Question

The maximum value of $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& 4\cos 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$ is

• A. 8
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is A 8

$\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& 4\cos 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$

$R_1+R_1-R_2,R_2\to R_2-R_3$

$=\left|\begin{array}{ccc}1& -1& 4\cos 2x-4\sin 2x\\ 0& 1& -1\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$

$c_1+c_1+c_2$

$=\left|\begin{array}{ccc}0& -1& 4\cos 2x-4\sin 2x\\ 1& 1& -1\\ 1& \cos 2x& 1+4\sin 2x\end{array}\right|$

$=1(1+4\sin 2x+1)+4(\cos 2x-\sin 2x)({\cos }^{2}x-1)$

$=2+4\sin 2x+(4\cos 2x-4\sin 2x)(-{\sin }^{2}x)$

$=2+4\sin 2x-4\cos 2x{\sin }^{2}x+4\sin 2x\sin 2x$

$=2+4\sin 2x-4\cos 2x\frac{(1-\cos 2x)}{2}+4\sin 2x\frac{(1-\cos 2x)}{2}$

$=2+4\sin 2x-2\cos 2x(1-\cos 2x)+2\sin 2x(1-\cos 2x)$

$=3+6s\in 2x-\sin 4x-2\cos 2x+\cos 4x$

At $x=\frac{11}{4}$

$y_{max}=3+6\times 1-0-0-1$

$=9-1$

$y_{max}=8$