Question

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

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

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

$(-3,-1)$

Explanation for the correct option:

Finding the ordered pair $(m,M)$

The given determinant: $\left|\begin{array}{l}\begin{array}{ccc}{\cos }^{2}x& 1+{\sin }^{2}x& \sin 2x\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ {\cos }^{2}x& {\sin }^{2}x& 1+\sin 2x\end{array}\end{array}\right|$

Applying the row operation $\begin{array}{rcl}{R}_1& \to & R_1-R_2\end{array}$

$\Rightarrow \left|\begin{array}{l}\begin{array}{ccc}-1& 1& 0\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ {\cos }^{2}x& {\sin }^{2}x& 1+\sin 2x\end{array}\end{array}\right|$

Applying the row operation $\begin{array}{rcl}{R}_3& \to & R_3-R_2\end{array}$

$\Rightarrow \begin{array}{l}\begin{array}{l}\left|\begin{array}{ccc}-1& 1& 0\\ 1+{\cos }^{2}x& {\sin }^{2}x& \sin 2x\\ -1& 0& 1\end{array}\right|\end{array}\end{array}$

Now taking determinant

$$\begin{gathered} \Rightarrow \begin{array}{rcl}& & -1({\sin }^{2}x-0)-1(1+{\cos }^{2}x+\sin 2x)\end{array} \\ \Rightarrow \begin{array}{rcl}& & -{\sin }^{2}x–{\cos }^{2}x-1-\sin 2x\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right]\end{array} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & -2-\sin 2x\end{array}$

Since we know the maximum value of $\sin \theta =1$

So minimum value of determinant $$\begin{gathered} m=-2-1 \\ m=-3 \end{gathered}$$

Since we know the minimum value of $\sin \theta =-1$

So minimum value of determinant $$\begin{gathered} M=-2-(-1) \\ M=-1 \end{gathered}$$

Therefore, $\begin{array}{rcl}\therefore (m,M)& =& (-3,-1)\end{array}$

Hence, option (A) is the correct answer.