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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ^ -1 ((3)/(x))+ ^ -1 ((4)/(x))=(π)/(2) , then x is equal to

If sin⁡−1(3x)+sin⁡−1(4x)=π2\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}, then xx is equal to

Options

  1. A33
  2. B55
  3. C77
  4. D1111

Correct answer

B. 55

Step-by-step solution

Given that, sin⁡−1(3x)+sin⁡−1(4x)=π2∴sin⁡−1(3x)=π2−sin⁡−1(4x)⇒sin⁡−1(3x)=cos⁡−1(4x)⇒sin⁡−1(3x)=sin⁡−1(x2−16x)⇒3x=x2−16x⇒9=x2−16⇒x2=25⇒x=±5⇒x=5\begin{aligned} & \sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2} \\ & \therefore \sin ^{-1}\left(\frac{3}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{x}\right) \\ & \Rightarrow \sin ^{-1}\left(\frac{3}{x}\right)=\cos ^{-1}\left(\frac{4}{x}\right) \\ & \Rightarrow \sin ^{-1}\left(\frac{3}{x}\right)=\sin ^{-1}\left(\frac{\sqrt{x^2-16}}{x}\right) \\ & \Rightarrow \frac{3}{x}=\frac{\sqrt{x^2-16}}{x} \\ & \Rightarrow 9=x^2-16 \Rightarrow x^2=25 \\ & \Rightarrow x= \pm 5 \\ & \Rightarrow x=5 \end{aligned} (∵−5\because-5 is not satisfied the given equation)
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