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, ${\sin }^{-1}\left({\alpha }^2-6\alpha +\frac{17}{2}\right)=\frac{\pi }{6}\Rightarrow {\alpha }^2-6\alpha +\frac{17}{2}=\frac{1}{2}\Rightarrow {\alpha }^2-6\alpha +8=0$
$\Rightarrow (\alpha -4)(\alpha -2)=0$
$\therefore \alpha =2,4$
$\text{ Hence, }\left|{\alpha }_1-{\alpha }_2\right|=|2-4|=2$