Free Inverse Trigonometric Functions 02 Practice Test - 12th Grade - Commerce
Question 1
The value of tan2(sec−13)+cot2(cosec−14) is
A.
20
B.
21
C.
23
D.
25
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)=
A.
10
B.
4π−10
C.
2π+10
D.
2π−10
SOLUTION
Solution : B
cos−1(cos10)=cos−1(cos(4π−10))=4π−10
Question 3
Find maximum value of x for which 2 tan−1x+cos−1(1−x21+x2) is independent of x.
A.
0
B.
2
C.
3
D.
1
SOLUTION
Solution : A
Let x=tan θ,−π2<θ<π22 θ+cos−1cos 2θ
(i)0≤2θ<π2cos−1cos 2θ=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:
A.
x∈R
B.
x≥1
C.
x≤−1
D.
−1≤x≤1
SOLUTION
Solution : D
Putting θ=tan−1x,−π2<θ<π2, we have
R.H.S.
=sin−12tan θ1+tan2 θ=sin−1sin 2θ=2 θ=2 tan−1x
When −π2≤2θ≤π2
i.e., −π4≤θ≤π4
i.e., −1≤tan θ≤1
i.e.,−1≤x≤1
Question 5
The value of cos−1(cos 5π3)+sin−1(cos 5π3) is
A.
0
B.
π2
C.
2π3
D.
10π3
SOLUTION
Solution : A
cos−1(cos 5π3)+sin−1(cos 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=
A.
6556
B.
2465
C.
1665
D.
5665
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 :
A.
(–4,4)
B.
(−∞,−4)∪(4,∞)
C.
ϕ
D.
None of these
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 :
A.
(−1−1√2)∪(1√2,1)
B.
(−1√2,1√2)−{0}
C.
(−1,1)−{0}
D.
None of these
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 :
A.
1
B.
4
C.
3
D.
2
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:
A.
0
B.
π4
C.
π6
D.
π2
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