Question

Maximum value of the expression $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|=$

• A. $4$
• B. $6$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $4$

$\Delta =\left|\begin{array}{ccc}1+\sin {}^{2}x& {\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|R_1\to R_1-R_2{R}_2\to R_2-R_3\Delta =\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|=1\left|\begin{array}{cc}1& -1\\ {\cos }^{2}x& 1+4\sin 2x\end{array}\right|+1\left|\begin{array}{cc}0& -1\\ {\sin }^{2}x& 1+4\sin 2x\end{array}\right|=1+4\sin 2x+{\cos }^{2}x+{\sin }^{2}x=2+4\sin 2xf\left(x\right)=\left|\begin{array}{ccc}1+\sin {}^{2}x& {\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|=2+4\sin 2x$

$\therefore f\left(x\right)$ will be maximum when $\sin 2x$ is maximum and we know that the maximum value of $\sin 2x$ is $1$.

Hence, ${f\left(x\right)}_{max}^{}=2+4=6$