Inverse Trigonometric FunctionseasyFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The number of points x∈[−π2,3π2]x\in\left[-\dfrac{\pi}{2},\dfrac{3\pi}{2}\right] satisfying the equation 1+sin⁡−1(sin⁡x)=π31+\sin^{-1}(\sin x)=\dfrac{\pi}{3}.
Solution
Answer: 2
Step 1: Rewrite: sin⁡−1(sin⁡x)=π3−1\sin^{-1}(\sin x)=\dfrac{\pi}{3}-1. Step 2: Case 1: x∈[−π2,π2]x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]. Here sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x)=x:
x=π3−1≈1.047−1=0.047x=\dfrac{\pi}{3}-1\approx 1.047-1=0.047
 Since π3−1∈[−π2,π2]\dfrac{\pi}{3}-1\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], this is valid. Step 3: Case 2: x∈[π2,3π2]x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]. Here sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x)=\pi-x:
π−x=π3−1  ⇒  x=π−π3+1=2π3+1\pi-x=\dfrac{\pi}{3}-1\;\Rightarrow\;x=\pi-\dfrac{\pi}{3}+1=\dfrac{2\pi}{3}+1
 Check: 2π3+1≈2.094+1=3.094\dfrac{2\pi}{3}+1\approx 2.094+1=3.094, and 3π2≈4.71\dfrac{3\pi}{2}\approx 4.71. So x∈[π2,3π2]x\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]. Valid. Step 4: Two solutions: x=π3−1x=\dfrac{\pi}{3}-1 and x=2π3+1x=\dfrac{2\pi}{3}+1. Answer: 22
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Inverse Trigonometric Functions · easy
Let f(x)= -1 (x 2 +kx+9-x) . If the range of f(x) lies in the interval (0, 2 ) for all va…
Inverse Trigonometric Functions · easy
If f(x) = -1 1 x 2+x+1 + -1 1 x 2+3x+3 + -1 1 x 2+5x+7 + to terms, then the value of |f'(…
Inverse Trigonometric Functions · hard
Two functions f(x) and g(x) are defined as f(x)= 3 | x-2 x 2 -10x+24 | and g(x)= -1 ( 2[x…
Inverse Trigonometric Functions · hard
If (30°+ x)= 13 14 and 0<x<1 , the value of x is a 3 b , where a,b are positive integers…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath