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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ^ -1 x+ ^ -1 (1-x)= ^ -1 x , then x ∈ to

If sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x, then x∈x \in to

Options

  1. A{1,0}\{1,0\}
  2. B{−1,1}\{-1,1\}
  3. C\left\{0, \frac{1}{2}\right\}
  4. D{2,0}\{2,0\}

Correct answer

C. \left\{0, \frac{1}{2}\right\}

Step-by-step solution

We have, sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x⇒sin⁡−1(1−x)=π2−sin⁡−1x−sin⁡−1x=π2−2sin⁡−1x[∵cos⁡−1x+sin⁡−1x=π2]⇒sin⁡−1(1−x)=sin⁡−11−sin⁡−12x1−x2⇒sin⁡−1(1−x)=sin⁡−1(11−4x2(1−x2)−0)⇒(1−x)=1−4x4+4x2=(1−2x2)2⇒(1−x)=1−2x2⇒2x2−x=0⇒x(2x−1)=0⇒x=0,12∴x∈{0,12}\begin{aligned} & \sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x \\ & \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}-\sin ^{-1} x-\sin ^{-1} x \\ & =\frac{\pi}{2}-2 \sin ^{-1} x \\ & {\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right]} \\ & \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1} 1-\sin ^{-1} 2 x \sqrt{1-x^2} \\ & \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1}\left(1 \sqrt{1-4 x^2\left(1-x^2\right)}-0\right) \\ & \Rightarrow (1-x)=\sqrt{1-4 x^4+4 x^2}=\sqrt{\left(1-2 x^2\right)^2} \\ & \Rightarrow (1-x)=1-2 x^2 \\ & \Rightarrow 2 x^2-x=0 \\ & \Rightarrow x(2 x-1)=0 \\ & \Rightarrow x=0, \frac{1}{2} \\ & \therefore x \in\left\{0, \frac{1}{2}\right\} \end{aligned}
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