Let $a \leq \tan ^{-1} x+\cot ^{-1} x+\sin ^{-1} x \leq b$. If $\alpha$ and $\beta$ denote the minimum and maximum possible values of a and b respectively, then :
Answer & explanation
Correct answer: option 1
$a \leq \tan ^{-1} x+\cot ^{-1} x+\sin ^{-1} x \leq b$
$a \leq \frac{\pi}{2}+\sin ^{-1} x \leq b\left\{\text { as } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right.$
so $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
so $0 \leq \frac{\pi}{2}+\sin -1 x \leq \pi$
$\alpha=a_{\min}=0 \quad \beta=b_{\min}=\pi$