Question

If the maximum and the minimum values of $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& 4\sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$ are $M$ and $m$ respectively, then $\frac{M}{m}$ is

• A. $2$
• B. $-3$
• C. $3$
• D. $-2$

Answer 1

The correct option is B $-3$

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

Applying $C_1\to C_1+C_2,$ we get
$\Delta =\left|\begin{array}{ccc}2& {\cos }^{2}x& 4\sin 2x\\ 2& 1+{\cos }^{2}x& 4\sin 2x\\ 1& {\cos }^{2}x& 1+4\sin 2x\end{array}\right|$

Applying $R_2\to R_2-R_1$ and then $R_3\to R_3-R_1,$ we get
$\Delta =\left|\begin{array}{ccc}2& {\cos }^{2}x& 4\sin 2x\\ 0& 1& 0\\ -1& 0& 1\end{array}\right|$

$=2+4\sin 2x$
$M=6$ and $m=-2$
Hence, $\frac{M}{m}=-3$