JEE PYQ: Inverse Trigonometric Functions Question 2
Question 2 - 2021 (16 Mar Shift 2)
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1} x$ is equal to:
(1) 2
(2) 1
(3) 3
(4) 0
Show Answer
Answer: (3)
Solution
$\sin^{-1}\left(\frac{3x}{5}\sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1 - \frac{9x^2}{25}}\right) = \sin^{-1} x$. Simplifying: $\frac{3x}{5}\sqrt{25 - 16x^2} + \frac{4x}{5}\sqrt{25 - 9x^2} = 5x$. For $x = 0$, it’s satisfied. For $x \neq 0$: $3\sqrt{25 - 16x^2} + 4\sqrt{25 - 9x^2} = 25$. Squaring and simplifying gives $x^2 = 1$, so $x = \pm 1$. Checking: all three values $x = 0, 1, -1$ satisfy the original equation. Number of solutions = 3.
BETA
