Question

If product of roots of the equation
$x^2-3kx+2e^{\sin x}-1=0$ is 7 then k=?

• A. ${\sin }^{-1}\left(\log 4\right)$
• B. ${\sin }^{-1}\left(\log 2\right)$
• C. ${\sin }^{-1}\left(\log 3\right)$
• D. None

Answer 1

The correct option is A ${\sin }^{-1}\left(\log 4\right)$

Product of the roots $=2e^{\sin x}-1=7$

$\Rightarrow 2e^{\sin x}=7+1=8$

$\Rightarrow e^{\sin x}=4$

$\Rightarrow {\log }_2{e}^{\sin x}={\log }_{2}4$

$\Rightarrow {\log }_2{e}^{\sin x}={\log }_{2}4$

$\Rightarrow \sin x{\log }_{2}e=2{\log }_{2}2$

$\Rightarrow \sin x{\log }_{2}e=2$

$\Rightarrow \sin x=\frac{2}{{\log }_{2}e}$

$\Rightarrow \sin x=2{\log }_{e}2$

$\Rightarrow \sin x={\log }_{e}4$

$\therefore x={\sin }^{-1}\left({\log }_{e}4\right)$