Free Objective Test 02 Practice Test - 11th and 12th
Question 1
The value of tan2(sec−13)+cot2(cosec−14) is
SOLUTION
Solution : C
Let Sec−13=α,cosec−14=β⇒tan2α+cot2β=9−1+16−1=23
Question 2
The value of cos−1(cos10)=
SOLUTION
Solution : B
cos−1(cos10)=cos−1(cos(4π−10))=4π−10
Question 3
Find maximum value of x for which 2tan−1x+cos−1(1−x21+x2) is independent of x.
SOLUTION
Solution : A
Let x=tanθ,−π2<θ<π22θ+cos−1cos2θ
(i)0≤2θ<π2.2cos−1cos2θ=2θ
So given expression =4θ=4tan−1x
(ii)−π2<θ≤0⇒−π<2θ≤0cos−1cos2θ=−2θ
So, given expression becomes independent of θ
For −π2<θ≤0⇒−∞<x≤0
Question 4
If 2tan−1x=sin−12x1+x2, then:
SOLUTION
Solution : D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x
When −π2≤2θ≤π2 i.e., −π4≤θ≤π4
i.e., −1≤x=tanθ≤1.
Question 5
The value of cos−1(cos 5π3)+sin−1(sin 5π3) is
SOLUTION
Solution : A
cos−1(cos 5π3)+sin−1(sin 5π3)=cos−1[cos (2π−π3)]+sin−1[sin(2π−π3)]=π3−π3=0.
Question 6
If sin−135+cos−1(1213)=sin−1 C,then C=
SOLUTION
Solution : D
Given sin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Using sin(A+B)=sin A cos B+cos A sin B⇒C=35×1213+√1−925√1−144169⇒C=5665.
Question 7
The set of values of p for which x2−px+sin−1(sin4)>0 for all real x is given by :
SOLUTION
Solution : C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀x ϵ R
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of p ϵ R.
∴ Set of values of p=ϕ
Question 8
Complete solution set of tan2(sin−1x)>1 is :
SOLUTION
Solution : A
tan2(sin−1x)>1
⇒tan(sin−1x)<−1 or tan(sin−1x)>1
⇒−π2<sin−1x<−π4 or π4<sin−1x<π2
⇒−1<x<−1√2 or 1√2<x<1
⇒xϵ(−1−−1√2)∪(1√2,1)
Question 9
Total number of positive integral value of `n' such that the equations cos−1x+(sin−1y)2=nπ24 and (sin−1y)2−cos−1x=π216 are consistent, is equal to :
SOLUTION
Solution : A
(sin−1y)2+cos−1x=nπ24
(sin−1y)2−cos−1x=π216
⇒(sin−1y)2=(4n+1)π232,cos−1x=π2(4n−1)32
⇒0≤(4n+1)32π2≤π24,0≤(4n−1)32π2≤π
⇒−14≤n≤74,14,≤n≤8π+14
Question 10
The value of sin−1{cot(sin−1√(2−√34)+cos−1√124+sec−1√2)} is:
SOLUTION
Solution : A
cos−1√124=cos−1√32=π6,sec−1√2=π4
sin−1√2−√34=sin−1√4−2√38=sin−1√3−12√2=π12
∴ sin−1√2−√34+cos−1√124+sec−1√2=π2
and sin−1(cotπ2)=sin−10=0
Question 11
If θ=tan−1d1+a1a2+tan−1d1+a2a3+⋯+tan−1d1+an−1an, where a1,a2,a3,⋯an are in A.P. with common difference d, then tanθ=
SOLUTION
Solution : A
θ=tan−1a2−a11+a1a2+tan−1a3−a21+a2a3+⋯+tan−1an−an−11+an−1an
=(tan−1a2−tan−1a1)+(tan−1a3−tan−1a2)+⋯+tan−1an−an−11+an−1an+⋯+(tan−1an−tan−1an−1)
=tan−1an−tan−1a1
=tan−1an−a11+a1an=tan−1(n−1)d1+a1an
∴tanθ=(n−1)d1+a1an
Question 12
cos[cos−1(−17)+sin−1(−17)]=
SOLUTION
Solution : B
cos{cos−1(−17)+sin−1(−17)}=cosπ2=0
Question 13
If cos−1√p+cos−1√1−p+cos−1√1−q=3π4, then the value of q is
SOLUTION
Solution : D
Let α=cos−1√p:β=cos−1√1−p
and γ=cos−1√1−q or cosα=√p:cosβ=√1−p
and cosγ=√1−q.
Therefore sinα=√1−p,sinβ=√p and sinγ=√q.
The given equation may be written as α+β+γ=3π4 or α+β=3π4−γ or cos(α+β)=cos(3π4−γ)
⇒cosαcosβ−sinαsinβ = cos{π−(π4+γ)}=−cos(π4+γ)
⇒√p√1−p−√1−p√p=−(1√2√1−q−1√2.√q)
⇒0=√1−q−√q⇒1−q=q⇒q=12.
Question 14
cot−1[(cosα)12]−tan−1[(cosα)12]=x, then sinx=
SOLUTION
Solution : A
tan−1[1√cosα]−tan−1[√cosα]=x
⇒tan−1⎡⎢⎣1√cosα−√cosα1+√cosα√cosα⎤⎥⎦=x⇒tan x=1−cosα2√cosα
∴sinx=1−cosα1+cosα=2sin2α22cos2α2=tan2(α2).
Question 15
If tan−1(x)+tan−1(y)+tan−1(z)=π, then 1xy+1yz+1zx=
SOLUTION
Solution : B
tan−1(x)+tan−1(y)+tan−1(z)=π
⇒tan−1x+tan−1y=π−tan−1z
⇒x+y1−xy=−z⇒x+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.