Question

Let $m$ and $M$ be respectively the minimum and maximum values of $\left|\begin{array}{cccc}& co{s}^{2}x& 1+si{n}^{2}x& sin2x\\ & 1+co{s}^{2}x& si{n}^{2}x& sin2x\\ & co{s}^{2}x& si{n}^{2}x& 1+sin2x\end{array}\right|$. Then the ordered pair $(m,M)$ is equal to

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

Answer 1

Answer: (A)

$\left|\begin{array}{cccc}& co{s}^{2}x& 1+si{n}^{2}x& sin2x\\ & 1+co{s}^{2}x& si{n}^{2}x& sin2x\\ & co{s}^{2}x& si{n}^{2}x& 1+sin2x\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\left|\begin{array}{ccc}-1& 1& 0\\ 1& 0& -1\\ co{s}^{2}x& si{n}^{2}x& 1+sin2x\end{array}\right|$
$=-1\left({\sin }^{2}x\right)-1\left(1+\sin 2x+{\cos }^{2}x\right)$
$=-\sin 2x-2$
$m=-3,M=-1$