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GUJCET · Maths · Inverse Trigonometric Functions
If \(2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\) then, \(x \in\) ____________ .
- A \(\left[\frac{1}{\sqrt{2}}, 1\right]\)
- B \([0,1]\)
- C \(\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)
- D \(\left[\frac{-1}{\sqrt{2}}, 1\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)
Step-by-step Solution
Detailed explanation
Let \( \sin^{-1} x = \theta \). For \( \sin^{-1}(\sin 2\theta) = 2\theta \), we must have \( -\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2} \).
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