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Solve the following.
Consider the function . Find the range of if and .
Consider the function $y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right),\ x \in \left[\dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$.
Find the range of $y$ if $x = 0,\ \dfrac{1}{2}$ and $\dfrac{-1}{2}$.
Answer
Answer
AI$y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$.
At $x = 0$: $y = \sin^{-1}0 = 0$.
At $x = \dfrac{1}{2}$: $y = \sin^{-1}\left(2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}\right) = \sin^{-1}\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{3}$.
At $x = -\dfrac{1}{2}$: $y = \sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{\pi}{3}$.
So $y = 0,\ \dfrac{\pi}{3},\ -\dfrac{\pi}{3}$ respectively.
From ISC 2025 Improvement Mathematics Paper 1, question 14(i).