
1
a
C
A.P. Þ ÐB = 60°
2
a + 10 - 9
20 a
B
2
60°
C
1
ab sin C
2
a+b
sin A + sin B
³ ab Þ
³ sin A ´ sin B
2
2
14. D =
10
60°–q
60°+q
9
A
337
Solution of Triangles
15. 3 cos A = cos( B - C ) - cos( B + C ) Þ 2 cos A = cos B - C ) = - cos( A + 2C )
2 = (tan A sin 2C - cos 2C )
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 2
æp Aö
æp Bö
æp Cö
2. r ¢ = 4 r sin ç - ÷ sin ç - ÷ sin ç - ÷
4
4
4
4
è
ø
è
ø
è4 4ø
r¢
æp Aö
æp Bö
æp Cö
= 4 sin ç - ÷ sin ç - ÷ sin ç - ÷
r
è4 4ø
è4 4ø
è4 4ø
A
B
C
+ sin + sin - 1
2
2
2
æp Aö
æp Bö
æp Cö
r1¢ = 4 r sin ç - ÷ cos ç - ÷ cos ç - ÷
4
4
4
4
è
ø
è
ø
è4 4ø
A
F
p–C
2 2
= sin
r1¢
æp Aö
æp Bö
æp Cö
= 4 sin ç - ÷ cos ç - ÷ cos ç - ÷
r
è4 4ø
è4 4ø
è4 4ø
= 1 - sin
p–B
2 2
E
p–A
2 2
B
D
C
A
B
C
+ sin + sin
2
2
2
Paragraph for Question Nos. 3 to 4
3. Ar ( DDEF )
A
4.
Ar( DABC )
2 R 2 sin A sin B sin C
A
B
C
=
= 8 sin sin sin £ 1
A
B
C
Ar( DDEF )
2
2
2
2 R 2 cos cos cos
2
2
2
F
E
I
B/2
C/2
B
C
C/2
B/2
æp Aö
æp Bö
æp Cö
= 2 R 2 sin ç - ÷ sin ç - ÷ sin ç - ÷
2
2
2
2
è
ø
è
ø
è2 2ø
A
B
C
= 2 R 2 cos cos cos
2
2
2
D
Paragraph for Question Nos. 5 to 6
Sol. c 2 = R Þ c = 82
2r = a + b - c
a + b = 98
Þ
2
2
2
a + b = c = (82)
Þ
a = 18 , b = 80
...(1)
2
...(2)
338
Solution of Advanced Problems in Mathematics for JEE
Paragraph for Question Nos. 7 to 8
Sol. Ð A1 =
p A
2 2
Ð A2 =
p 1
p 1æ p Aö
- ( Ð A1 ) = - ç - ÷
2 2
2 2è2 2 ø
A
p A
+
4 4
p 1
Ð A3 = - (Ð A2 )
2 2
3p A
=
8
8
=
Ð An =
C1
B
B1
A1
C
n
pæ
1 1 1
ö ( -1) A
ç 1 - + - + ........ ÷ +
2è
2 4 8
ø
2n
Paragraph for Question Nos. 9 to 10
1
´ BD ´ h1
D1 2
h
h
Sol.
=
= (1 - x) 1 ´ 3 = (1 - x) yz
1
D
h2 h3
´ AB ´ h2
2
1
´ EC ´ h4
D2 2
=
= x(1 - y)(1 - z )
1
D
´ AC ´ h5
2
A
h1
D
Paragraph for Question Nos. 11 to 13
cö
æ
Sol. log ç 1 + ÷ + log a - log b = log 2
a
è
ø
a + c = 2b
( c - a) x 2 + 2 bx + ( a + c) = 0 has equal roots, then
a 2 + b2 = c 2
Paragraph for Question Nos. 14 to 16
Sol.
BE
ED
EC
ED
=
,
=
A sin B
A
sin C
sin
sin
2
2
E
F
h2
B
Þ
h3
C
339
Solution of Triangles
Þ
BE + EC = a =
ED
(sin B + sin C )
A
sin
2
A
´ 2R
2
ED =
b+ c
2 bc
A
la =
cos
b+ c
2
A
æ
ö
a sin ´ 2 R ÷
ç 2 bc
A
2
ç
÷
cos +
2
b+ c
çb+ c
÷
ç
÷
è
ø
2 sin B sin C
sin B sin C
la =
=
A
Aö
æ
2 sin B sin C + 2 sin 2
sin 2 ç B + ÷
2
2
è
ø
a sin
Þ
Þ
A
A/2
B
A/
2
E
A/2
A/2
C
C B
D
Exercise-4 : Matching Type Problems
2. (A) 3 0 {2 0 + 2 -1 + 2 -2 ¼¼¼¥} = 1{2}
1
{2}
3
1
3 -2 {2 0 + 2 -1 + 2 -2 ¼¼¼¥} = {2}
3
M
M
M
M
3 -1 {2 0 + 2 -1 + 2 -2 ¼¼¼¥} =
¥
Hence,
2 ´1
=3
1
13
(B) b 2 + c 2 - a 2 = 2 bc cos A = 54
bc cos A = 27 = a 3 Þ a = 3
b 2 + c 2 63
=
=7
9
9
(C) Circumcentre of DABC is ( -1, 0).
Point A lie on the circle ( x + 1) 2 + y 2 = 4 Þ x 2 + y 2 + 2 x - 3 = 0
(D) (cos q sin q + 6) = 6 (sin q - cos q) Þ 36 + sin 2 q cos 2 q + 12 sin q cos q = 36(1 - 2 sin q cos q)
Let sin q cos q = t
340
Solution of Advanced Problems in Mathematics for JEE
t 2 + 84t = 0 Þ t = 0
If sin q = 0 Þ cos q = -1 Þ q = p
p
If cos q = 0 Þ sin q = 1 Þ q =
2
3. r1 r2 + r3 r2 + r1 r3 = S 2
Þ S = 42
1
1
1 1
+
+
=
Þ r =8
r1 r2 r3 r
D
Þ D = 336
S
D
D
D
D
4. Use r = , r1 =
, r2 =
, r3 =
s
s-a
s-b
s-a
A
B
C
(C) r = 4 R sin sin sin
2
2
2
r=
and similarly r1 , r2 , r3
Exercise-5 : Subjective Type Problems
2. Ð O 1 EO 2 = 90° , E is the orthocentre of D O 1 EO 2
A
x
3
=
;x =1
sin 30° sin 120°
D
E
x
B
1
3.
r ( AD + AE ) = 5
2
1
r( BF + BD) = 10
2
Þ
4.
( r1 r2 r3 ) 2
D
r cot
=
C
E
I
B
2
s
s ö
æ s
=ç
´
´
÷
3
6
s
a
s
b
s
cø
è
D
r
( s - a) + ( s - b) + ( s - c)
³ [( s - a)( s - b)( s - c)]1 3
3
D 1D 2D 3
30°
Ö3
A
B
B
+ r cot
2
2 =2
A
A
r cot + r cot
2
2
C
cos
2 =3
Applying C and D,
A-B
sin
2
BF + BD
=2 Þ
AD + AE
30°
F
C
341
Solution of Triangles
Þ
s3
³ 27
( s - a)( s - b)( s - c)
Minimum value = 1
AB
AM
5. In DABM ,
=
sin 150° sin 7°
AC
AM
In DACM ,
=
sin(97°-q) sin q
°
23
7°
q
30°
B
C
AH BH CH R
R2
+
+
= ( a cos A + b cos B + c cos C ) =
(sin 2 A + sin 2 B + sin 2C )
AD BE CF D
D
=
7.
83°
M
sin q = 2 sin 7° sin(97°- q)
sin q = sin q - cos(104°- q)
cos(104°- q) = 0
q = 14°
Þ
Þ
Þ
Þ
6.
A
c
=
sin C
4R 2
bc sin A
sin A sin B sin C =
=2
D
D
AA1
A
Aö
æ
sin ç B + ÷
2ø
è
A
AA1 cos = sin B + sin C
(Q R = 1)
2
A
B
C
AA1 cos + BB 1 cos + CC 1 cos
2
2
2 =2
Þ
sin A + sin B + sin C
B
C
A/2
c
A1
8. ax 2 + bx + c = 0 has equal roots, then
b 2 = 4 ac
2
sin A sin C a c a + c
+
= + =
sin C sin A c a
ac
…(1)
2
2
=
= 4 + 2 cos B
9. cot
A
B
C
is AP
, cot , cot
2
2
2
In DABC ,
A
B
C
A
B
C
cot + cot + cot = cot × cot × cot
2
2
2
2
2
2
A
C
Þ
cot × cot = 3
2
2
AM ³ GM
b + 2 ac cos B
ac
342
Solution of Advanced Problems in Mathematics for JEE
cot
A
C
+ cot
2
2 ³ cot A × cot C
2
2
2
Þ cot
B
³ 3
2
10. ( R 2 - 4 Rr + 4 r 2 ) + ( 4 r 2 - 12 r + 9) = 0
( R - 2 r) 2 + (2 r - 3) 2 = 0
3
Þ
r = ; R = 2r
2
DABC is an equilateral triangle.
11. In DBCP ,
3
PC
=
sin 60° sin q
PC = 2 3 sin q
A
3
B
3
60° 3
a
q
60°
P
12. b + c =
2 ab cos C + 2 3 ab sin C ( a 2 + b 2 - c 2 ) + 12
=
2b
2b
13. R = 3, D = 6
PDDEF = DE + EF + DF = R(sin 2 A + sin 2 B + sin 2C )
= 4R sin A sin B sin C
æ b c
ö 1
= 4Rç
sin A ÷ = (2 D ) = 4
è 2R 2R
ø R
❑❑❑
C
343
Inverse Trigonometric Functions
25
INVERSE TRIGONOMETRIC
FUNCTIONS
Exercise-1 : Single Choice Problems
p
æp
ö
æp
ö
2. (cot -1 x) ç - cot -1 x ÷ + 2 cot -1 x - cot -1 x + 3 ç - tan -1 x ÷ - 6 > 0
2
2
2
è
ø
è
ø
- (cot -1 x) 2 + 5 cot -1 x - 6 > 0
(cot -1 x) 2 - 5 (cot -1 x) + 6 < 0
(cot -1 x - 3)(cot -1 x - 2) < 0
2 < cot -1 x < 3
(Q cot -1 x is decreasing)
cot 3 < x < cot 2
3. 1 + tan 2 (tan -1 2) + 1 + cot 2 (cot -1 3) = 1 + 2 2 + 1 + 3 2 = 15
a
4.
æ (n + 1) 2 + (n + 1) - ((n + 1) 2 - (n + 1)) ö
÷
4
2
÷
1
+
(
n
+
1
)
(
n
+
1
)
è
ø
å tan -1 çç
n =1
5. cot -1 ( cos a ) - tan -1 ( cos a ) = x
p
- 2 tan -1 cos a = x
2
p x
- = tan -1 cos a
4 2
x
1 - tan
2
cos a =
x
1 + tan
2
Þ
tan
x 1 - cos a
=
2 1 + cos a
Þ sin x = tan 2
a
2
344
Solution of Advanced Problems in Mathematics for JEE
1ö æ
1ö ö
æ æ
ç çn + ÷ - çn - ÷ ÷
æ
ö
æ
ö
2ø è
2ø ÷
4
1
÷ = tan -1 ç è
÷÷ = tan -1 ç
6. T n = tan -1 çç
2
2
ç
÷
ç
1
1ö ÷
æ
ö
æ
è 4n + 3 ø
è n + ( 3 4) ø
ç 1 + çn + ÷çn - ÷ ÷
2 øè
2øø
è
è
1ö
1ö
æ
æ
T n = tan -1 ç n + ÷ - tan -1 ç n - ÷
2ø
2ø
è
è
1ö
p
æ
æ 1ö
æ 1ö
S n = tan -1 ç n + ÷ - tan -1 ç ÷ Þ S ¥ = - tan -1 ç ÷
2
2
2
è
ø
è ø
è2ø
7. cos -1 (1 - x) + m cos -1 x =
np
2
Domain x Î[0 , 1]
cos -1 (1 - x) + m cos -1 x > 0
(Qm > 0)
There is no solution.
p
8. 2 tan -1 (2 x - 1) = cos -1 x
2x - 1³ 0
1³ x > 0
1
x³
2
Only one solution
9. Put x = 2 sin q , y = 3 cos q
y
x
sin q cos q
+
-2 =
+
- 2 Î [-3 , - 1]
2 2 3 2
2
2
sin q cos q
\
+
- 2 = -1 only
2
2
O
–p
10. (cos -1 x) 2 - (sin -1 x) 2 > 0 Þ (cos -1 x + sin -1 x)(cos -1 x - sin -1 x) > 0
cos -1 x - sin -1 x > 0
p
p
p
- 2 sin -1 x > 0 Þ - £ sin -1 x <
2
2
4
Þ
Þ
2
11. f( x) = x + 7 x + k( k - 3) = 0
Þ
f(0) < 0 (Q k Î (0 , 3))
a and b are of opposite sign.
æ 1ö
æ 1ö
tan -1 a + tan -1 ç ÷ + tan -1 b + tan -1 çç ÷÷ = 0
èaø
èb ø
12. f( x) = a + 2 b cos -1 x
D f : [-1, 1]
f( x) is decreasing function.
Þ f( -1) = 1
Þ a + 2 bp = 1
Þ -1 £ x <
+1
–1
1
2
345
Inverse Trigonometric Functions
and f(1) = -1
Þ a = -1
2
13. Let tan -1 x = t
5p 2
æp
ö
Þ t2 + ç -t÷ =
8
è2
ø
-p
3p
-p
or
Þ t=
Þ tan -1 x =
4
4
4
Þ x = -1
y
14.
–5p/2
–2p
–3p/2
–p/2
O
15. 1 £ sin -1 (cos -1 (sin -1 (tan -1 x))) £
p/2
2p
3p/2
p
2
sin 1 £ cos -1 (sin -1 (tan -1 x)) £ 1
cos(sin 1) ³ sin -1 (tan -1 x) ³ cos 1
sin(cos(sin 1)) ³ tan -1 x ³ sin(cos 1)
tan(sin(cos(sin 1))) ³ x ³ tan(sin(cos 1))
1
16. x + = -2 sin(cos -1 y) Þ x = -1 and y = 0
x
17. tan -1 1 + tan -1 2 + tan -1 3
æ 5 ö
tan -1 1 + p + tan -1 ç
÷=p
è1-6ø
18. Let
æ -p p ö
tan -1 x = q , q Î ç
, ÷
è 2 2ø
2q + cos -1 cos 2q Þ 2q £ 0
q £ 0 Þ tan -1 x £ 0 Þ x £ 0
y
p/2
19.
0
p/2
p 3p/2
5p/2
2p
–p/2
16 ( x 2 + y 2 ) - 48 px + 16 py + 31p 2 = 0
x 2 + y 2 - 3 px + py +
31p 2
=0
16
x
5p/2
x
346
Solution of Advanced Problems in Mathematics for JEE
2
2
3p ö
pö
9p 2
æ
æ
çx÷ +çy + ÷ =
2 ø
2ø
16
è
è
22.
sin -1 (sin 8) = 3 p - 8 = t
tan -1 (tan 8) = 8 - 3p = - t
f (t ) + f ( - t ) = l
2 =l
23. Graphs of y = 3 sin -1 x and y = p(1 - x) are
3p
2
1
–1
1
y = p(1–x)
– 3p
2
Clearly one point of intersection
24. D f : [-1, 1]
p
f( x) max = + 6 at x = 1
2
p
f( x) min = - - 2 at x = -1
2
¥
1
æ
ö
æ 1ö
æ 1ö
æ 1 ö
27. tan -1 ç ÷ + tan -1 ç ÷ + tan -1 ç ÷ +¼ =
tan -1 ç
÷
2
è3ø
è7ø
è 13 ø
è r + r + 1ø
r =1
å
¥
=
å(tan -1 ( r + 1) - tan -1 ( r))
r =1
1ù
é
1- ú
1
1
1
4
-1
-1 (1 4) + (2 9)
-1 æ 1 ö
-1 æ 1 ö
-1 ê
28.
cos x = tan
= tan ç ÷ = ´ 2 tan ç ÷ = cos ê
ú
2
1 - ( 1 4) ´ ( 2 9)
è2ø 2
è2ø 2
ê1 + 1 ú
êë
4 úû
2 ö
æ
ö
æ
ç using 2 tan -1 x = cos -1 ç 1 - x ÷ for x ³ 0 ÷ = 1 cos -1 3 .
ç1+ x2 ÷
ç
÷ 2
5
è
ø
è
ø
p
p
p
p
29. tan 2 (sin -1 x) > 1 Þ - < sin -1 x < - or < sin -1 x <
2
4
4
2
æ1+ 2 ´ 4ö
-1 æ 1 + 4 ´ 8 ö
-1 æ 1 + 8 ´ 16 ö
30. cot -1 ç
÷ + cot ç
÷ + cot ç
÷ +¼
4
2
8
4
è
ø
è
ø
è 16 - 8 ø
347
Inverse Trigonometric Functions
= cot -1 (2) - cot -1 ( 4) + cot -1 ( 4) - cot -1 (8) + cot -1 (8) - cot -1 (16) +¼
= cot -1 (2)
32. sin -1 (1 + x) is defined for x < 0 and sin -1 x =
p
- cos -1 x " - 1 £ x £ 1.
2
The given equation is sin -1 x + sin -1 (1 + x) = cos -1 x
which can be written as
p
p
- cos -1 x + - cos -1 (1 + x) = cos -1 x
2
2
Þ
p - cos -1 (1 + x) = 2 cos -1 x
Þ
cos -1 ( -1 - x) = 2 p - cos -1 (2 x 2 - 1)
Þ
cos -1 ( -1 - x) + cos -1 (2 x 2 - 1) = 2 p
Þ
cos -1 ( -1 - x) = cos -1 (2 x 2 - 1) = p
Þ
-1 - x = 2 x 2 - 1 = -1
Þ
x =0
which implies that the total number of solutions sin -1 x + sin -1 (1 + x) = cos -1 x is only one.
33. (sin -1 x) 3 - (cos -1 x) 3 + (sin -1 x)(cos -1 x)(sin -1 x - cos -1 x) =
p3
16
(sin -1 x - cos -1 x){(sin -1 x) 2 + (cos -1 x) 2 + (2 cos -1 x sin -1 x)} =
(sin -1 x - cos -1 x)(sin -1 x + cos -1 x) 2 =
p3
16
p2 p3
=
4
16
p
p
2 sin -1 x - =
2 4
3p
2 sin -1 x =
4
3
p
sin -1 x =
8
(sin -1 x - cos -1 x)
x = sin
æ 1 + x 2 - 1ö
÷
35. f( x) = tan -1 ç
ç
÷
x
è
ø
1+ x2 -1
=y
x
3x
x
or cos
8
8
p3
16
348
Solution of Advanced Problems in Mathematics for JEE
x
y¢ =
=
1 2x
- ( 1 + x 2 - 1)
2 1+ x2
x2
1+ x2 -1
x 2( 1 + x 2 )
> 0 always
x®¥
y®1
x ® -¥
y ® -1
tan
-1
( -1 ® 1)
æ p pö
ç - , ÷ - {0}
è 4 4ø
40. cos -1 x + cot -1 x = l " x Î [-1, 1]
é p 7p ù
lÎê ,
ë 4 4 úû
41. x 3 + bx 2 + cx + 1 = 0
f( -1) = b - c < 0
f(0) = 1 > 0
Þ -1 < a < 0
a = -B
B Î(0 , 1)
æ 2 sin B ö
y = -2 tan -1 (cosec B ) - tan -1 ç
÷
è cos 2 B ø
2 cos B ö
2 sin B
æ
= - ç p + tan -1
= -p
÷ - tan -1
2
1 - cosec B ø
cos 2 B
è
p
42. f( x) = + cot -1 {- x}
2
p
p
< cot -1 {- x} £
4
2
43. sin -1 (sin 3) + tan -1 (tan 3) + sec -1 (sec 3)
( p - 3) + ( 3 - p ) + 3 = 3
44. (2np , 0) n Î I
45. f( x) = sin -1 ([ x] - 1) + 2 cos -1 ([ x] - 2)
-1 £ [ x] - 1 £ 1
-1 £ [ x] - 2 £ 1
Þ 0 £ [ x] £ 2
Þ 1 £ [ x] £ 3 Þ [ x] = 1 or 2
349
Inverse Trigonometric Functions
Exercise-2 : One or More than One Answer is/are Correct
2. cos -1 x = tan -1 x Þ x Î[0 , 1]
æ 1- x2 ö
÷ = tan -1 x
tan -1 ç
ç
÷
x
è
ø
x2 = 1- x2 Þ x4 + x2 -1=0
Þ
5 -1
2
4
2ö
æ
æ
æ 17 ö ö
3. tan ç cos -1 + tan -1 ÷ = tan ç tan -1 ç ÷ ÷
5
3ø
è
è 6 øø
è
x2 =
a = 17 , b = 6
17 ö
æ
-1
5. sin -1 ç x 2 - 6 x +
÷ = sin k
2 ø
è
1
where -1 £ k £ 1
y = x2 - 6x +
17
2
0
1
–
2
6. (sin -1 x - cos -1 x)((sin -1 x) 2 + (cos -1 x) 2 + 2 sin -1 x cos -1 x) =
Þ
sin -1 x - cos -1 x =
p
4
Þ cos -1 x =
p
8
Þ x = cos
p
8
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 2
1. a = 2p
b = -3
2. a = 0
b=3
p3
16
x
3
350
Solution of Advanced Problems in Mathematics for JEE
Exercise-4 : Matching Type Problems
n
3. (A) 33n = [2 + (n - 1) 2 ] Þ n = 9
2
é p 7p ù
(B) x Î [-1, 1] Þ cos -1 x + cot -1 x Î ê ,
ë 4 4 úû
(C) cos q =|1 + sin q| Þ cos q ³ 0
Sq. both sides,
Þ cos 2 q = 1 + sin 2 q + 2 sin q
sin q = 0 or sin q = -1
Number of solution = 3
(D) a = x( x - 1)
Possible values of a are 6, 12, 20, 30.
4. (A)
(B)
tan -1 (3) + tan -1 ( -3) = 0
y
1 + sin x = 2 cos 2 x
Þ 2 sin 2 x + sin x - 1 = 0
(C)
(D)
(2 sin x - 1)(sin x + 1) = 0
p p
tan x = 4 2
0
p
p
2
3p
2
2p
x
f( x) = x 3 + x 2 + 4 x + 2 sin x
f ¢( x) = 3 x 2 + 2 x + 4 + 2 cos x > 0
and f(0) = 0
Exercise-5 : Subjective Type Problems
1. 5 - 2 p > x 2 - 4 x
x 2 - 4 x + (2 p - 5) < 0
2 - 9 - 2p < x < 2 + 9 - 2p Þ l = 9
B
r
2.
sin =
2 IB
A
C
IB = 4 R sin sin
2
2
Bö r
A
C
æ
sin ç 90° - ÷ = 1 Þ BI 1 = 4 R sin cos
2
BI
2
2
è
ø
1
( II 1 ) 2 = ( BI ) 2 + ( BI 1 ) 2 = 16 R 2 sin 2
A
2
A
I
B
B/2
r
C
90°–B/2
r1
…(1)
I1
90°–C/2
351
Inverse Trigonometric Functions
Aö
æ
I 2 I 3 cos ç 90° - ÷ = a
2
è
ø
(by using pedal triangle)
I 2 I 3 = 4 R cos
A
2
( I 2 I 3 ) 2 = 16 R 2 cos 2
A
2
…(2)
From (1) & (2) we get l = 16
æ 1ö
æ3ö
3. 2 tan -1 ç ÷ - sin -1 ç ÷
5
è ø
è 5ø
æ 5 ö
æ3ö
tan -1 ç ÷ - sin -1 ç ÷
è 12 ø
è 5ø
æ
æ 5 ö
æ3ö
æ3ö
æ 5 öö
tan -1 ç ÷ - tan -1 ç ÷ = - ç tan -1 ç ÷ - tan -1 ç ÷ ÷
è 12 ø
è4ø
è4ø
è 12 ø ø
è
æ 16 ö
æ 63 ö
= - tan -1 ç ÷ = - cos -1 ç ÷
è 63 ø
è 65 ø
l = 65
Þ
æ
ö
1
ç
÷
¥
æ
ö
2
-1
-1 ç
÷
2
÷÷ =
5.
2 tan çç
2 tan
2
ç n2 n
÷
n
+
n
+
4
è
ø
n =0
n =0
ç
+ + 1÷
4
è 4
ø
¥
å
å
æ æ n 1ö n ö
ç ç + ÷- ÷
2ø 2 ÷
-1 ç è 2
=
2 tan
ç n æ n 1ö
÷
n =0
ç ç + ÷ + 1÷
è 2è2 2ø
ø
¥
å
¥
=
æ
n
1
n =0
6. cos -1 (|3 log 26 (cos x) - 7|) = cos -1 (|log 26 (cos x) - 1|)
|3 log 26 (cos x) - 7| =|log 26 (cos x) - 1|
Let log 26 (cos x) = t
Þ
Þ
n ö
å 2 çè tan -1 æçè 2 + 2 ö÷ø - tan -1 æçè 2 ö÷ø ÷ø
|3t - 7| =|t - 1|
t = 3 and t = 2
cos x = 6 - 3 and 6 - 2
❑❑❑
352
Solution of Advanced Problems in Mathematics for JEE
26
VECTOR & 3DIMENSIONAL
GEOMETRY
Exercise-1 : Single Choice Problems
1. Perpendicular distance from origin
p
d=
d2 =
\
a 2 + b2 + c 2
p2
a 2 + b2 + c 2
1 ® ®
2. Area of triangle = | a ´ b| = 3
2
® ®
® ®
p
12
Þ | a|| b|sin = 6 Þ | a|| b| =
3
3
® ®
® ®
a × b =| a|| b|cos
®
®
®
p
=2 3
3
® ®
®
®
®
®
4. | c - a|2 = 8 Þ | c|2 - 2 c × a + | a|2 = 8 Þ | c|2 - 2| c| + 1 = 0 Þ | c|2 = 1
®
®
®
a ´ b = 2i$ + 2 j$ + k$
Also,
®
®
®
®
® ®
|( a ´ b ) ´ c| =| a ´ b|| c|sin
\
®
Þ | a ´ b| = 3
p
1 3
= 3 ×1× =
6
2 2
4
5
4
cos q 2 =
5
5.
cos q 1 =
cos 2 q 1 + sin 2 q 2 = 1
Þ
®
®
®
®
7. l( a ´ b ) × ( a ´ b ) = 4 3
® ®
l( a 2 b 2 - ( a × b ) 2 ) = 4 3
p
æ
ö
ç abcos = 1÷ Þ
3
è
ø
b=1
353
Vector & 3Dimensional Geometry
l( 4 ´ 1 - (1) 2 ) = 4 3
4 3
3
$
8. x(3i$ + 2 j$ + 4 k) + y(2i$ + 2 k$) + z( 4i$ + 2 j$ + 3 k$) = a( xi$ + yj$ + zk$)
l=
(3 - a ) x + 2 y + 4 z = 0
2 x - ay + 2 z = 0
Þ
4 x + 2 y + (3 - a )z = 0
For non-trivial solution
3 -a 2
4
2
4
9.
-a
2 =0
2 3 -a
® ®
® ®
a×a
a× b
® ®
a× c
® ®
® ®
® ®
®
®
b×a
b× b
b × c =[a
b
® ®
® ®
® ®
c×a
c×b
c×c
®
®
®
c ]2
®
® ®
10. | c|2 = 4( a ´ b ) 2 + 9 b 2 = 4( a 2 b 2 - ( a × b ) 2 ) + 9 b 2 = 192
®
®
®
®
® ®
® ®
c + 3 b = 2 a ´ b Þ c 2 + 9 b 2 + 6 b × c = 4( a 2 b 2 - ( a × b ) 2 )
Þ
6 × 4 × 192 cos q = -288 Þ cos q =
®
®
®
®
®
- 3
2
®
® ®
® ®
® ®
11. | a - 2 b|2 + | b - 2 c|2 + | c - 2 a|2 = 5a 2 + 5b 2 + 5c 2 - 4( a × b + b × c + c × a )
® ® ® ® ® ®
æ -3 ö
= 15 - 4( a × b + b × c + c × a ) £ 15 - 4 ç
÷ = 21
è 2 ø
® ®
Q
12.
Þ
® ®
® ®
-3
2
® ®
®
®
® ®
p
16| a|| b|sin = 3| a|2 + 3| b|2 + 6| a|| b|
2
a× b+ b× c + c ×a ³
3 a 2 - 10 ab + 3 b 2 = 0 Þ (3 a - b)( a - 3 b) = 0
—®
Now
Þ
\
—®
®
®
®
®
—®
—®
OC × AB = ( a + b ) × ( b - a ) =| OC || AB|cos q
b2 - a 2
a 2 + b2 a 2 + b2
= cos q =
cos q =
4
5
9a 2 - a 2
9a 2 + a 2
(using b = 3 a)
354
Solution of Advanced Problems in Mathematics for JEE
q
1 - cos q
tan =
=
2
1 + cos q
\
4
5 =1
4 3
1+
5
1-
—®
1 —® —®
13. AM = ( AB + AC )
2
14.
A
B
2 l 3
3 3 5 =0
l 2 2
C
M
2
Þ l - 3l + 2 = 0
® ® ® ê ® ® ®
® ® ® ú
15. ( a + b - c ) × ê( b + c - a ) ´ ( c + a - b )ú
ë
û
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
®
( a + b - c ) × ( b ´ c + b ´ a + c ´ a - c ´ b - a ´ c + a ´ b ) = 2( a + b - c ) × ( b ´ c + c ´ a )
®®®
®®®
®®®
= 2 ([ a b c ] + [ b c a ]) = 4 [ a b c ]
$
$
$
$
$
$
$
16. ( a$ ´ b) ´ ( a$ + b) = ( a$ × ( a$ + b)) b - ( b × ( a$ + b)) a$ = (1 + a$ × b)( b$ - a$ )
®
®
®
—®
—®
®
—®
—®
17. Angle between planes is angle between n 1 and n 2 , where n 1 = AB ´ AC and n 2 = AD ´ AC
®
n 1 = -2i$ + 4 j$ - 3 k$,
®
18. a1 = x 1 i$ + y 1 j$ + z 1 k$ ,
®
n 1 = 6i$ + 3 j$ - 6 k$
®
®
a 2 = x 2 i$ + y 2 j$ + z 2 k$
and
a 3 = x 3 i$ + y 3 j$ + z 3 k$
are
mutually
perpendicular unit vectors, then
22. On solving,
®
®
®
x1
y1
z1
[ a1
a2
a3 ] = x 2
x3
y2
y3
z2 = ±1
z3
Ax = C and Bx = D
é 1ù
é3 ù
ê
ú
x= 2
x = ê 1ú
ê ú
ê ú
êë3 úû
êë2 úû
P = (1, 2 , 3), Q = (3 , 1, 2)
x -1 y -2 z -3
PP ¢ :
=
=
=l
1
1
1
lies on plane
(l + 1, l + 2, l + 3)
\
Similarly
3l + 6 = 9 Þ l = 1
P ¢ = (3 , 4 , 5)
Q ¢ = ( 5, 3 , 4)
Q
® Ù Ù Ù
n=i+j+k
P
(1
)
,3
,2
Plane
x+y+z=9
P¢
Q¢
355
Vector & 3Dimensional Geometry
Now check the options.
—®
23. AM = (a - 1) i$ + j$
—®
BM = (a - 1) i$
a -1 1 0
$
$
$
CM = (a - 3) i + 2 j + 2 k are coplanar, then a - 2 0 0 = 0
a -3 2 2
—®
—®
24. Normal vector is parallel to PQ
P(1,–2,3)
x1 - 1 y1 + 2 z 1 - 3
=
=
=l
1
-1
1
Þ
x 1 = l + 1, y 1 = -2 - l, z 1 = 3 + l
Mid point of PQ is lie on the plane
2
Þ
l=
3
æ 5 - 8 11 ö
Qç ,
, ÷
è3 3 3 ø
Q(x1, y1, z1)
|a$ - b$| = 1
25.
cos q =
Þ
1
2
3
Volume of parallelopiped = [a$ b$ a$ ´ b$] = sin 2 q =
4
26. Equation of line PQ
x -3 y -7 z -1
=
=
=l
1
2
-6
Point Q(3 + l , 7 + 2l , 1 - 6l)
If it lies on plane 3 x + 2 y + 11z = 9 , then
25
l=
59
®
®
®
27. V 1 = [ a
b
c]
®
®
®
V2 =[ a + b - 2 c
®
®
®
3 a -2 b + c
®
®
®
®
a - 4 b + 2 c ] = 15[ a
®
®
b
c]
28. Line represented by x + ay - b = 0, cy + z - d = 0 is parallel to
(i$ + aj$) ´ ( cj$ + k$) = ai$ - j$ + ck$
Line represented by - x + a ¢y + b¢ = 0, c ¢y - z + d ¢ = 0 is parallel to
(i$ - a ¢ j$) ´ ( c ¢ j$ - k$) = a ¢ i$ + j$ + c ¢ k$
356
Solution of Advanced Problems in Mathematics for JEE
If these two lines are perpendicular, then
aa ¢ + cc ¢ = 1
29. Equation of line PQ
®
r = (2i$ - 2 j$ + 3 k$) + m (i$ + 5 j$ + k$)
Þ Co-ordinate of Q(2 + m , 5 m - 2 , 3 + m)
If point Q lies on plane, then
10
m=
27
—®
10 $ 50 $ 10 $
PQ = m i$ + 5 m j$ + m k$ =
i +
j+
k
27
27
27
®
30.
®
®
®
®
®
( a ´ b) ´ c = a ´ ( b ´ c )
® ® ®
® ® ®
® ® ®
® ® ®
( a × c ) b - ( b × c ) a = ( a × c ) b - ( a × b) c
® ® ®
® ® ®
( a × b) c = ( b × c ) a
Þ
y
(0,1)
®
r = xi$ + yj$
31. Let
®
(– 7, 0)
( 7, 0)
r × ( r + 6i$) = 7
x 2 + ( y + 3) 2 = 16
Þ
Area of quadrilateral = 8 7
®
®
®
(0,–7)
®
1 |( p - q ) ´ ( r - q )|
=4
2
1 ® ®
| q ´ r|
2
33.
®
Also,
®
®
P(p)
®
O
p + k1 q + k 2 r = 0
®
Þ
x
O
®
®
®
p = -k1 q - k 2 r = 0
®
Q(q)
®
R(r)
Þ
k1 + k 2 + 1 = 4
Þ
k1 + k 2 = 3
34. Let length, breadth and height of rectangular box be a , b, c respectively.
®
P = ai$ + ck$
Z
(0,0,c)
®
R = bj$ + ck$
®
a
b
c
O = i$ + j$ + k$
2
2
2
—® —®
b
c ö æa
b
c ö
æa
| OQ|| OR |cos q = ç i$ + j$ + k$ ÷ × ç i$ - j$ - k$ ÷
2
2 ø è2
2
2 ø
è2
P
(a,0,c)
(0,0,0)
Q
x (a,0,0)
R(0,b,c)
(0,b,0)
357
Vector & 3Dimensional Geometry
1
3
1
cos f = 3
cos q = -
Þ
Similarly,
®
®
®
®
®
®
®
36. r = a (m ´ n ) + b( n ´ l ) + c ( l ´ m )
where
® ® ®
® ®
® ®
® ®
[ l m n ] = 4,
r × l = 4a,
r × m = 4b,
r × n = 4c
which imply that
a+ b+ c
®
®
®
®
=
r ×( l + m + n )
1
4
®®®
1 ®®®
37. The volume tetrahedron is given by k = [ a b c ] Þ [ a b c ] = 6 k
6
The volume of parallelepiped is given by
®
® ®
®
®
®
®®
®
®
®
®®
®
®
®
[ a - b b + 2 c 3 a - c ] = [ a b + 2 c 3 a - c ] + [- b b + 2 c 3 a - c ]
®®
®
®
®
®
®
®
®®
®
®
®
®
®
®
= [ a b 3 a - c ] + [ a 2 c 3 a - c ] + [- b b 3 a - c ] + [- b 2 c 3 a - c ]
®®
®
®
®
®
®®®
®®®
= [ a b - c ] + [- b 2 c 3 a ] = - [ a b c ] - 6 [ a b c ]
®®®
= -7[ a b c ]
Volume is 42 k.
38. We know that the equation of the plane passing through the line of intersection of planes
p1 = 0 and p 2 = 0 is
p1 + lp 2 = 0
That is,
…(1)
( x + 2 y + z - 10) + l (3 x + y - z - 5) = 0
Since, this plane passes through the origin (0 , 0 , 0) satisfies this equation. This implies that
( -10) + l( -5) = 0
Þ
l = -2
Substituting the value of l in Eq. (1), we get
( x + 2 y + z - 10) - 2 (3 x + y - z - 5) = 0
-5 x + 3 z = 0
Þ
5x - 3z = 0
39. Let the point P( x p , y p , z p ) be the required point.
That is,
The distance of the point from x-axis is
The distance from the point (1, - 1, 2) is
y 2p + z 2p .
358
Solution of Advanced Problems in Mathematics for JEE
( x p - 1) 2 + ( y p + 1) 2 + ( z p - 2) 2
Þ
y 2p + z 2p = ( x p - 1) 2 + ( y p + 1) 2 + ( z p - 2) 2
Þ
x 2p - 2 x p + 2 y p - 4 z p + 6 = 0
Therefore, the locus of point P is
x 2 - 2 x + 2 y - 4z + 6 = 0
Exercise-2 : One or More than One Answer is/are Correct
3. Point P on line L1
Ù
Ù
Ù
i + 7j – 5k
P(2 + l , 1 + 7l , - 2 - 5l)
(2,1,–2)
Point P on line L 2
P
(4,–3,0)
P( 4 + r , - 3 + r , - r) Þ l = -1, r = -3
Acute angle between L1 and L 2
13
cos q =
15
Equation of plane containing L1 and L 2 is x + 2 y + 3 z + 2 = 0
4. a$ = b$ + ( b$ ´ c$)
a$ × b$ = 1 and a$ × c$ = b$ × c$
Ù
Ù
Ù
Ù
Ù
Ù
i+j–k
|a$ - b$| =|b$ ´ c$| Þ sin q = 0
(Q q = b$ c$)
|a$ + b$ + c$|2 = 3 + 2( a$ × b$ + b$ × c$ + a$ × c$) = 5 + 4( b$ × c$)
i – j + mk
5. If these two lines are coplanar, then
1 -1 m
1 m 2 =0
2 0 5
Ù
Ù
Ù
(3i – 2j – 4k)
Ù
Ù
Ù
(5i – 2j + k)
2
2m - 5m - 1 = 0
Þ
®
®
®
6. i$ ´ [( a - j$) ´ i$] + j$ ´ [( a - k$) ´ j$] + k$ ´ [( a - i$) ´ k$] = 0
®
2 a - (i$ + j$ + k$) = 0
Þ (2 x - 1)i$ + (2 y - 1) j$ + (2 z - 1)k$ = 0
Þ x = y =z =
®
®
7. [ a ´ b
®
®
c´d
®
®
®
®
®
®
1
2
®
®
®
®
®
®
®
®
e ´ f ] = ( a ´ b ) × [( c ´ d ) ´ ( e ´ f )] = ( c ´ d ) × [( e ´ f ) ´ ( a ´ b )]
®
®
®
®
®
®
= ( e ´ f ) × [( a ´ b ) ´ ( c ´ d )]
®
®
®
®
® ®
®
® ®
®
= ( a ´ b ) × [( c ´ d ) × f ] e - [( c ´ d × e ) ´ f ]
Ù
Ù
Ù
i + mj + 2k
359
Vector & 3Dimensional Geometry
®
®
®
®
®
®
®
®
®
®
®
®
=[ c
d
f ][ a
b
e] -[ c
d
e ][ a
b
f]
Similarly, solve other 2.
®
®
®
®
®
®
8. 3 ( a - b ) + ( b - c ) + 2 ( c - d ) = 0
—®
—®
BC + 2 CD —®
= BA
1+ 2
®
b = 2 c$ + la$
10.
®
æ 1ö
| b |2 = 4 + l2 + 4l ç ÷ = 16 Þ l = -4 , 3
è4ø
11. L1 : x = y = z
x -1 y + 1 z
L2 :
=
=
1
-1
-1
1
Shortest distance =
2
Equation of plane containing line L 2 and parallel to L1
y - z + 1=0
1
Distance of origin from this plane =
2
®
12.
®
®
®
r ×( a + b + c ) = 0
®®®
Þ [ a b c ](sin x + cos y + 2) = 0
sin x = -1 and cos y = -1
Þ
®
®
®
®
®
®
® ®
®
® ® ®
®
®
13. ( a ´ b ) ´ ( c ´ d ) = [( a ´ b ) × d ] c - ( a ´ b × c ) d = r c + s d
®
®
®
®
®
®
where
r =[a
b
c ] and s = - [ a
b
c ] as c and d are non-collinear.
®
®
®
®
®
Similarly,
h = -[ b
c
d ] and k = [ a
c
®
14. Here,
®
®
®
®
d]
®
a = i$ + 2 j$, b = 2i$ + aj$ + 10 k$ and g = 12i$ + 20 j$ + ak$
1 2 0
[a b g ] = 2
a 10 = a 2 - 24 a + 240 > 0 , for all a
12 20 a
®®®
\
® ®
\
®
a , b and g are non-coplanar or linearly independent for all a.
Hence, (a, b, c) is the correct answer.
360
Solution of Advanced Problems in Mathematics for JEE
®
19. Let r = xi$ + y$j + zk$
®
If r ´ i$ = $j + k$
Þ
- yk$ + z$j = $j + k$
Þ
xk$ - zi$ = i$ + k$
®
Þ
r = xi$ - $j + k$
®
If r ´ $j = i$ + k$
®
r = i$ + y$j - k$
20. (A) See dot product
(C) y = ln( e -2 + e x )
e y - e -2 = e x
21. ( -3 - 4l , 6 + 3l , 2l) = ( -2 - 4m , 7 + m , m)
Exercise-3 : Comprehension Type Problems
Paragraph for Question Nos. 1 to 3
1. AB = BC
(2,0,0)
p.v. of H = j$ + r (i$ - j$ + k$)
A
D
—® —®
Also,
AH × BC = 0
H
Þ [( r - 2)i$ + (1 - r) j$ + rk$] × ( - j$ + 2 k$) = 0
r - 1 + 2r = 0
Þ
2. p.v. of H =
Þ r=
(0,1,0)
B
1
3
i$ 2 j$ k$
+
+
3 3 3
p.v. of centroid =
2 $ j$ 2 k$
i + +
3
3
3
®
3(p. v .) of centroid - p. v . of H
p.v. of S =
2
1
y coordinate of S =
6
3. Let P º ( a , b, c)
Þ ( a - 2) 2 + b 2 + c 2 = a 2 + ( b - 1) 2 + c 2 = a 2 + b 2 + ( c - 2) 2 = a 2 + b 2 + c 2
æ 1 ö
Þ
P = ç 1, , 1÷
è 2 ø
PA =
(1,0,1)
3
2
(0,0,2)
C
361
Vector & 3Dimensional Geometry
Paragraph for Question Nos. 4 to 6
®
®
®
4. PQ = ( b - a ) cos q
®
=
®
B(b)
(where q angle between AB and plane)
®
A(a)
®
|( b - a ) ´ n|
®
®
®
®
® ®
r×n=d
®
5. Equation of line AP is r = a + l n
®
Q
P
| n|
®
A(a)
®
For point P( a + l n ) × n = d
® ®
l=
® ®
P
d- a×n
r×n=d
®
| n |2
\
A¢
® ®ö
® ®
æ®
æ
ö
®
ç
ç d - a × n ®÷
d- a×n ÷
¢
is
P ç a+
\
A
a
+
2
n
÷
ç ®
÷
®
2
ç
÷
ç | n|2
÷
|
n
|
è
ø
è
ø
6. Distance =|BQ - AP|=
® ®
® ®
®
b×n -d
a×n -d
( b - n) × n
®
| n|
-
®
| n|
=
®
®
®
| n|
Paragraph for Question Nos. 7 to 9
7. B(3 + 2l , - 1 - 3l , 2 - l)
d r of L 2 < 2l , - 3l + 3, 1 - l >
L1
B
®
L 2 is parallel to plane r × (2i$ + j$ - k$) = 5
4l - 3l + 3 - 1 + l = 0
2l = -2 Þ l = -1
\
\
A
L2
(3,–4,1)
B(1, 2 , 3)
®
So, equation of L 2 is r = (3i$ - 4 j$ + k$) + l(i$ - 3 j$ - k$)
8. Equation of plane contain L1 & L 2 is
x -3
0
-1
i . e. ,
y + 1 z -2
3
1 =0
6
2
( x - 3)(6 - 6) + ( y + 1)(0 + 1) + ( z - 2)(0 + 3) = 0
y + 3z - 5 = 0
9. Any point of L1 is (3 + 2l , - 1 - 3l , 2 - l)
if on plane p, then
362
Solution of Advanced Problems in Mathematics for JEE
2 (3 + 2l) + 1( -1 - 3l) - 1(2 - l) = 5
2l = 2 Þ l = 1
\ Q(5, - 4 , 1)
if on xy plane, then
2 -l =0 Þ l =2
\
R(7 , - 7 , 0)
1 —®
Volume of tetrahedran = [ OA
6
—®
—®
OQ
OR ] =
3 -4 1
1
7
5 -4 1 =
6
3
7 -7 0
Paragraph for Question Nos. 10 to 11
Sol. Use crammer rule,
Intersect at a unique point Þ D ¹ 0
Do not have any common point of intersection.
Þ D = 0 and atleast any one of D x , D y , D z is non-zero (condition of no solution)
Paragraph for Question Nos. 12 to 14
®
®
®
Sol. | a | =|b | =| c | = r
æ ® ®ö
® ça +b÷ ®
a +ç
÷+c
ç 2 ÷
è
ø
PV of E :
3
®
®
®
®
3 a+ b + 2 c
e=
6
®
®
®
®
a+b+c
PV of G : g =
3
® ®
®ö æ ® ®
æ
ö
—® —®
ç 3 a + b + 2c ÷ ç a + b ® ÷
12. OE × CD = 0 Þ ç
- c ÷ =0
÷ ×ç
6
ç
÷ ç 2
÷
è
ø è
ø
®
®
®
Þ
a ×( b - c ) = 0
\
OA ^ BC
\
DABC must be isosceles with base BC.
\
| AC| =| AB|
—®
—®
—®
—®
®
C(c)
®
B(b)
O
®
A(a)
363
Vector & 3Dimensional Geometry
æ ® ® ® ® ® ® ö æ ® ® ®ö
ç 3 a + b + 2c a + b + c ÷ ç a + b
÷
13. GE × CD = 0 Þ ç
- c ÷ =0
÷ ×ç
6
3
2
ç
÷ ç
÷
è
ø è
ø
—® —®
®
®
—®
®
—®
Þ ( a - b ) × c = 0 Þ AB ^ OC
\
\
\
ABC must be isosceles with base AB.
Circumcentre and centroid lie on median through C.
Orthocenter also lie on median through C.
—® —® —®
—®
—® —®
14. [ AB AC AB ´ AC ] = ( AB AC ) 2
®
®
®
®
®
®
(a ´ b + b ´ c + c ´ a)2
—® —® —® —®
—® —®
ì -1 ® ® ® ® ® ® ü
[ AE AG AE ´ AG ] = ( AE ´ AG ) 2 = í ( a ´ b + b ´ c + c ´ a )ý
î18
þ
=
2
1 —® —® 2
( AB ´ AC )
324
Paragraph for Question Nos. 15 to 16
15. D(3 , - 1, 2)
AB lies along (0 , 1, 2)
CD lies along (3 , - 2 , 0)
Equation of plane containing AB line
x -1 y -1 z -1
0
2
1
-2
2
0
= 2 ( x - 1) + 2 ( y - 1) - ( z - 1) = 0
Containing CD line 2 ( x - 1) + 2 ( y - 1) - ( z - 2) = 0
16. r = (3 , - 1, 2) + d (1, 0 , 0)
Equation of ABC plane is x = 1.
Paragraph for Question Nos. 17 to 18
®ö
æ ®
æ ®ö
ç3b +2a ÷
ç2b ÷
17. R ç
÷
÷ and Q ç
5
ç 5 ÷
ç
÷
è
ø
è
ø
®ö
æ ®
æ ®ö
®
ç3b +2a ÷
ç2b ÷
mç
÷ l a + 1ç
÷
5
ç
÷
ç 5 ÷
è
ø
è
ø
=
m+1
l+1
Þ
2m
3m
l
2
10
and
=
=
Þm=
5 (m + 1) l + 1
5 (m + 1) 5 (l + 1)
9
®
B(b)
y
Q
R
O
P
®
A(a)
x
364
Solution of Advanced Problems in Mathematics for JEE
18. Ar ( DOPA) =
®
®
®ù
1 —® —®
1é 2
3 ® ®
OP ´ OA = ê (3 b + 2 a ) ´ a ú =
( b ´ a)
2
2 ë 19
û 19
®ö
®ù
éæ ®
®
®
1 —® —® —® —®
1 êç 3 b + 2 a ÷ ® 2
2bú
Ar ( PQBR ) = OQ ´ OB - OP ´ OR = êç
(3 b + 2 a ) ´
÷´b 2
2 ç
5
19
5 ú
÷
êëè
úû
ø
=
3 ® ®
(a ´ b)
19
Exercise-4 : Matching Type Problems
1. (A) Line
x -1 y + 2 z
=
=
-2
3
-1
®
is
along
®
the
vector
a = -2i$ + 3 j$ - k$
®
®
and
line
®
r = (3i$ - j$ + k$) + t(i$ + j$ + k$) is along the vector b = i$ + j$ + k$. Here a ^ b .
Also,
3 - 1 -1 - ( -2) 1 - 0
-2
3
-1 ¹ 0
1
1
1
(B) The direction ratios of the line x - y + 3 z - 4 = 0 = 2 x + y - 3 z + 5 = 0 are
i$ j$ k$
1 -1 2 = i$ + 7 j$ + 3 k$
2
1
3
Hence, the give two lines are parallel.
(C) The given lines are
®
( x = t - 3 , y = 2t + 1, z = -3t - 2) and r = (t + 1) i$ + (2t + 3) j$ + ( -t - 9)k$ ,
x + 3 y -1 z + 2
x -1 y -3 z + 9
and
=
=
=
=
1
-2
-3
1
2
-1
The lines are perpendicular as (1)(1) + ( -2)(2) + ( -3)( -1) = 0
-3 - 1 1 - 3 -2 - ( -9)
Also,
1
-2
-3
=0
or
1
2
-1
Hence, the lines are intersecting.
®
®
3
(D) The given lines are r = (i$ + 3 j$ - k$) + t(2i$ - j$ - k$) and r = ( -i$ - 2 j$ + 5k$) + s (i$ - 2 j$ + k$).
4
1 - ( -1) 3 - ( -2) -1 - 5
2
-1
-1 = 0
1
-2
3 4
365
Vector & 3Dimensional Geometry
Hence, the lines are coplanar and hence intersecting (as the lines are not parallel).
®
®
2. (A) If a , b and c are mutually perpendicular, then [ a ´ b b ´ c c ´ a ] = [ a
® ®
®
®
® ®
® ®
®
b
®
c ]2
® ® ®
= (| a|| b|| c|) 2 = 16
®
®
®
®
(B) Given a and b are two unit vectors, i . e. ,| a| =| b| = 1 and angle between them is
®
sin q =
®
| a ´ b|
Þ sin
® ®
| a|| b|
®
Now [ a
®
®
®
b+ a ´ b
®
p ® ®
=| a ´ b| ;
3
®®®
®®
® ®
3
=| a ´ b|
2
®®
®®
®®
b] = [ a b b] + [ a a ´ b b] = 0 + [ a a ´ b b]
®
®
®
®
®
®
®
®
3
4
®
®
®
= ( a ´ b ) × ( b ´ a ) = -| a ´ b|2 = ®
p
.
3
®
® ®
(C) If b and c are orthogonal, b × c = 0
®
®
®
Also, it is given that b ´ c = a
®
®
®
Now [ a + b + c
®
®
a+ b
®
®
b + c ] =[a
®
®
a+ b
b + c]+[b + c
®
®
®
=[a
b
®
®
®
® ®
®
®
a+ b
®
®
b + c]
®
c ] = a × ( b ´ c ] = a × a =| a|2 = 1
®
(because a is a unit vector)
®
®
®
(D) [ x
y
a] = 0
® ®
®
Therefore, x , y and a are coplanar.
® ®®
[ x , y b] = 0
® ®
®
Therefore, x , y and b are coplanar.
®®®
Also,
[a b c] =0
® ®
®
Therefore, a , b and c are coplanar.
From (i), (ii) and (iii)
® ®
®
® ®®
x , y and c are coplanar. Therefore, [ x , y c ] = 0
366
Solution of Advanced Problems in Mathematics for JEE
Exercise-5 : Subjective Type Problems
1. Line L is the shortest distance line of given lines.
2. [a$ b$ c$] = [b$ ´ c$ c$ ´ a$ a$ ´ b$] = [a$ b$ c$]2
[a$ b$ c$] = 1
Þ
( b$ + c$) × ( a$ ´ b$) [a$ b$ c$]
Projection of b$ + c$ on a$ ´ b$ =
=
|a$ ´ b$|
|a$ ´ b$|
3. Let l = m = n =
—®
®
1
2
®
®
®
®
®
®
®
®
®
4. OC = a 2 ( a + b ) 2 + b 2 ( a ´ b ) 2 + 2ab [ a × ( a ´ b ) + b × ( a ´ b )]
2
æ 3ö
1ö
÷ +0
1 = a ç 1 + 1 + 2 × 1 × 1 ÷ + b 2 × 1 × 1çç
÷
2ø
è
è 2 ø
2æ
Þ
—® —®
Also,
—®
—®
OB × OC =| OB||
× OC |cos
p
3
1
1
1
+ a ×1 =
Þ a=
2
2
3
8
From (1) and (2),
b2 =
9
a ×1×1×
Þ
®
®
æé 0
-1 / 2 ù ö
÷
5. v n +1 - v n = çç ê
0 úû ÷ø
è ë1 / 2
n +1
®
v0
®
®
-1 / 2 ù ®
é 0
v 2- v 1 = ê
v0
0 úû
ë1 / 2
®
®
-1 / 2 ù
é 0
v 3- v 2 = ê
0 úû
ë1 / 2
®
3 ®
®
-1 / 2 ù
é 0
v n - v n -1 = ê
0 úû
ë1 / 2
v0
n ®
v0
Adding all the equations,
®
®
…(1)
®
v n - v 0 = ( A + A 2 + A 3 +¼¼ A n ) v 0
®
®
-1 / 2 ù
é 0
where A = ê
Þ v n = ( I + A + A 2 +¼¼ ) v 0
ú
0 û
ë1 / 2
…(2)
367
Vector & 3Dimensional Geometry
6. Let
B=A-
1 2 1 3
1 4
A + A A +¼¼
3
9
27
AB
A2 1 3
1 4
=+ A A +¼¼
3
3
9
27
Aö
æ
çI + ÷B = A
3ø
è
-
1
B = (3 I + A) -1 A
3
n
7. det M n =
æ
1
®
®
1
ö
1
®
®
®
1
å çè (2 k + 1)! - (2 k + 2)! ÷ø = 1! - (2n + 2)!
k =0
8.
| a + b| = 3
Þ
Þ
Squaring both sides
® ®
1
a× b =
2
®
®
c = a + 2 b-3 a ´ b
® ®
Þ
® ®
a × c =2 & b × c =
® ® ®
5
2
® ® ®
p =|( a × a ) b - ( b × c ) a|
®
p = 2 b-
9.
5®
a
2
2
p=
21
Þ [ p] = 2
2
®
®
®
®
®
®
®
r = ( a ´ b ) sin x + ( b ´ c ) cos y + 2 ( c ´ a )
® ®
®®®
r × a = [ b c a ]cos y
® ®
®®®
r × b = 2[ c a b ]
® ®
®®®
r × c = sin x [ a b c ]
Þ sin x + cos y + 2 = 0
Þ
sin x = -1 and cos y = -1
p
x =y =p
2
368
Solution of Advanced Problems in Mathematics for JEE
10. New equation of plane : 4 x + 7 y + 4 z + 81 + l(5 x + 3 y + 10 z - 25) = 0
Þ
Þ
®
( 4 + 5l) 4 + (7 + 3l) 7 + ( 4 + 10l) 4 = 0
l = -1
Equation of plane : x - 4 y + 6 z - 106 = 0
106
distance =
= 212
53
®
®
18. w ´ m = n
®
®
®
® ®
n ×( w ´ m ) = n × n = 1
❑❑❑