Question:
The number of real solutions of the equation
\[
\sin^{-1}(2-x) - 2\sin^{-1}x = \pm \frac{\pi}{2}
\]
is:
The number of real solutions of the equation
\[
\sin^{-1}(2-x) - 2\sin^{-1}x = \pm \frac{\pi}{2}
\]
is:
Show Hint
Always intersect domains first in inverse trigonometric equations; most problems reduce to a single valid point.
Updated On: Jun 18, 2026
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The Correct Option is B
Solution and Explanation
Concept:
We use domain restrictions of inverse trigonometric functions.
Step 1: Find domain constraints.
For \(\sin^{-1}(2-x)\) to be defined: \[ -1 \le 2-x \le 1 \Rightarrow 1 \le x \le 3 \] For \(\sin^{-1}x\) to be defined: \[ -1 \le x \le 1 \] Thus intersection gives: \[ x = 1 \]
Step 2: Verify the value.
Substitute \(x=1\): \[ \sin^{-1}(1) - 2\sin^{-1}(1) = \frac{\pi}{2} - \pi = -\frac{\pi}{2} \] Since RHS is \(\pm \frac{\pi}{2}\), condition is satisfied. \[ \boxed{1} \]
Step 1: Find domain constraints.
For \(\sin^{-1}(2-x)\) to be defined: \[ -1 \le 2-x \le 1 \Rightarrow 1 \le x \le 3 \] For \(\sin^{-1}x\) to be defined: \[ -1 \le x \le 1 \] Thus intersection gives: \[ x = 1 \]
Step 2: Verify the value.
Substitute \(x=1\): \[ \sin^{-1}(1) - 2\sin^{-1}(1) = \frac{\pi}{2} - \pi = -\frac{\pi}{2} \] Since RHS is \(\pm \frac{\pi}{2}\), condition is satisfied. \[ \boxed{1} \]
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