# be respectively the minimum and maximum values of

Question:

Let $m$ and $M$ be respectively the minimum and maximum values of

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

Then the ordered pair $(m, M)$ is equal to :

1. (1) $(-3,3)$

2. (2) $(-3,-1)(3)(-4,-1)$

3. (3) $(-4,-1)$

4. (4) $(1,3)$

Correct Option: , 2

Solution:

$C_{1} \rightarrow C_{1}+C_{2}$

Let $f(x)=\left|\begin{array}{ccc}2 & 1+\sin ^{2} x & \sin 2 x \\ 2 & \sin ^{2} x & \sin 2 x \\ 1 & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$

$R_{1} \rightarrow R_{1}-2 R_{3} ; R_{2} \rightarrow R_{2}-2 R_{3}$

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

$f^{\prime}(x)=-2 \cos 2 x=0$

$\Rightarrow \cos 2 x=0 \Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}$

$f^{\prime \prime}(x)=4 \sin 2 x$

So, $f^{\prime \prime}\left(\frac{\pi}{4}\right)=4>0$ (minima)

$m=f\left(\frac{\pi}{4}\right)=-2-1=-3$

$f^{\prime \prime}\left(\frac{3 \pi}{4}\right)=-4<0$ (maxima)

$M=f\left(\frac{3 \pi}{4}\right)=-2+1=-1$

So, $(m, M)=(-3,-1)$