Q106 / 180 · KCET 2015 · Mathematics

If \( \alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta \), then

  • A
    \( \alpha=-\frac{\Pi}{2} \quad \beta=\frac{\Pi}{2} \)
  • B
    \( \alpha=-\frac{\Pi}{2} \quad \beta=\frac{3 \Pi}{2} \)
  • Correct answer
    C
    \( \alpha=0 \quad \beta=I \)
  • D
    \( \alpha=0 \quad \beta=2 \Pi \)

Solution

Given that $\alpha \leq 2 \sin ^{-1} \chi+\cos ^{-1} \chi \leq \beta$ So, $2 \sin ^{-1} \chi+\cos ^{-1} \chi$ $=\sin ^{-1} \chi+\left(\sin ^{-1} \chi+\cos ^{-1} \chi\right)=\sin ^{-1} \chi+\frac{\pi}{2}$ As we know $-\frac{\Pi}{2} \leq \sin ^{-1} \chi \leq \frac{\pi}{2}$ Thus, $0 \leq 2 \sin ^{-1} \chi+\cos ^{-1} \chi \leq \Pi$ Therefore, $\alpha=0$ and $\beta=\pi$

Hence, the correct answer is option: 'C'.

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