Question

The maximum value of $f\left(x\right)=\left|\begin{array}{ccc}{\sin }^{2}x& 1+{\cos }^{2}x& \cos 2x\\ 1+{\sin }^{2}x& {\cos }^{2}x& \cos 2x\\ {\sin }^{2}x& {\cos }^{2}x& \sin 2x\end{array}\right|,x\in R$ is :

• A. $\sqrt{7}$
• B. $\sqrt{5}$
• C. $5$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\sqrt{5}$

$\Delta =\left|\begin{array}{ccc}{\sin }^{2}x& 1+{\cos }^{2}x& \cos 2x\\ 1+{\sin }^{2}x& {\cos }^{2}x& \cos 2x\\ {\sin }^{2}x& {\cos }^{2}x& \sin 2x\end{array}\right|$

$C_1\to C_1+C_2$
$\Delta =\left|\begin{array}{ccc}2& 1+{\cos }^{2}x& \cos 2x\\ 2& {\cos }^{2}x& \cos 2x\\ 1& {\cos }^{2}x& \sin 2x\end{array}\right|$

$R_1\to R_1-R_2$
$\Delta =\left|\begin{array}{ccc}0& 1& 0\\ 2& {\cos }^{2}x& \cos 2x\\ 1& {\cos }^{2}x& \sin 2x\end{array}\right|$

$=(-1)[2\sin 2x-\cos 2x]=\cos 2x-2\sin 2x$
Maximum value $=\sqrt{1^2+(-2{)}^2}=\sqrt{5}$