Inverse trigonometric functions &
General Solution of Trigonometric Equations.
------------------ ----------------------------- ----------------1. Sin (
)=
a) √
b)
c)
√
d)
Ans b.
Solution :
a) is rejected.
Method 1. Ans a: √17 > 1
) =
d is rejected.
w.k.t Sin ( sin
If sin θ = 8/17 then cos θ = 15/17
GE ≠ 15/17 . c is rejected.
b is the answer.
OR
= θ then sin θ =
Let sin
sin θ =
Cos θ =√1
1
then G E =sin
15 +8 = 17
=
=
=
=
=
=
√
------------------ ----------------------------- ----------------2.
a)
The value of Tan (
b)
Ans: b.
Solution:
Tan (
-
) =
c)
d)
Let tan
- tan
= θ then tan θ =
) = Tan (
–
- θ) =
=
–
=
.
------------------ ----------------------------- ----3. The domain of the function
f(x) =
a) 1 x 2 b) -1 x 2
Ans: a.
Solution : we know that sin
-1
1
2
is
c) 2
x
4 d) 1
x
is defined when - 1
2
2
1 x
4
1 x 2
------------------ ----------------------------- ------
5
x
1
4. If cos ( 3
) = tan (
) then x =
a) 0
b) 1
c) -1
d) 0,1, -1
Ans: d .
Solution : 1) inspection method GE is satisfied by all values .
Hence ans is d. oR
=θ
cos θ = x
ii) cos
GE. cos3 θ = x
4 cos θ – 3 cos θ = x
4x - 3x = x
4( x - x )= 0
x( x - 1) = 0
x = 0,1, -1.
------------------ ----------------------------- ----------------- = + β and
-β ) =√
+
=
then
a) 1 + a b) a + b
c) 1 + ab d) 1+ b
Ans: c.
Solution:
α - β ) = √1 b
α - β = cos
√1 b ) = sin
sin (α - β) = b
Now sin α + cos β = sin α + 1 sin β
1 + sin α sin β = 1 + sin (α + β) sin (α - β) = 1 + a b.
------------------ ----------------------------- ---------------
5.
If
6. If
+
then x satisfies
b) 2x - 3x = 0
a) 2x + 3x + 1 =0
d) 2x + x + 1 =0
c) 2x + x - 1 =0
Ans: b
Solution: ( x= 0 satisfies GE and b also. Hence ans b)
Or
sin
+ cos
1
sin
1
- sin
+ cos
sin
cos
1
- 2 sin
= -2 θ
(sin θ = x)
1 – x = cos (-2 θ ) = cos 2 θ = 1 – 2 sin2 θ
1 – x = 1 – 2x
x = 2x
2x - x = 0
x = 0 or x =
x = 0 is the solution
( x = does not satisfies the equation )
------------------ ----------------------------- ----------------- ------------
7.
If sec-1x = cosec-1 y then sin-1( ) + sin-1( ) =
a) Π
b) π / 2
c)
- π /2
Ans: b.
Solution: Given sec-1x = cosec-1 y
cos
) = sin
)
d) - π
sin-1( ) + sin-1( ) = sin-1( ) + cos-1( ) =
.
------------------ ----------------------------- ----------------- -----------8. If tan ax – tan bx = 0 ( a ≠ b) then the values of x form a
series in
a) AP
b) G P
c) H P
d) AGP.
Ans: a.
Solution : tan ax = tan bx
ax = n π + bx
ax – bx = n π
x ( a – b) = n π
. put n = 1 ,2, x=
,
etc. which are in AP.
x =
------------------ ----------------------------- ----------------- -----------9. If
( ) +
a)
0
( )=
b)
Ans: C
Solution: w.k.t sin
)
√1
(
√
) then x =
√
c)
x )+ sin
sin ( ) + sin ( ) = sin (
√
√
( y ) = sin
( x 1
1
1
)+
d) π / 2 .
+ y
))
x =( √5 + 4 √2 )/ 9
= sin ( ( √5 + 4 √2 )/ 9 ) )
Ans c
------------------ ----------------------------- ----------------- -----------10. The value of
5+
=
a)
b)
c)
Ans: c
Solution: GE = π + tan (
= π + tan (
) - cot
.
d) .
) - cot
= π - tan ( ) - cot
= π – [tan ( ) + cot
] =π =
Ans c
------------------ ----------------------------- ----------------- -----------
11. Two angles of a triangle are
third angle is
b)
c)
a)
Ans: b.
Solution: A + B + C = Π where A = cot
B= cot
A+ B = tan
+ tan
and
, then the
d)
.
2 = tan
3 = tan
.
= tan
] = tan
1 ] =
.
.
=
C = π - ( A + B) = π ------------------ ----------------------------- ----------------- ------------ -----12.
a)
: .
:
1 x
1 x
cos cos
cos sin x
sin cos x
√1
sin sin √1 x
GE
sin
√1 x
cos
√1 x
π
cos x
sin x
.
2
cos 0
cos
or Put x
0 then GE
sin
π
sin
1
cos
1
.
sin
2
x
sin
2
------------------ ----------------------------- ----------------- ------------ --=
13. If
then the values of x are
a) 0 or ½
b) 1 or 1/3
c) -1 or -1/3 d) -1 ,1/2
Ans: a
Solution type 1: Inspection method: GE is satisfied by a)
1 x
type 2: GE = cos x – sin x = sin
= – sin x - sin x = sin
1 x
sin
1
x =
– 2 sin
x
( 1 – x ) = sin – 2 sin x ) = cos 2 sin x )
= 1 – 2 sin ( sin x) = 1 – 2x
2x
x
2x
x =0
1 – x = 1 – 2x
x ( 2x – 1) = 0
x= 0 , ½
------------------ ----------------------------- ----------------- ------------ ----
14.
If
+
=
then x =
b) 1
c)
d) 0
a) ± √3
√
Ans: b .
Solution. Inspection method: ± √3 is not a solution since
cos √3 is not defined. GE is satisfied by x = 1
------------------ ----------------------------- ----------------- -------+
+
=
15. The value of
c) cot 1 d) cot 1
a) 0
b) cot 26
Ans: a .
Solution: w.k.t cot a - cot b = cot
.
= cot 5 - cot 4
cot 21 = cot
Similarly cot 13 = cot 4 - cot 3
8 = cot 3 - cot 5
And cot
GE = cot 5 - cot 4 + cot 4 - cot 3 + cot
0
------------------ ----------------------------- -------------16.
…
If
.
+
then x =
for 0< | | < √
3) 0
d)
a) 1
b)
Ans: a)
Solution: clearly when x = 1 GE =
Or x
3 - cot
5 =
=
ans: a)
=x
=
=
x (2 x
x (2+x)
x = 0 , 1. But x ≠ 0.
- ----------- ----------------------------- ----------------- -----------=
17. The value of
2
b)
c)
d)
a)
Ans: b. searching method . put x = 1 then
GE = 2tan √2 - 1] = 2
= .
Put x = 1 in choices,
1 = tan 1 = Hence answer must be either a or
Hence cot
b
Put x = -1 then
GE = 2tan
)]=
√2 - tan (π
= 2tan
√2
+ 1 ] = - 2tan
= -2 ( ) =
√2
-1]
Put x= -1 in a and b
Then
cot
1 = π
=
b
tan
1)=
. Hence answer is b
------------------ ----------------------------- ----------------- ----------√
18. If 2
=
, then x is in
a) [-1, 1 ] b) [- , 1]
c) [, ] d) [ 0, 1] .
√
√
Ans: c.
Solution w.k. t. the range of RHS is [
2 sin
sin
x
x
sin
sin
-
√
,
]
x
x
√
------------------ ----------------------------- ----------------- ---=2
is
19. The solution set of the equation
a) { 1, 2}
b) { -1, 2} c) {-1, 1, 0} d) {1, ½ , 0}
Ans: c)
Solution : clearly
is not defined.
reject a) and b) .
consider d)
=
but
2
.
=
=
≠
hence reject d)
Only possibility is ans c.
or
=2
=
x=
x(
) = 2x
x=0
x( 1 – x ) = 0
x = 0, ±1
------------------ ----------------------------- ----------------- -----------+
20. If in a triangle ABC , C is 900, then
a)
b)
c)
Ans: c)
Solution:
ABC is a right angled triangle .
Put a= 3 , b= 4 and c= 5 . Then C= 90
+ tan
= tan
Then tan
+ tan
= tan
d)
+
tan
= tan
=
+ tan
= where m = 1.
Ans: c.
------------------ ----------------------------- ----------------- ------------
=
21.
The value of tan[
a) a
) +
)] =
b)
c) b
d)
Ans: b) .
Solution Inspection method. Step1. Put a = 1 , b= 0
1)] = 1.
Then GE = tan [tan
When we put a=1, b= 0 in choices answer is either a or b
Step2. a=0, b= 1
1)+ tan
1) ] = tan [ ] =∞
Then GE = tan [tan
When we put a=0, b= 1 in choices
Ans b)
( 1/a = 1/0 = ∞ ).
---- -------------- ----------------------------- ----------------- -----------....an is an AP with common difference d , then
22. If , ,
tan {
+
a)
b)
+.......+
} =
c)
d)
Ans : b )
....an are in AP
Solution: Given , ,
- a1 = a3 – a2 = ...... an - a n-1 = d
+ tan
Consider tan
= + tan
= tan
a
= tan
a
+.......+ tan
- tan
- tan
a
+ tan
a
a - tan a
+ tan a - tan
= tan
= tan
GE: tan tan
+.......+ tan
+ ...........
a
=
=tan
) =
Hence answer is b.
------------------ ----------------------------- ----------------- ------------
23. If
+
=
then x =
a) 0
b) 1
c) √2 -1
Ans: d).
Solution: Inspection method:
Clearly x = o is not a solution since tan
And cot
0 =
LHS =
≠
.
When x= 1 LHS of GE =
X = I is not a solution
+
≠
When x= √2 -1
, LHS of GE =
+
d)
1
0 =0
=
√2 -1 is not a solution.
x = -1 is only a solution
------------------ ----------------------------- ----------------- 24. If
+
+
= , then
∑ X √1 x =
a x y z
b) 2xyz
c) 2( x+ y + z)
d) x y z
Ans b
Solution: It is a standard result .
Or
Inspection method.
Step 1. put x= y = 1 and z = 0. Then
∑ sin x = π and choice b) and d) matches with this value.
and z = 1 . Then
Step 2. Put x =
√
∑ sin x = π and choice b) matches with this value
b is the answer.
------------------ ----------------------------- --------then y =
25. If tan( x + y ) = 33 and x =
a)
b) tan 30
c) tan
d) tan
Ans: d)
y =tan 3 3 - tan 3
Solution: x + y = tan 3 3
= tan
=tan
.
answer d).
------------- ----------------------------- ----------
= tan
26. 2
with │x│
b) cos 2x c) cos x
d) tan
1 x
Ans: c
Solution : put x = 1 then GE :
&
When x = 1
, a) and b) are not defined due to
c)
= 0 and d)
= .
Hence Ans is c.
------------------ ----------------------------- ----------------27. The general solution of sin5x = k
where k is a real root of
+2x - 1 = 0 is given by
2 +
1
b x=
+
1
a) x =
a)
sin
=
2x
(c) x =
1
+
1
(d) x = n π +
Ans (a).
Solution:
Consider 2x - x +2x - 1 = 0
x (2x -1) +1 ( 2x – 1) = 0
x +1 ) ( 2x – 1) = 0
Since x is real x = ½ .
1
5x = n π +
x=
sin 5x = k = ½ = sin ( )
α =n
1
+
α=
1
Ans (a)
------------------ ----------------------------- ----------------28. If 5cos 2 θ + 2
a)
( b)
+ 1 = 0 , θε (- π , π ) ,then θ =
, cos
c) cos
(d)
, π - cos
Ans: (d ).
Solution: GE = 5cos 2 θ + 2 cos
+1=0
5( 2 cos θ -1) + ( 1 + cos θ ) + 1 =0
10 x + x - 3 =0 where x = cos θ
10 x + 6x - 5x - 3 =0
2x ( 5x + 3 ) -1 ( 5x + 3 ) =0
cos θ = ½ = cos
Solution is
,
x=
and cos θ =
, π - cos-1 (3/5)) .
,
= cos( π - cos-1 (3/5))
29. If
The general solution of
a) 2n π ±
(b)
(c). 2n π ±
= 3 is given by θ =
nπ±
(b)
nπ±
Ans: (d)
=3
Solution: GE =
tan θ =±√3 = tan( ± )
tan θ = 3
θ= nπ±
------------------ ----------------------------30. If Sec x cos5x + 1 =0 where 0<x < 2 π , then x is equal to :
a) ,
b)
d) none of these
c)
Ans: c
Solution: GE :
+1 = 0
cos5x + cosx = 0
2 cos 3x cos 2x =0
Cos 3x = 0 or cos2x =0
3x =
X=
or
or
2x =
or
or
or
x=
or inspection method.
------------------ ----------------------------- ----------------- -----------31. The solution of the equation
2x = 1+
x is
a) x = ( 2n + 1 ) , n Є Z (b) x = n π , n Є Z
.(C) x= ( 2n + 1 )
,nЄZ
(d) no solution.
Ans: d
Solution:GE: sin 2x = 1+ cos x
0 The max value of LHS is 1 and minimum Vaue
Here cos x
of RHS is 1 . This is possible only when
cos x = 0 and sin 2x
=1
x = and 2x =
X=
.This is not possible . Hence solution does not exists.
------------------ ----------------------------- ----------------- ------------
32. The most general value of θ satisfying the equation
+ √ tan θ - 1) 2 = 0 are given by
a) n π ±
b) n π + ( -1) n
c).
(d) 2n π +
2n π +
Ans: c.
Solution: a
b =0
a = 0 and b =0
1
2 sin θ = 0 and √3 tan θ - 1 = 0
sin θ = - ½ and tan θ =
G.S. is θ = m π + ( -1)
m
√
and θ = mπ +
. Both holds good
only when m is odd
m = 2n + 1
= 2n π +
G.S. is θ = ( 2n + 1)π
------------------ ----------------------------- ----------------- -----------33. The most general solution of θ satisfying
Tan θ + tan(
a) n π
±
+θ) = 2
,nЄZ .
.c) 2n π ±
b)
,nЄZ
is / are
2n π +
d) )
,nЄZ
2n π +( -1) n
,nЄZ
Ans: a
Solution: GE: Tan θ + tan(
Tan θ + tan 135
Tan θ +
θ
+θ) = 2
=2
=2
Tan θ + tan θ - 1 + tan θ = 2 + 2 tan θ
tan θ = 3
tan θ = ±√3
θ=nπ±
,nЄI
------------------ ----------------------------34. The general solution of x for which cos2x , , and sin 2x are
in Ap , are given by
a) n π , n π +
b) n π , n π +
c) n π +
,
of these.
Ans: b .
Solution: a,b,c are in AP
a + c = 2b
Cos2x + sin 2x = 1
Sin2x = 1 – cos2x = 2 sin2x
2 sinx cos x - 2 sin2x = 0
2 sinx( cos x – sinx) =0
sinx = o or cosx = sinx
d) none
sinx = 0 or tanx = 1
X = n π or x = n π +
, n Є Z.
------------------ ----------------------------- ----------------35. If √ sec θ + tan θ =1 then the general solution of θ=
b) 2n π +
a) n π +
c) 2n π -
d) 2n π ±
Ans : c.
GE :
√
+
√
cos θ -
cos( θ +
G.S. is θ +
cos θ – sin θ = √2
=1
sin θ = 1
√
) = cos( 0 )
=2nπ
θ =2nπ -
.
------------------ ----------------------------- ----------------36. The equation sin
cos
+ 1 =0
has
a) one solution
b) two solutions
c) infinite solutions d) no solution.
Ans: d
GE: sin3x = -2 < - 1 . Hence solution does not exists.
------------------ ----------------------------37. The general solution of
a) x = n π b) x = ( 4n + 1)
c) x = ( 4n + 1)
√
d) n π +
(
+
) = 2 is
.
Ans: b.
Solution clearly x = satisfies the GE.
------------------ ----------------------------38. If – π x Π ,– π y Π and
cos x + cosy = 2 then general solution is x =
a)
2nπ+y
b) 2 n π - y
c) n π + y
d) n π + ( -1) n y.
Ans a.
Solution : the Max. value of cosx is 1 .
Hence cos x + cosy = 2
cosx = 1 and cosy =1
x = 0 and y = 0 hence x – y = 0
Hence cos( x – y) = cos 0
X–y=2nπ
x=2nπ+y
------------------ ----------------------------- -----------------
39. The equation 3
x + 10cosx – 6 =0 is satisfied if
a) x = n π ± cos
b) x = 2n π ± cos
c) x = n π ± cos
b) x = 2n π ± cos
Ans: b.
Solution GE.: 3( 1 - cos x ) + 10cosx – 6 =0
3 – 3t + 10 t - 6 =0 where t = cosx
3t - 10 t +3 =0
3t - 9t - t +3 =0
3t( t – 3 ) -1 ( t – 3) =0
( 3t -1) (t-3)=0
cosx = t = 3 > 1 No solution ,
Cosx = 1/3 = cos α where α = cos
X = 2n π ± cos
i.e. b).
------------------ ----------------------------- ----------------40. If tan2x = tan
a)
c)
, then the value of x =
√
b)
√
d) None of these
Ans: a)
Solution: GE: tan2x = tan
2x = n π +
X=
2x - nπ x– 2 = 0
√
------------------ ----------------------------- ----------------41. If the equation cos3x
x + sin3x
x= 0., then x =
a) ( 2n+1)
b) ( 2n - 1)
c)
d) n π , .
Ans: a:)
Solution: put n = 0 in answers a,b,c and d , Substitute x=0, x=
, x=
. GE is not satisfied when x =0 Hence c and d are not
correct answers.
GE is not satisfied when x =
.
answer.
Hence a is the answer.
Or
Hence b is not the correct
cos3xcos x + sin3x sin x= 0.
cos3x [ ( cos3x + 3 cosx )] + sin3x[ ( 3sinx - sin3x)] =0
( cos 3x - sin 3x ) +3 ( cos3x cosx + sin3x sinx )} =0
{cos6x + 3 cos2x} =0
2x = A
cos A
{cos3A + 3 cosA} =0
cos2x = 0
0
cos 2x
0
x = ( 2n + 1)
2x = ( 2n + 1)
------------------ ----------------------------- ----------------- -----------–
and
42. If x ≠
of x are given by
a) 2n π +
c)
2n π +
= 1 , then all solutions
b) ( 2n + 1) π
1
-
d) None of these.
Ans : d.
Solution: Since x ≠
, cos x ≠ 0,1,-1. Hence only possibility is
2=0
sin x – 3 sinx
sin x – 2 sinx – sinx 2 0
( sinx - 1 ) ( sinx – 2) =0
Sin x = 1 or 2 which is not possible as x ≠
and 2 > 1
Hence d is the answer.
------------------ ----------------------------- ----------------x+
x +..........∞ = 4 + 2√ with 0<x <
43. If 1 + sin x +
, then x =
π and x ≠
b)
a)
c)
or
d)
or
Ans d.
Solution : put sinx = r then
GE: 1 + r + r + ........ ∞ = 4 + 2√3 ( in GP.)
With | | < 1 since 0<x < π and x ≠
S
=
= 4 + 2√3
= 4 + 2√3
1 – sinx =
1 – sinx =
√
√
x
√
√
=
√
.
√
sinx = 1 -
√
=
=
√
x = 60 or 120 Ans: d.
------------------ ----------------------------- ----------------- ------= cos2x – 1 is
44. The general solution of of
c) n π
d) (2n + 1) π
a) 2n π b)
Ans: c
Solution: when n=1 a) π
b)
c) 2 π d is 3 π
Among these GE is not satisfied by b)
b is not the correct answer.
Since GE is satisfied for π , 2π , 3 π also.
General solution is θ = n π .
Or
tan x
= cos2x – 1
= - 2sin x
= - ( 1 – cos2x)
sin x = - 2sin x cos x
sin x ( 1 + 2cos x ) = 0
sin x = 0
sinx = 0
x= n π where n Є Z
------------------ ------------------------------------------------45. Cot θ =sin2 θ ( θ ≠ n π ),if θ =
b) ,
c) only
d) only
a) ,
Ans b)
Solution : when θ = , in GE
LHS = RHS = 1 . GE is satisfied by θ =
solution
θ= ,
.
Or
GE
= 2 sin θ cos θ
cos θ = 2 sin θ cos θ
cos θ ( 1 - 2 sin θ ) = 0
Cos θ =0
θ=
sin θ = ½
sin θ = ±
θ=±
------------------ ------------
√
also . and
is not a
46. If sin( cotθ ) =cos( tanθ) then the value of θ =
a)n π +
b ) 2n π ±
c) n π -
d) 2n π ±
Ans: a)
Solution: we know that
π cotθ = π tanθ
sin = cos
cotθ = tanθ
tan θ = 1
θ =nπ +
.
------------------ ----------------------------- ----------------- -----------47. If
tan θ + 3 cot θ = 5 sec θ then θ =
a) n π + ( -1)n
, nЄ Z . b) n π + ( -1)n
, nЄ Z
c) n π + ( -1)n+ 1
, nЄ Z or n π + ( -1)n
d) n π + ( -1)n
, nЄ Z
, nЄ Z or n π + ( -1)n
, nЄ Z
Ans: b).
Solution.
+3
GE:
=
sin θ + 3 cos θ = 5 sin θ
1 + 2 cos θ = 5sin θ which is satisfied for x =
only.
answer is b).
------------------ ----------------------------- ----------------- -48. If tanx + tan4x + tan7x = tanx tan4x tan7x then x =
a) n π / 3
b) n π / 4
c) n π / 6
d) n π / 12
Ans d).
Solution : w. k.t if tan x + tany + tanz = tanx tany tanz then ]
tan( x + y + z ) = 0.
tan12x = 0
12x = n π
x = n π /12.
------------------ ----------------------------- ----------------49. The general solution of sinx + sin7x = sin4x in ( 0, ) are
a)
,
b)
,
c)
,
d)
,
Ans d)
Solution: GE : sin7x + sin x – sin4x =0
2 sin4x cos 3x – sin 4x =0
sin4x( 2 cos3x - 1) =0
sin4x =0
4x = n π or x = n π / 4 = π / 4 in ( 0,
cos 3x =
= cos ( ) in ( 0,
3x =
x=
x= ,
)
Ans d
------------------ ----------------------------- -----------------
)
|,0 x
50. The number of solutions of cosx = |
a) 3
b) 2
c) 4
d)
Ans: a.
Solution: clearly 1
0
cosx = 1 + sinx
( | | = x if x
0)
cos x – sin x = 1
By inspection x = 0, 2π , 3 π /2 are 3 solutions in [0,3π]
Ans a
------------------ ----------------------------- ----------------- ----51. The set of values of x for which
a) φ
b) { n π +
,nЄZ}
c)
–
3 π is
5
= 1 is
d) { 2n π +
,nЄZ}
Ans a.
Solution: Clearly tan x = 1 = tan ( )
X=nπ+
,nЄZ .
But for no values of x , sin2x is not defined , so that the equation
has
no
solution.
------------------ ----------------------------- ----------------52. The general solution of cos7 θ cos 5 θ = cos3θ cosθ
is
a)
,
b)
,
c)
,
d)
,
Ans c.
Solution :
GE: cos7 θ cos 5 θ = cos3θ cosθ
( cos12 θ + cos2 θ ) = ( cos 4 θ + cos2 θ )
cos12 θ – cos 4 θ = 0
-2 sin 8θ sin 4θ =0
,
8θ = n
or 4θ = n
θ=
Ans c
-------------- ----------------------------- ----------------- -----------53. If sec2x + + cosec2x = 4 , then the value of x is
b)
c)
d)
a)
Ans d.
Clearly by inspection method
d)
is the answer.
------------------ ----------------------------- ----------------- ------
54. The equation 2 sin θ cos θ = x 2 +
a) one real solution
c) two real solutions
ans b)
b) no solution
d) θ = n π .
Solution : GE : sin 2 θ = x 2 +
If x
has
. clearly sin 2 θ > 0.
1 solution does not exists.
If 0<x< 1 then x
< 1 but
>1 hence x 2 +
>1
Solution does not exists.
no solution
------------------ ----------------------------- ----------------- -----------55. If A and B are acute angles such that sinA = sin2B and
2 cos2A = 3 cos2B then A is
a)
b)
c)
d)
Solution : Clearly by inspection method A=
Since when A=
, B =
.
from sinA = sin2B . These values
satisfies 2 cos2A = 3 cos2B also.
a=
is the ans.
------------------ ----------------------------- ----------------- ------+ cos
56. Let n be a positive integer such that sin
,then
a) n = 4
c) n = 2
Ans d.
Solution: clearly by inspection n = 6 .
Or
sin
+ cos
sin
sin
b) n= 1,2,3,4,.......
=
=
=
√
-1 =
squaring 1 + 2 sin
=
√
d) n = 6
cos
=
which is true when n = 6.
n =6 only.
------------------ ----------------------------- ----------------- -----------then the value of x in ( 00,
57. If sin 400 = k and cosx = 1 – 2
3600 ) are
b) 80o and 280o
a) 40o and 140o
c) 40o and 220o
d) 80o and 260o
Ans: b.
Solution : Now cosx = 1 – 2 sin 2400 = cos800 .
x = 800 and 3600 - 800 = 2800 .
x = 800 , 2800 in ( 00, 3600 )
------------------ ----------------------------- ----------------58. If 3 cos2x - 2√ cos x sinx - 3 sin2x = 0 then x =
+
b) +
c)
d) n π +
a)
Ans: a)
Solution : GE: 3 cos2x - √3 sin2x =0
3 cos2x = √3 sin2x
= √3
x=
2x = n π +
+
tan2x = √3 .
.
------------------ ----------------------------- ----------------- ------+
59. The most general solution of
=0
a) 2n π + , n ЄZ , b) n π + , n ЄZ
c) 2n π -
, n ЄZ
Ans: b.
Solution: GE: log
log
tanx
d) n π -
, n ЄZ
tanx = - log
= - log
tanx = log
tanx
log
Tanx = 1
x = n π + , n ЄZ
cotx
= log
tanx
sinx = cosx
ans :b .
------------------ ----------------------------- ----------------- ------60. The general solution of 3 tan( θ – 15 o)= tan ( θ + 15 o ) is
±
b) n ±
c)
+ ( -1) n
d) 2n a) θ =
Ans: c
Solution: 3 tan( θ – 15 o)= tan ( θ + 15 o )
3
–
–
=
3 sin θ – 15 cos θ
2 sin θ – 15 cos θ
sin2 θ + sin ( -300)
Sin 2 θ = 1
2θ
15 = sin θ 15 cos θ 15
15 + sin ( -300) =0
+ sin ( -300) =0
= n π + ( -1) n
θ=
+ ( -1) n
------------------ ----------------------------- ----------------- -------
61. The most general solution of √ tan θ +1 = 0 & √ secθ 2 =0 is
a) 2n π -
b) 2n π +
c) n π + ( -1)n
d) 2n π +
Ans b)
Solution: Given
tan θ =
and cos θ =
√
√
θ in IV
quadrant
take θ = 2 π –
=
general solution is θ = 2n π +
Ans b.
------------------ ----------------------------- ----------------- -----------62. If cosx + cosy = 1 and cosxcosy = ¼ then the general
solutions are
, y = 2k π ±
b) 2n π ±
, y = 2k π ±
a) x = 2n π ±
c) x = 2n π ±
, y = 2k π ±
b) n π ±
Ans: b.
Solution: clearly by inspection method x =
,y=kπ±
, y=
is a solution .
Hence ans is b).
Or
cosx
Cosx – cosy =
cosy
4 cosxcosy
Now cosx + cosy = 1 and cosx
cosx = ½ and cosy = ½
x = 2n π ±
, y = 2k π ±
=
1
4
=0
- cosy = 0
ans : b.
------------------ ----------------------------- ----------------- ------=
63.
a
b)
-
c)
d)
Ans : c
Solution: put x= 0 then GE :
.
In choices only a and c results in
Hence answer is either a or c.
Put x = 300 . then GE: tan
√
= tan
√3 =
In choices a reduces to 150 and c reduces to 600. Hence c is the answer.
------------------ ----------------------------- ----------------- ------+ cos-1(
64 . Tan[
a) 2x
b)
) ] + Tan[
c)
2x
reject
0
) ] with x ≠ 0
x
2 √1
d)
Ans: b)
Solution : put x = 1 in GE: = 2
Put x =
cos-1(
-
d.
0
in GE: tan( 45 + 30 ) + tan ( 450 – 300 )
= tan ( 750 ) + tan 150 = 2
√3 + 2 - √3
= 4
In choices b) gives 4 . hence b is the correct answer.
------------------ ----------------------------- ----------------- -----65. If sin-1( ) +
a) 4
) =
b) 5
, then x =
c) 1
Ans: d.
d)
3
) = sin-1( )
Solution: cosec
GE : sin-1( ) + sin-1( ) =
sin-1( ) = cos-1( ) clearly x =
sin-1( ) = cos-1( )
3.
------------------ ----------------------------- ----------------- ------66. The value of cos( 2
+
) at x = 1/5 is
a)
√
b)
√
d) 2√6
c)
Ans: b
Solution: cos( 2cos x + sin x )
+ cos x ) = - sin (cos
= cos (
= - sin (sin
At
x = 1/5
ans is
-
√1
x)
x ) = - √1
1
=-
=-
x
√
.
------------------ ----------------------------- ----------------- --67. The general solution of cosecx + cotx = √ is
b) x =
a) x = 2n π ±
c) x =
-
+
d) x = 2n π ±
-
Ans: d.
Solution: GE: cosecx + cotx = √3
1 + cosx = √3 sinx
cosx + √3 sinx = - 1
cos( x +
) = cos
cosx +
√
sinx = -
G.S. is x = 2n π ±
-
ans :d.
------------------ ----------------------------- ----------------68. If α, β and γ are the roots of the equation x3 + m x2 +
3x + m = 0
, then the general solution of
+
+
=
b) n π
c)
d) depend upon the value of
a) ( 2n + 1)
x.
Ans: b
Solution : consider x3 + m x2 + 3x + m = 0
∑ α = -b/a = -m , ∑ α = c/a = 3 ; α β γ = -d/a = -m
tan
Now
=tan
α + tan
β + tan
–
γ = tan
–
= tan-1(0) =0.
–
G.S. is
tan α +tan β +tan γ = n .where n Є Z .
Ans b
------------------ ----------------------------- ----------------- -----------69. The equation 3 cosx + 4 sinx = 6 has
a) finite solution
b) infinite solution
c) one solution
d) no solution .
Ans: d
Here
r=
√
=
> 1 where a= 3, b= 4 c = 6.
Hence no solution.
------------------ ----------------------------- ----------------- -----------70.
The value of x satisfying the equation
sinx +
=
√
is given by
a) 100
b) 300
d) 600
Solution :
satisfies the equation .
because
c) 450
Ans: d
Clearly by inspection x = 600
sin 600 +
=
√
+
√
=
√
------------------ ----------------------------- ----------------- -----------71. If log 5(1 + sinx) + log 5(1 - sinx) = 0 then x in ( 0, ) is
b) 0
c)
d No value
a)
Ans d
Solution: Clearly If log 5 [ (1 + sinx)( 1 – sinx)] =0
cos2x =1
cosx = ± 1
1 – sin2x = 0
x = 0,
0, )
no value.
------------------ ----------------------------- ----------------- ------------
72. If y = cosx and
a)
b)
Solution:
5x =
y = sin 5x then x =
c)
d)
Ans: c
clearly sin5x = cosx
-x
6x =
sin5x = sin (
x =
- x)
ans C
------------------ ----------------------------- ----------------- -----------73. The value of
a)
))+
b)
c)
)) =
d) 0
Ans: d
:
cos
sin
cos
sin
=
2
) ) = cos
) = sin
‐
.
cos 2π
sin 2π
)) =
) =
-
GE: = 0
Ans: d
------------------ ----------------------------- ----------------- -----------74. If θ =
)) , then one of the possible value
of θ is
a)
b)
c)
Ans: a)
Solution:
600 ≡ 600 + 7200 = 1200
600 = sin sin 120
θ = sin sin
= sin sin 180
60 ) = sin sin
d)
60
) = 60
------------------ ----------------------------- ----------------- -----------.
=
75. The value of sin( 2
a) 0.96 b) 0.86
c) 0.94
d) 0. 84.
Ans a)
Solution:
sin( 2 sin 0.8 = sin ( 2 θ ) where θ = sin
0.8 =
= 0.096 .
= 2 sin θ cos θ = 2 . . =
Ans a)
------------------ ----------------------------- ----------------- ------------
=