# Free Objective Test 02 Practice Test - 11th and 12th

The value of tan2(sec13)+cot2(cosec14) is

A. 20
B. 21
C. 23
D. 25

#### SOLUTION

Solution : C

Let Sec13=α,cosec14=βtan2α+cot2β=91+161=23

The value of cos1(cos10)=

A. 10
B. 4π10
C. 2π+10
D. 2π10

#### SOLUTION

Solution : B

cos1(cos10)=cos1(cos(4π10))=4π10

Find maximum value of x for which 2tan1x+cos1(1x21+x2) is independent of x.

A. 0
B. 2
C. 3
D. 1

#### SOLUTION

Solution : A

Let x=tanθ,π2<θ<π22θ+cos1cos2θ
(i)02θ<π2.2cos1cos2θ=2θ
So given expression =4θ=4tan1x
(ii)π2<θ0π<2θ0cos1cos2θ=2θ
So, given expression becomes independent of θ
For π2<θ0<x0

If 2tan1x=sin12x1+x2, then:

A. xR
B. x1
C. x1
D. 1x1

#### SOLUTION

Solution : D

Putting θ=tan1x,π2<θ<π2, we have
R.H.S. sin12tanθ1+tan2θ=sin1sin2θ=2θ=2tan1x
When π22θπ2 i.e., π4θπ4
i.e., 1x=tanθ1.

The value of cos1(cos 5π3)+sin1(sin 5π3) is

A. 0
B. π2
C. 2π3
D. 10π3

#### SOLUTION

Solution : A

cos1(cos 5π3)+sin1(sin 5π3)=cos1[cos (2ππ3)]+sin1[sin(2ππ3)]=π3π3=0.

If sin135+cos1(1213)=sin1 C,then C=

A. 6556
B. 2465
C. 1665
D. 5665

#### SOLUTION

Solution : D

Given sin1C=sin135+cos11213C=sin(sin135+cos11213)Using sin(A+B)=sin A cos B+cos A sin BC=35×1213+19251144169C=5665.

The set of values of p for which x2px+sin1(sin4)>0 for all real x is given by :

A. (4,4)
B. (,4)(4,)
C. ϕ
D. None of these

#### SOLUTION

Solution : C

x2px+sin1(sin4)>0 for all real x.
x2px+sin1(sin4)>0
x2px+(π4)>0x ϵ R
D=p24(π4)<0    p2+164π<0
Since 164π>0,p2+164π cannot be negative for any value of p ϵ R.
Set of values of p=ϕ

Complete solution set of tan2(sin1x)>1 is :

A. (112)(12,1)
B. (12,12){0}
C. (1,1){0}
D. None of these

#### SOLUTION

Solution : A

tan2(sin1x)>1
tan(sin1x)<1 or tan(sin1x)>1
π2<sin1x<π4 or π4<sin1x<π2
1<x<12  or 12<x<1
xϵ(112)(12,1)

Total number of positive integral value of `n' such that the equations cos1x+(sin1y)2=nπ24 and (sin1y)2cos1x=π216 are consistent, is equal to :

A. 1
B. 4
C. 3
D. 2

#### SOLUTION

Solution : A

(sin1y)2+cos1x=nπ24
(sin1y)2cos1x=π216
(sin1y)2=(4n+1)π232,cos1x=π2(4n1)32
0(4n+1)32π2π24,0(4n1)32π2π
14n74,14,n8π+14

The value of sin1{cot(sin1(234)+cos1124+sec12)} is:

A. 0
B. π4
C. π6
D. π2

#### SOLUTION

Solution : A

cos1124=cos132=π6,sec12=π4
sin1234=sin14238=sin13122=π12
sin1234+cos1124+sec12=π2
and sin1(cotπ2)=sin10=0

If θ=tan1d1+a1a2+tan1d1+a2a3++tan1d1+an1an, where a1,a2,a3,an  are in A.P. with common difference d, then tanθ=

A. (n1)d1+a1an
B. (n1)da1+an
C. ana1an+a1
D. nd1+a1an

#### SOLUTION

Solution : A

θ=tan1a2a11+a1a2+tan1a3a21+a2a3++tan1anan11+an1an
=(tan1a2tan1a1)+(tan1a3tan1a2)++tan1anan11+an1an++(tan1antan1an1)
=tan1antan1a1
=tan1ana11+a1an=tan1(n1)d1+a1an
tanθ=(n1)d1+a1an

cos[cos1(17)+sin1(17)]=

A. 13
B. 0
C. 13
D. 49

#### SOLUTION

Solution : B

cos{cos1(17)+sin1(17)}=cosπ2=0

If cos1p+cos11p+cos11q=3π4, then the value of q is

A. 1
B. 12
C. 13
D. 12

#### SOLUTION

Solution : D

Let α=cos1p:β=cos11p
and γ=cos11q or cosα=p:cosβ=1p
and cosγ=1q.
Therefore sinα=1p,sinβ=p and sinγ=q.
The given equation may be written as α+β+γ=3π4 or α+β=3π4γ or cos(α+β)=cos(3π4γ)
cosαcosβsinαsinβ =  cos{π(π4+γ)}=cos(π4+γ)
p1p1pp=(121q12.q)
0=1qq1q=qq=12.

cot1[(cosα)12]tan1[(cosα)12]=x, then sinx=

A. tan2(α2)
B. cot2(α2)
C. tan α
D. cot(α2)

#### SOLUTION

Solution : A

tan1[1cosα]tan1[cosα]=x
tan11cosαcosα1+cosαcosα=xtan x=1cosα2cosα
sinx=1cosα1+cosα=2sin2α22cos2α2=tan2(α2).

If tan1(x)+tan1(y)+tan1(z)=π, then 1xy+1yz+1zx=

A. 0
B. 1
C. 1xyz
D. xyz

#### SOLUTION

Solution : B

tan1(x)+tan1(y)+tan1(z)=π
tan1x+tan1y=πtan1z
x+y1xy=zx+y=z+xyz
x+y+z=xyz
Dividing by xyz, we get
1yz+1xz+1xy=1.
Note: Students should remember this question as a formula.