Question

Let $m$ and $M$ be respectively the minimum and maximum values of
$\left|\begin{array}{ccc}{\cos }^{2}x& 1+{\sin }^{2}x& \sin 2x\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ {\cos }^{2}x& {\sin }^{2}x& 1+\sin 2x\end{array}\right|$
Then the ordered pair $(m,M)$ is equal to:

• A. $(–3,–1)$
• B. $(–4,–1)$
• C. $(1,3)$
• D. $(–3,3)$

Answer 1

The correct option is A $(–3,–1)$

$\left|\begin{array}{ccc}{\cos }^{2}x& 1+{\sin }^{2}x& \sin 2x\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ {\cos }^{2}x& {\sin }^{2}x& 1+\sin 2x\end{array}\right|$
$R_1\to R_1-R_2,R_3\to R_3-R_2$
$\left|\begin{array}{ccc}-1& 1& 0\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ -1& 0& 1\end{array}\right|$
$=-1\left({\sin }^{2}x\right)-1(1+{\cos }^{2}x+\sin 2x)$
$=-{\sin }^{2}x-{\cos }^{2}x-1-\sin 2x$
$=-2-\sin 2x$
$\therefore$ Minimum value when $\sin 2x=1$
$m=-2-1=-3$
$\therefore$ Maximum value when $\sin 2x=-1$
$M=-2+1=-1$
$\therefore (m,M)=(-3,-1)$