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Consider the function Find the range of if . Prove: . Evaluate: . If , calculate the value of ' '.
Consider the function $y = \sin^{-1}(2x\sqrt{1 - x^2}), x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$
(i)[1.0]
Find the range of $y$ if $x \in \left[0, \frac{1}{\sqrt{2}}\right]$.
(ii)[1.0]
Prove: $2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})$.
Given: $x = \sin\theta$
To show: 2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})
(iii)[2.0]
Evaluate: $\tan^{-1}\left(2\cos\left(2\sin^{-1}\frac{\sqrt{3}}{2}\right)\right)$.
(iv)[2.0]
If $4\sin^{-1}x + \cos^{-1}x = \pi$, calculate the value of '$x$'.
Answer
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From ISC 2025 Improvement Mathematics Paper 1, question 14.