Question

If the maximum value of $f\left(x\right)=\sqrt{2}\left|\begin{array}{ccc}\sin 2x& 1& 1\\ \cos 2x& 2& 3\\ \cos 2x& 3& 5\end{array}\right|$ is $\alpha ,$ then the value of $(\alpha +2)$ is

Answer 1

Expanding along $C_1:$
$\Rightarrow f\left(x\right)=\sqrt{2}\left[\sin 2x\right(10-9)-\cos 2x(5-3)+\cos 2x(3-2\left)\right]$
$\Rightarrow f\left(x\right)=\sqrt{2}(\sin 2x-\cos 2x)$
$\Rightarrow$ Range of $f\left(x\right)$ is $[-2,2]$
$(\because |\sin 2x-\cos 2x|\le \sqrt{2})$
$\therefore f(x{)}_{max.}=\alpha =2$