Question:medium
Let the functions \(f\) and \(g\) be defined by \(f(x)=3\sin x,\; -\frac{\pi}{2}\le x\le \frac{\pi}{2}\) and \(g(x)=6-3x^2,\; x\in\mathbb{R}\). Then \(f^{-1}(g(x))=\)
Let the functions \(f\) and \(g\) be defined by \(f(x)=3\sin x,\; -\frac{\pi}{2}\le x\le \frac{\pi}{2}\) and \(g(x)=6-3x^2,\; x\in\mathbb{R}\). Then \(f^{-1}(g(x))=\)
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Always simplify the argument inside the inverse function by factoring out constants. Here, dividing \(6 - 3x^2\) by 3 is the crucial simplification step.
Updated On: Jun 25, 2026
- \(\sin^{-1}(2 - x^2)\)
- \(3\sin^{-1}(6 - 3x^2)\)
- \(3\sin^{-1}(2 - x^2)\)
- \(\sin^{-1}(6 - 3x^2)\)
- \(\frac{1}{3}\sin^{-1}(6 - 3x^2)\)
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The Correct Option is A
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