Inverse Trigonometric Functions(11th And 12th > Mathematics ) Questions and Answers
Question 1.
If x=sin(2tan−12),y=sin(12tan−143) then
x = 1 – y
x2=1–y
x2=1+y
y2=1−x
Explanation:-
Answer: Option D. -> y2=1−x : D Let tan−12=α⇒x=sin2α=45y=sin(β2)=√1−cosβ2=√1−352=1√5
Question 2.
sin−1|cosx|−cos−1|sinx|=a has at least one solution if a ∈
0
[0,π2]
[π2,3π2]
(0,π)
Explanation:-
Answer: Option A. -> 0 : A sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
Question 3.
cos−1x=tan−1√1−x2x, then:
X∈R
−≤x≤1,x≠0
0<x≤1
Noneofthese
Explanation:-
Answer: Option C. -> 0<x≤1 : C Putting θ=cos−1x in R.H.S., we have R.H.S =tan−1√1−x2x=tan−1√1−cos2θcosθ(0≤θ≤π)=tan−1sinθcosθ=tan−1tanθ=θ=cos−1x when −π2<θ<π2 i.e., when 0≤θ<π2 i.e., when 0<cosθ≤1 i.e., 0<x≤1.
Question 4.
sin−1(sin10)=
10
3π−10
π+6
2π−6
Explanation:-
Answer: Option B. -> 3π−10 : B 3π<10<3π+π2∴10 rad ∈Q3∴−π2<3π−10<0∴3π−10∈Q4Also sin(3π−10)=sin10Hence sin−1(sin10)=sin−1sin(3π−10)=3π−10
Question 5.
The value of x for which sin(cot−1(1+x))=cos(tan−1x) is:
12
1
0
−12
Explanation:-
Answer: Option D. -> −12 : D sin(cot−1(1+x))=cos(tan−1x)⇒cot−1(1+x)=[π2±tan−1x]⇒π2−tan−1(1+x)=π2±tan−1x⇒1+x=±x x=−12 Which, on verification, satisfies the equation.
Question 6.
3cos−1x−πx−π2=0 has : Â
One solution
Infinite solutions
No solution
None of these
Explanation:-
Answer: Option A. ->
One solution
: A 3cos−1x−πx+Ï€2 Clearly graphs of y=3cos−1x and y=Ï€x+Ï€2 in the domain of cos−1x i.e., in [-1, 1] intersect only once, therefore there is only one solution of the given equation.Â
Question 7.
If cos−1x+cos−1y+cos−1z=3π, then xy+yz+zx=
0
1
3
-3
Explanation:-
Answer: Option C. ->
3
: C Given cos−1x+cos−1y+cos−1z=3π ∵0≤cos−1x≤π ∴0≤cos−1y≤π and 0≤cos−1z≤π Here cos−1x=cos−1y=cos−1z=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(tan−112−tan−113)=
56
76
16
17
Explanation:-
Answer: Option D. -> 17 : D tan[tan−112−tan−113]=tan[tan−112−131+16] =tantan−1(16×67)=17
Question 9.
tan[2tan−1(15)−π4]=
177
−177
717
−717
Explanation:-
Answer: Option D. -> −717 : D tan[2tan−1(15)−π4]=tan[tan−1251−125−tan−1(1)] =tan[tan−1512−tan−1(1)]=tantan−1(512−11+512)=−717
Question 10.
12cos−1(1−x1+x)=
cot−1√x
tan−1√x
tan−1x
cot−1x
Explanation:-
Answer: Option B. -> tan−1√x : B Let x=tan2θ⇒θ=tan−1√x Now, 12cos−1(1−x1+x) =12cos−1(1−tan2θ1+tan2θ) =12cos−1cos2θ=2θ2=θ=tan−1√x.