EklavyATS-2023-Part Test 1(Math)
Questions
1.
There are ten pairs of shoes in a cup board out of which 4 shoes are picked up at random one
after the other. The probability that there is at least one pair is
(a)
(b)
(c)
(d)
2.
4
11
3
11
33
107
99
323
Fifteen coupons are numbered 1, 2, 3, . . . .15. Seven coupons are selected at random one at a
time with replacement. The probability that the largest number appearing in the process is 9,
is
(a)
(b)
(c)
()
()
()
9
6
16
8
7
15
3 7
5
(d) none of these
3.
A bag contains 4 red, 5 white and 6 black balls. Three balls are selected from this bag
simultaneously. The probability that exactly one of the colour is missing in the selected balls, is
equal to
(a)
(b)
(c)
301
455
366
455
261
455
(d) None of these
4.
If two numbers a and b are chosen randomly from the set consisting of numbers 1, 2, 3, 4, 5, 6
with replacement, then probability that lim
x→0
(a)
1
3
1
(b) 4
(c)
(d)
1
9
2
9
( )
ax + b x
2
2
x
= 6 is
5.
The area of the region consisting of points (x, y) satisfying |x ± y| ⩽ 2 and x 2 + y2 ⩾ 2 is
(a) 8 − 2π sq. units
(b) 4 − 2π sq. units
(c) 1 − 2π sq. units
(d) 2π sq. units
6.
Area of the region in which point p(x, y), {x > 0} lies; such that y ⩽
| ()|
( )
( √)
y
tan − 1 x
(a)
16
(b)
8π
3
3
√16 − x 2 and
π
⩽ 3 is
π
+8 3
(
)
(d) (√3 − π )
(c) 4√3 − π
7.
Value of ∫ 51
(a)
(b)
(c)
(d)
8.
8
(√x + 2√x − 1 + √x − 2√x − 1)dx is
3
16
3
32
3
34
3
π/4
π/4
π/4
0
0
0
Let I1 = ∫ x 2008(tanx) 2008dx, I2 = ∫ x 2009(tanx) 2009dx and I3 = ∫ x 2010(tanx) 2010dx. Then the
correct order sequence, is
(a) I2 < I3 < I1
(b) I1 < I2 < I3
(c) I3 < I1 < I2
(d) I3 < I2 < I1
9.
Im , n = ∫ 10x m(ln x) n dx =
(where m, n ∈ N )
(a)
(b)
(c)
(d)
n
I
n +1 m, n −1
−m
I
n +1 m, n −1
−n
I
m+1 m, n −1
m
I
n +1 m, n −1
3
10.
∫
6
2
√x + √x
(
3
x 1 + √x
)
dx =
(where C is a constant of integration)
( )
6
−6
(a) 1 tan − 1 x + x
+C
2
2
(b)
(
x 24
)
( )
− log 1 + x 24 + tan − 1 x 3 + C
2
( )
√
(c) 6tan − 1 x 6 + 3x 12 − 6log e 1 + x 12 + C
(
)
(d) 3 3 x + 6tan − 1 6 x − 3log 1 + 3 x + C
√
√
√
11.
If ∫
sec 2x − 2010
(b)
dx =
()
P(x)
π
+ C, then value of P 3 is
sin
x
sin
x
(where C is a constant of integration)
(a) 0
2010
2010
1
√3
(c) √3
(d) None of these
12.
Let f : R → R is defined as f(x) =
4
{
|x − [x]|,
[x] is odd
|x − [x + 1]|,
[x] is even
function, then ∫ f(x)dx is equal to
(a)
−2
5
2
3
(b) 2
(c) 5
(d) 3
13.
1
∫
x6 − x3
(
0 2x 3 + 1
(a) 0
1
(b) −
6
1
(c) − 12
1
(d) −
36
)
3
dx is equal to
, where [ . ] denotes greatest integer
14.
The differential equation of all the circles in the first quadrant which touch both the
coordinates axes is
(
(
( )) (
)
(a) (x − y) 2 1 + y ′ 2 = x + yy ′ 2
( )) ( )
(c) (x − y) (1 + y ) = (x + yy )
(b) (x − y) 2 1 + y ′ 2 = x + y ′ 2
2
′ 2
′
(d) none of these
(
15.
x
) (
x
)
The solution of the differential equation xdy y2e xy + e y = ydx e y − y2e xy , is
(where λ is a constant of integration)
( )
= ln (e + λ )
(a) xy = ln e x + λ
(b)
x
2
y
xy
( )
x
(c) xy = ln e y + λ
(
(d) xy2 = ln e xy + λ
16.
)
dy
sin 2x
A function y = f(x) satisfying the differential equation dx . sinx − ycosx + 2 = 0 is such that
x
y → 0 as x → ∞, then
(a) lim f(x) = 1
(b)
(c)
x→0
π/2
π
∫ f(x)dx is less than 2
0
π/2
0
∫ f(x)dx is greater than unity
(d) f(x) is an odd function
17.
dx
The solution of differential equation 3 dy =
x
x3 − y
is/are CORRECT? {c is any arbitrary constant}
(a) ℓ + m + n = 11
(b) ℓ + n = 9
(c) ℓ + 2m = 10
(d) m + n = 4
is x ℓ = mx n y + c, then which of the following
18.
Two lines drawn through the point P(4, 0) divide the area bounded by the curves
y=
πx
√2sin 4 and x − axis, between the lines x = 2 and x = 4, in three equal parts.
2√2
(a) One of the line has slope ⇒ m = −
1
3π
(b) One of the line has slope ⇒ m =
2
− 4√2
3π
2√2
(c) the sum of the slopes of the two lines is (−
)
π
(d) the slope of one of the line is two times the slope of other.
19.
2
If f(2 − x) = f(2 + x) and f(4 − x) = f(4 + x) for all x and f(x) is a function for which ∫ f(x) dx = 5,
0
50
then ∫ f(x) dx is equal to:
0
(a) 125
46
(b) ∫ f(x) dx
−4
51
(c) ∫ f(x) dx
1
52
(d) ∫ f(x) dx
2
20.
If a function f(x) satisfies the relation f(x) =
(a) f(0) < 0
(b) f(x) is a decreasing function
(c) f(x) is an increasing function
ex
1
+ ∫ e xf(t) dt, then:
0
1
(d) ∫ f(x) dx > 0
0
21.
π / 2 sin ( 2n − 1 ) x
If A n = ∫
0
sin x
(a) A n + 1 = A n
(b) B n + 1 = B n
(c) A n + 1 − A n = B n + 1
(d) B n + 1 − B n = A n + 1
( )
π / 2 sin nx 2
dx; B n = ∫
0
sin x
dx, for n ∈ N, then:
22.
{ }
x u
∫ ∫ f(t) dt du is equal to:
0 0
x
(a) ∫ (x − u) f(u) du
0
x
(b) ∫ u f(x − u) du
0
x
(c) x∫ f(u) du
0
x
(d) x∫ u f(u − x) du
0
23.
A fair coin is tossed n times. Let X = the number of times head occurs. If P(X = 4), P(X = 5) and
P(X = 6) are in A.P., then the value of n can be
(a) 7
(b) 10
(c) 12
(d) 14
24.
In a single throw with 3 ordinary unbiased dice, the chance of:
1
(a) throwing a “4 − 5 − 6” is equal to
216
(b) throwing a total of 11 is
1
8
(c) all the 3 dice showing up with different faces is
1
4
9
(d) all the 3 dice showing up alike faces is 36
25.
dy
The solution of
dx
=
√y − x is given by
(where c, c 0 are arbitrary constants)
| √y − x − 1 |
(b) x + c = 2√y − x + 2log (√y − x + 1 )
(a) x + c = 2√y − x + 2log
x
(c)
− y−x
√y − x + c0e 2 √ = 1
(d)
− y−x
√y − x + 1 = c0e 2 √
x
26.
A signal which can be green or red with probability
() ()
4
5
and
1
5
respectively is received by the
station A and transmitted to station B. The probability that a station receives signal correctly
is
()
3
4
. If signal received by station B is green, then the probability that the orginal signal was
p
green is q (in lowest form), then p + q is equal to:
27.
The area of the region in the xy-plane defined by the inequalities x − 2y2 ⩾ 0, 1 − x − |y| ⩾ 0 is
A then 12A is
28.
Area bounded by the curves y = e x, y = log ex and the lines x = − 1, y = 0, y = 1 is A then find
[A], where [.] is the greatest integer function.
29.
30.
Area of the region which consists of all the points satisfying the conditions
|x − y| + |x + y| ⩽ 8 and xy ⩾ 2, is 4(a – ln 8) then a equal to
Let y = f(x) satisfies the differential equation xy(1 + y)dx = dy. If f(0) = 1 and f(2) =
find the value of k.
31.
(
)(
If ∫ x 9 + x 6 + x 3 2x 6 + 3x 3 + 6
)
1/ 3
(
dx = a 2x 9 + 3x 6 + 6x 3
integration, then the value of 48a must be
32.
)
4/ 3
1
e2
k − e2
, then
+ C, where C is a constant of
1
If the value of definite integral ∫ 207C7x 200. (1 − x) 7dx is equal to k , where k ∈ N, then the value
0
k
of 26 is _______.
33.
−4
−5
34.
1
) (
)
√
−2
(
) (
Let In = ∫ x n 1 − x 2 dx then find the value of lim
0
35.
(
−1
)
Let J = ∫ 3 − x 2 tan 3 − x 2 dx and K = ∫ 6 − 6x + x 2 tan 6x − x 2 − 6 dx . Then J + K is _______.
n→∞
In
In − 2
.
If three small squares are chosen at random on a chess-board and the probability that they
are in a diagonal line is p, then 744p =
Answer Key
1. D
2. D
3. A
4. C
5. A
6. B
7. D
8. D
9. C
10. D
11. C
12. D
13. D
14. A
15. C
16. A,B,C
17. A,B,C
18. A,B,C,D
19. A,B,D
20. A,B
21. A,D
22. A,B
23. A,D
24. B,D
25. A,C
26. 43.00
27. 7.00
28. 2.00
29. 7.00
30. 2.00
31. 2.00
32. 8.00
33. 0.00
34. 1.00
35. 7.00
Solutions
1. (D)
Out of 20 shoes 4 can be taken in 20P 4 ways.
Ways of getting no pair = 20 × 18 × 16 × 14
20 × 18 × 16 × 14
Probability of no pair =
20P
4
224
= 323
224
99
Probability of at least one pair = 1 − 323 = 323
2. (D)
Total ways = 15 7
For favourable ways, we must select seven coupons from 1 to 9 such that 9 is selected at least
once.
Thus total number of favourable ways = number of ways of selecting seven coupons from 1 to 9 –
number of ways of selecting seven coupons from 1 to 8
= 97 − 87
⇒ Required probability =
97 − 87
157
3. (A)
i) Selected balls can be W W B or W B B (when red colour is missing).
Corresponding probability =
5C
2.
6C
1+
5C
15C
1.
6C
2
3
=
135
15C
3
ii) Selected balls can be R R W or R W W (when black colour is missing).
Corresponding probability =
4C
2.
5C
1+
4C
15C
1.
5C
2
3
=
70
15C
3
iii) Selected balls can be R B B or R R B (when white colour is missing).
Corresponding probability =
Thus, required probability =
4. (C)
lim
x→0
lim
⇒ e x→0
( )
a x + bx
x
x
bx − 1
x
15C
=6
⇒ e log a + log b = 6
⇒ e log ab = 6
⇒ ab = 6
(a, b) = (1, 6), (6, 1), (2, 3), (3, 2)
Required probability =
4
6×6
=
6C
2+
4C
3
135 + 70 + 96
=6
( )( )
+
1.
15C
2
2
ax − 1
4C
1
9
3
2.
6C
1
301
= 455
=
96
15C
3
5. (A)
Shaded region is the required one.
1
∴ Required Area = 4 × 2 × 2 × 2 − π.2 = 8 − 2πsq. unit
6. (B)
Required area is the area of shaded region (APOQ)
= area of Δ OAQ+ area of sector (OAP)
=
=
1
2
(
× 4 × 4√3 +
8π
3
+ 8√3
π ( 4×4)
6
)
7. (D)
∵
√x + 2√x − 1 = √(√x − 1)
2
+ 1 2 + 2√x − 1 =
√x − 1 + 1
√ (√x − 1 )2 + 12 − 2√x − 1 = |√x − 1 − 1 |
Then ∫ 51 (√x + 2√x − 1 + √x − 2√x − 1 )dx = ∫ 51 (√x − 1 + 1 + | √x − 1 − 1 |)dx
= ∫ 21 (√x − 1 + 1 + | √x − 1 − 1 |)dx + ∫ 52 ( √x − 1 + 1 + |√x − 1 − 1 |)dx
= ∫ 21 (√x − 1 + 1 + 1 − √x − 1 ) dx + ∫ 52 (√x − 1 + 1 + √x − 1 − 1 )dx
And √x − 2√x − 1 =
= ∫ 212 dx + 2∫ 52√x − 1 dx
=2+
4
[
]
5
4
34
(x − 1) 3 / 2 2 = 2 + (8 − 1) =
3
3
3
8. (D)
( )
For x ∈ 0,
π
4
x 2008(tanx) 2008 > x 2009(tanx) 2009 > x 2010(tanx) 2010
∴
π/4
π/4
π/4
0
0
0
∫ x 2008(tanx) 2008dx > ∫ x 2009(tanx) 2009dx > ∫ x 2010(tanx) 2010dx
⇒ I1 > I2 > I3
⇒ I3 < I2 < I1
9. (C)
Im , n = ∫ 10x m(logx) n dx
(Integrating by parts, taking (logx) n as first function)
=
[
]
xm + 1
(logx) n
⋅ m+1
n
1
−
∫n(logx) n − 1
0
1
xm + 1
⋅ x ⋅ m + 1 dx
n
= 0 − m + 1 ∫ 10x m(logx) n − 1 = − m + 1 Im , n − 1
10. (D)
3
Let, I = ∫
6
√x2 + √x
(
3
x 1 + √x
)
dx
Put x = t 6 ⇒ dx = 6t 5dt
(t +t ) 5
t +1
I=∫
6t dt = 6∫
dt
1+t
t (1+t )
4
6
3
2
2
= 6∫ t dt + 6∫
= 3t 2 + 6∫
1−t
1 + t2
1
1+t
dt
dt − 6∫
2
t
1 + t2
(
dt
)
= 3t 2 + 6tan − 1t − 3log 1 + t 2 + C
3
6
(
= 3 √x + 6tan − 1 √x − 3log 1 +
3
√x
)
+C
11. (C)
∫
sec 2x − 2010
sin 2010x
dx = ∫ sec 2x(sinx) − 2010dx − 2010∫
Applying by parts on I1, we get
I1 =
=
tan x
( sin x ) 2010
tan x
( sin x ) 2010
tan xcos x
+ 2010∫
+ 2010∫
tan x
( sin x ) 2011
dx
dx + C
+C
( sin x ) 2010
P(x)
⇒ I1 − I2 =
+C=
+C
( sin x ) 2010
( sin x ) 2010
⇒ P(x) = tan(x)
() ()
π
π
P 3 = tan 3 =
√3
12. (D)
f(x) =
f(x) =
f(x) =
4
{
{
{
|x − [x]|,
[x] is odd
|x − [x + 1]|,
[x] is even
|{x}|,
[x] is odd
|{x} − 1|,
[x] is even
{x},
[x] is odd
1 − {x},
[x] is even
1
∫ f(x) dx = 6. 2 (1.1) = 3
−2
1
( sin x ) 2010
dx = I1 − I2
13. (D)
( )
( )
1−
1
Let I = ∫
0
2x +
1
Let 2x +
( )
2
2−
x3
x2
1
x3
1
x
dx
3
2
=t
dx = dt
3 1 dt
⇒I=∫
∞t
2
3
[ ]
1 t−2
= 2 −2
=
−1
4
3
∞
[ ]
1
9
−0
1
= − 36
14. (A)
Let centre of circle be (h, h)
So, equation of circle is
(x − h) 2 + (y − h) 2 = h 2 . . . . . (1)
x 2 + y2 − 2xh − 2yh + h 2 = 0
⇒ 2x + 2yy ′ − 2h − 2hy ′ = 0
x + yy ′
⇒ h=
1+y ′
Putting in equation (1)
⇒
(
x−
) (
x + yy ′ 2
1+y ′
+ y−
On simplifying, we get
(
) (
x + yy ′ 2
1+y ′
( ) ) = (x + yy )
(x − y) 2 1 + y ′
2
′ 2
=
( x + yy )
(1+y )
′
′
2
2
)
15. (C)
(
x
) (
x
xdy y2e xy + e y = ydx e y − y2e xy
)
x
⇒ xy2e xydy + y3e xydy = (ydx − xdy)e y
x
( ydx − xdy ) e y
⇒ e xy(xdy + ydx) =
x
⇒ e xyd(xy) = e y d
Integrate
y2
()
x
y
x
e xy = e y + λ
( )
x
⇒ xy = ln e y + λ
16. (A,B,C)
sinxdy − ycosxdx
1
= − 2 dx
x
sin 2x
⇒ d
( )
y
sinx
= −
1
x2
dx
sinx
⇒ y=
+ csinx
x
∵ y → 0, x → ∞ ⇒ c = 0
sinx
f(x) =
x
∴
sin x
lim
x
x→0
2
sin x
∵ π <
=1
<1
x
π / 2 sin x
∴ 1< ∫
x
0
π
< 2
17. (A,B,C)
3x 2
dx
dy
=
x3
x3 − y
Put x 3 = t ⇒ 3x 2
dt
t
dx
dy
=
dt
dy
⇒ dy = t − y ⇒ (t − y)dt = t dy
⇒ t dt = ydt + tdt ⇒
6
3
⇒ x = 2x y + c
t2
2
= t. y +
c
2
18. (A,B,C,D)
πx
Area bounded by y = √2. sin 4 and x-axis between the lines x = 2, and x = 4,
Δ =
4
πx
√2 ∫ sin 4 dx =
2
4√2
πx
− π . cos 4
4√2
|
4
2
= π sq. units.
Let the drawn lines are L 1 : y − m 1(x − 4) = 0 and L 2 : y − m 2(x − 4) = 0, meeting the line x = 2 at the
(
)
(
points A and B respectively Clearly A = 2, − 2m 1 , B = 2, − 2m 2
Δ
4√2
1
Now Δ ACD = 3 ⇒ 3π = 2 .2. − 2m 1
2√2
⇒ m 1 = − 3π
2Δ
AlsoΔ BCD = 3 ⇒
⇒ m2 =
− 4√2
3π
8√2
1
= 2 .2 − 2m 2
3π
2√2
Required sum = − π
)
19. (A,B,D)
f(2 − x) = f(2 + x), f(4 − x) = f(4 + x)
⇒ f(4 + x) = f(4 − x) = f(2 + 2 − x) = f(2 − (2 − x)) = f(x) ⇒ 4 is a period of f(x)
Also f(2 − x) = f(2 + x) ⇒ graph of f(x) is symmetric about x = 2
2
4
0
2
2
4
6
8
0
2
2
4
6
⇒ ∫ f(x) dx = ∫ f(x) dx
Also period of f(x) is 4
⇒ ∫ f(x) dx = ∫ f(x) dx = ∫ f(x) dx = ∫ f(x) dx = . . . . . . .
50
∫ f(x) dx = 25∫ f(x) dx = 125
0
46
0
46 + 4
−4
52
−4+4
50
2
51
2
1
∫ f(x) dx =
∫
50
f(x) dx = ∫ f(x) dx
52
0
2
50
50
0
51
2
1
0
50
0
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx = ∫ f(x) dx
50
50
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx + ∫ f(x) dx − ∫ f(x) dx
1
0
50
= ∫ f(x) dx +
1
51 − 48
∫
1
f(x) dx − ∫ f(x) dx
0
50 − 48
0
50
3
1
50
0
2
0
0
= ∫ f(x) dx + ∫ f(x) dx − ∫ f(x) dx ≠ ∫ f(x) dx
20. (A,B)
1
1
0
0
(
f(x) = e x + ∫ e xf(t) dt = e x + ke x where k = ∫ f(t) dt
1
(
)
∴ k = ∫ e t + ke t dt = e + ke − 1 − k
∴ k=
0
e−1
2−e
(
)
e−1
ex
Thus f(x) = e x 1 + 2 − e = 2 − e
1
Obviously, f(0) = 2 − e < 0
Also,
f ′ (x)
ex
= 2 − e < 0 for ∀x ∈ R
⇒ f(x) is a decreasing function
1
1 ex
Also, ∫ f(x) dx = ∫ 2 − e dx =
0
0
[ ]
ex
2−e
1
0
e−1
= 2−e < 0
3
1
2
0
∵ ∫ f(x) dx ≠ ∫ f(x) dx
)
21. (A,D)
π / 2 sin ( 2n + 1 ) x − sin ( 2n − 1 ) x
An + 1 − An = ∫
sin x
0
0
⇒ An + 1 = An
π / 2 sin 2 ( n + 1 ) x − sin 2nx
Bn +1 − Bn = ∫
sin 2x
0
π/2
= ∫ 2cos2nx dx = 0
π / 2 sin ( 2n + 1 ) x
dx = ∫
sin x
0
dx = A n + 1
22. (A,B)
{ }
x u
∫ ∫ f(t) dt du
0 0
Integrating by parts, choose ‘1’ as the second function
{ }
u
x
0
0
= u∫ f(t) dt
x
x
x
x
0
0
0
− ∫ u f(u) du = x∫ f(t) dt − ∫ u f(u) du
x
x
x
0
0
0
= ∫ x f(u) du − ∫ u f(u) du = ∫ (x − u) f(u) du = ∫ u f(x − u) du
0
23. (A,D)
()
()
()
1 n
P(X = 4) = n C4 2
1 n
P(X = 5) = n C5
2
1
P(X = 6) = n C6 2
n
P(X = 4), P(X = 5) and P(X = 6) are in A.P.
n
( )
C4 + n C6 = 2 n C5
1
+
4!
( n −4) ( n −5)
6!
(
( n −4)
= 2 5!
)
30 + n 2 − 9n + 20 = 12n − 48
n2
− 21n + 98 = 0
⇒ n = 7 or 14
24. (B,D)
(a)
(b)
(c)
(d)
3!
=
63
10C
6C
1
36
2 −3
(C)
63
3 ×3!
6
63
6
3
1
4
2
5
= 9
= 36
1
= 8
25. (A,C)
dy
dx
=
√y − x
Put y − x = t
dy
dx
dt
− 1 = dx
dt
√t
⇒ 1 + dx =
dt
⇒
√t − 1
⇒ ∫
= dx
dt
√t − 1
= ∫ dx
Put √t = z ⇒ t = z 2
dt = 2zdz
2zdz
⇒ ∫ z − 1 = ∫ dx
(
⇒ 2∫ 1 +
1
z −1
)
dz = ∫ dx
⇒ 2z + 2ln|z − 1| = x + c
| √t − 1 | = x + c
⇒ 2√y − x + 2ln |√y − x − 1 | = x + c
x
c
or ln |√y − x − 1 | = 2 − √y − x + 2
⇒ 2√t + 2ln
⇒
⇒
x
− y−x
√y − x − 1 = c1e 2 √
x
− y−x
√y − x + c0e 2 √ = 1
26. (43.00)
Let G denotes the event that original signal is green
A denotes the event that station A receives correct signal
B denotes the event that station B receives correct signal
E denotes the event that signal received by station B is green
P
()
G
E
=
P(G ∩ E)
P(E)
( ) ( ) ( )
¯¯
¯
¯
¯¯
P(E) = P(GAB) + P GAB + P GAB + P GAB
4 3 3
4 1 1
1 3 1
1 1 3
( )
40
46
= 5 . 4 . 4 + 5 . 4 . 4 + 5 . 4 . 4 + 5 . 4 . 4 = 80
¯¯
P(G ∩ E) = P(GAB) + P GAB = 80
P
()
G
E
=
P(G ∩ E )
P(E )
=
40
46
=
⇒ p + q = 20 + 23 = 43
20
23
=
p
q
27. (7.00)
1/ 2
Area = 2 ∫
0
√
x
2
dx +
1
2
×1×
1
2
=
1
3
+
1
4
7
=
12
28. (2.00)
0
A=∫
e x dx
−1
=
[ ]
0
ex − 1
e
+ Area of rectangle OBCD − ∫ log ex dx
[
1
= 1 − e + e − [e − e + 0 + 1]
1
= e − e ∈ (2, 3)
So, [A] = 2
]1
+ e × 1 − xlog ex − x e
1
29. (7.00)
The expression |x − y| + |x + y| ⩽ 8, represents the interior region of the square formed by the
lines x = ± 4, y = ± 4 and xy ⩾ 2. represents the region lying inside the hyperbola xy =2
Required area
( )
4
2
Δ = 2∫
4
4 − x dx = 2(4x − 2lnx) 1 / 2
1/ 2
= 4(7 − 3ln2) sq. units.
= 4(7 − ln8) sq. units.
30. (2.00)
dy
dx
∫
= xy(1 + y)
dy
= ∫ x dx
( 1+y) y
x2
2y
1+y
=e2
⇒ f(2) =
(∵
f(0) = 1)
e2
2 − e2
31. (2.00)
(
)(
) dx
I = ∫ x (x + x + x )(2x + 3x + 6 ) dx
I = ∫ (x + x + x )(2x + 3x + 6x ) dx
Let 2x + 3x + 6x = t ⇒ 18(x + x + x ) dx = dt
I = ∫t
dt = t
+C=
(2x + 3x + 6x )
I = ∫ x 9 + x 6 + x 3 2x 6 + 3x 3 + 6
8
8
5
9
1
18
5
6
1/ 3
1
2
6
2
9
1/ 3
3
3 1/ 3
6
3
1 4/ 3
24
⇒ a = 24 ⇒ 48a = 2
1/ 3
8
1
24
5
9
2
6
3 4/ 3
+C
32. (8.00)
Let I = \int\limits_0^1 {{}^{207}{C_7}.\underbrace {{x^{200}}}_{II}} .\underbrace {{{\left( {1 - x}
\right)}^7}}_I\;dx
I = {}^{207}{C_7}\left[ {\left. {\underbrace {{{\left( {1 - x} \right)}^7} \cdot \frac{{{x^{201}}}}
{{201}}}_{{\text{zero}}}} \right|_0^1 + \frac{7}{{201}}\int\limits_0^1 {{{\left( {1 - x} \right)}^6} \cdot
{x^{201}}dx} } \right] = {}^{207}{C_7} \cdot \frac{7}{{201}}\int\limits_0^1 {{{\left( {1 - x} \right)}^6}
\cdot {x^{201}}} dx
Integrating by parts again 6 more times
= {}^{207}{C_7} \cdot \frac{{7!}}{{201.202.203.204.205.206.207}}\int\limits_0^1 {{x^{207}}} dx
= \frac{{\left( {207} \right)!}}{{7!\left( {200} \right)!}} \cdot \frac{{7!}}{{201.202...207}} \cdot \frac{1}
{{208}} = \frac{{\left( {207} \right)!}}{{\left( {207} \right)!}} \cdot \frac{1}{{208}} = \frac{1}{k}\;\;\;
\Rightarrow \;\;k = 208
\Rightarrow \frac{k}{{26}} = 8
33. (0.00)
We have J = \int\limits_{ - 5}^{ - 4} {\left( {3 - {x^2}} \right)} \;\tan \left( {3 - {x^2}} \right)\;dx
Put \left( {x + 3} \right) = t, we get:
J = \int\limits_{ - 2}^{ - 1} {\left( {3 - {{\left( {t - 3} \right)}^2}} \right)} \tan \left( {3 - {{\left( {t - 3}
\right)}^2}} \right)\;dt
= \int\limits_{ - 2}^{ - 1} {\left( { - 6 + 6t - {t^2}} \right)} \tan \left( { - 6 + 6t - {t^2}} \right)\;dt
= - \int\limits_{ - 2}^{ - 1} {\left( {6 - 6x + {x^2}} \right)} \tan \left( {6x - {x^2} - 6} \right)dx = - K
\Rightarrow J + K = 0
34. (1.00)
{I_n} = \int\limits_0^1 {{x^n}{{\left( {1 - {x^2}} \right)}^{1/2}}} dx
= \int\limits_0^1 {{x^{n - 1}} \cdot x{{\left( {1 - {x^2}} \right)}^{1/2}}} dx = \left( { - {x^{n 1}}\frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{3}} \right)_0^1 + \int\limits_0^1 {\left( {n - 1} \right){x^{n 2}}} \frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{3}dx
= 0 + \left( {\frac{{n - 1}}{3}} \right)\int\limits_0^1 {{x^{n - 2}}} \left( {1 - {x^2}} \right)\sqrt {1 - {x^2}}
\;dx
= \left( {\frac{{n - 1}}{3}} \right)\left[ {\int\limits_0^1 {{x^{n - 2}}} \sqrt {1 - {x^2}} \;dx - \int\limits_0^1
{{x^n}} \sqrt {1 - {x^2}} \;dx} \right]
3{I_n} = \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}
3{I_n} + \left( {n - 1} \right){I_n} = \left( {n - 1} \right){I_{n - 2}}
\frac{{{I_n}}}{{{I_{n - 2}}}} = \frac{{\left( {n - 1} \right)}}{{\left( {n + 2} \right)}}\;\;\; \Rightarrow
\;\;\mathop {\lim }\limits_{n \to \infty } \frac{{{I_n}}}{{{I_{n - 2}}}} = 1
35. (7.00)
In a chess-board, there are 8 \times 8 = 64 small squares and any three small squares can be
chosen at random in
n\left( S \right) = {\,^{64}}{C_3} = 64 \times 31 \times 21 = 41664 ways.
While choosing 3 small squares so that they lie on a diagonal line, we observe that they lie on lines
such as {{\text{L}}_1},{{\text{L}}_2},\,...,\,{{\text{L}}_5},{\text{BD}} and lines on the other side of
{\text{BD}}, and an equal number of opportunities on the either side of the other diagonal
{\text{AC}}. Hence, the total number of favourable cases is
n\left( E \right) = 2{\,^8}{C_3} + 4\left[ {^7{C_3} + {\,^6}{C_3} + {\,^5}{C_3} + {\,^4}{C_3} + {\,^3}{C_3}}
\right]
= 2 \times 56 + 4\left( {35 + 20 + 10 + 4 + 1} \right)
= 112 + 280 = 392
Hence p = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{392}}{{41664}} = \frac{7}{{744}}
\Rightarrow 744p = 7