Free Objective Test 01 Practice Test - 11th and 12th 

Question 1

If x=sin(2tan12),y=sin(12tan143) then

A. x = 1 – y
B. x2=1y
C. x2=1+y
D. y2=1x

SOLUTION

Solution : D

Let tan12=αx=sin2α=45y=sin(β2)=1cosβ2=1352=15
 

Question 2

sin1|cosx|cos1|sinx|=a has at least one solution if a

A. 0
B. [0,π2]
C. [π2,3π2]
D. (0,π)

SOLUTION

Solution : A

sin1|cosx|cos1|sinx|=π2cos1|cosx|π2+sin1|sinx|=asin1|sinx|cos1|cosx|=aa=0x

Question 3

sin1(sin10)=

A. 10
B. 3π10
C. π+6
D. 2π6

SOLUTION

Solution : B

3π<10<3π+π210 rad Q3π2<3π10<03π10Q4Also sin(3π10)=sin10Hence sin1(sin10)=sin1sin(3π10)=3π10

Question 4

cos1x=tan11x2x, then:
 

A. XR
B. x1,x0
C. 0<x1
D. None of these

SOLUTION

Solution : C

Putting θ=cos1x in R.H.S., we have   
R.H.S =tan11x2x=tan11cos2θcosθ(0θπ)=tan1sinθcosθ=tan1tanθ=θ=cos1x when π2<θ<π2
i.e., when 0θ<π2
i.e., when 0<cosθ1 i.e., 0<x1

Question 5

The value of x for which sin(cot1(1+x))=cos(tan1x) is:
 

A. 12
B. 1
C. 0
D. 12   

SOLUTION

Solution : D

sin(cot1(1+x))=cos(tan1x)cot1(1+x)=[π2±tan1x]π2tan1(1+x)=π2±tan1x1+x=±x
x=12 Which, on verification, satisfies the equation.

Question 6

3cos1xπxπ2=0 has :
 

A. One solution
B. Infinite solutions
C. No solution
D. None of these

SOLUTION

Solution : A



3cos1xπx+π2
Clearly graphs of y=3cos1x and y=πx+π2 in the domain of cos1x i.e., in [-1, 1] intersect only once, therefore there is only one solution of the given equation. 

Question 7

If cos1x+cos1y+cos1z=3π, then xy+yz+zx=
 

A. 0
B. 1
C. 3
D. -3

SOLUTION

Solution : C

Given cos1x+cos1y+cos1z=3π
0cos1xπ
0cos1yπ and 0cos1zπ
Here cos1x=cos1y=cos1z=π
x=y=z=cosπ=1
xy+yz+zx=(1)(1)+(1)(1)+(1)(1)
=1+1+1=3.

Question 8

The value of tan(tan112tan113)=

A. 56
B. 76
C. 16
D. 17

SOLUTION

Solution : D

tan[tan112tan113]=tan[tan112131+16]
=tan tan1(16×67)=17

Question 9

tan[2tan1(15)π4]=

A. 177
B. 177
C. 717
D. 717

SOLUTION

Solution : D

tan[2tan1(15)π4]=tan[tan1251125tan1(1)]
=tan[tan1512tan1(1)]=tan tan1(51211+512)=717

Question 10

12cos1(1x1+x)=

A. cot1x
B. tan1x
C. tan1x
D. cot1x

SOLUTION

Solution : B

Let x=tan2θθ=tan1x
Now, 12cos1(1x1+x)
=12cos1(1tan2θ1+tan2θ)
=12cos1cos2θ=2θ2=θ=tan1x.

Question 11

4tan115tan11239 is equal to

A. π
B. π2
C. π3
D. π4

SOLUTION

Solution : D

Since 2tan1x=tan12x1x2
4tan115=2[2tan115]=2tan1251125
=2tan11024=tan120241100576=tan1120119
So, 4tan115tan11239=tan1120119tan11239
=tan112011912391+120119.1239=tan1(120×239)119(119×239)+120
tan11=π4.

Question 12

If 3sin12x1+x24cos11x21+x2+2tan12x1+x2=π3 then x=

A. 3
B. 13
C. 1
D. None of these

SOLUTION

Solution : B

3sin12x1+x24cos11x21+x2+2tan12x1+x2=π3
Putting x=tanθ
3sin1(2tanθ1+tan2θ)4cos1(1tan2θ1+tan2θ)
+2tan1(2tanθ1tan2θ)=π3
3sin1(sin2θ)4cos1(cos2θ)
+2tan1(tan2θ)=π3
3(2θ)4(2θ)+2(2θ)=π36θ8θ+4θ=π3
θ=π6tan1x=π6x=tanπ6=13

Question 13

The value of sin[2tan1(13)]+cos[tan1(22)]=

A. 1615
B. 1415
C. 1215
D. 1115

SOLUTION

Solution : B

sin[2tan1(13)]+cos[tan1(22)]
=sin[tan123119]+cos[tan1(22)]
=sin[tan134]+cos[tan122]
=tan122=cos113
Also tan134=sin135
=35+13=1415.

Question 14

sin(12cos145)=

A. 110
B. 110
C. 110
D. 110

SOLUTION

Solution : A

Let  0xπ
So, cos145=xcos x=45     ....(i)
Now sin(12cos145)=sin(x2)    ....(ii)
From (i), cos x=4512sin2x2=45
2sin2x2=145=15sinx2=110.

Question 15

The value of cos1(cos12)sin1(sin14) is

A. 2
B. 8π26
C. 4π+2
D. None of these

SOLUTION

Solution : D

cos1(cos12)sin1(sin14)4π12+5π14=9π26