Question

If $\alpha_1$ and $\alpha_2$ are two values of $\alpha$ for which $3 \sin ^{-1}\left(\alpha^2-6 \alpha+\frac{17}{2}\right)=\frac{\pi}{2}$, then $\left|\alpha_1-\alpha_2\right|$ equals

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Clearly, $\sin ^{-1}\left(\alpha^2-6 \alpha+\frac{17}{2}\right)=\frac{\pi}{6} \Rightarrow \alpha^2-6 \alpha+\frac{17}{2}=\frac{1}{2} \Rightarrow \alpha^2-6 \alpha+8=0$
$\Rightarrow (\alpha-4)(\alpha-2)=0 $
$\therefore \alpha=2,4 $
$\text { Hence, }\left|\alpha_1-\alpha_2\right|=|2-4|=2$