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Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise Multiple Choice Questions Question 36 maths Textbook Solution.

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Answer:  \alpha=0, \beta=\pi

Hint: \frac{-\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}

Given: \alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta


We have \frac{-\pi}{2}+\frac{\pi}{2} \leq \sin ^{-1} x+\frac{\pi}{2} \leq \frac{\pi}{2}+\frac{\pi}{2}

              \begin{aligned} &0 \leq \sin ^{-1} x+\left(\sin ^{-1} x+\cos ^{-1} x\right) \leq \pi \\ &0 \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \pi \end{aligned}

\Rightarrow \alpha=0, \beta=\pi

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