Question

If the minimum and the maximum values of the function $f:\left[\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right]\to R$ defined by $f\left(\theta \right)=\left|\begin{array}{ccc}-{\sin }^2\left(\theta \right)& -1-{\sin }^2\left(\theta \right)& 1\\ -{\cos }^2\left(\theta \right)& -1-{\cos }^2\left(\theta \right)& 1\\ 12& 10& -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $\left(m,M\right)$ is equal to

• A. $\left(0,4\right)$
• B. $\left(-4,0\right)$
• C. $\left(-4,4\right)$
• D. $\left(0,2\sqrt{2}\right)$

Answer 1

The correct option is B

$\left(-4,0\right)$

Explanation of the correct option.

Compute the required value.

Given : $f\left(\theta \right)=\left|\begin{array}{ccc}-{\sin }^2\left(\theta \right)& -1-{\sin }^2\left(\theta \right)& 1\\ -{\cos }^2\left(\theta \right)& -1-{\cos }^2\left(\theta \right)& 1\\ 12& 10& -2\end{array}\right|$

$C_1\to C_1-C_2,C_3\to C_3+C_2$

$\Rightarrow$ $f\left(\theta \right)=\left|\begin{array}{ccc}1& -1-{\sin }^2\left(\theta \right)& -{\sin }^2\left(\theta \right)\\ 1& -1-{\cos }^2\left(\theta \right)& -{\cos }^2\left(\theta \right)\\ 2& 10& 8\end{array}\right|$

$C_2\to C_2-C_3$

$\Rightarrow$ $f\left(\theta \right)=\left|\begin{array}{ccc}1& -1& -{\sin }^2\left(\theta \right)\\ 1& -1& -{\cos }^2\left(\theta \right)\\ 2& 2& 8\end{array}\right|$

$\Rightarrow$ $f\left(\theta \right)=1(2{\cos }^2\left(\theta \right)-8)+\left(8+2{\cos }^2\left(\theta \right)\right)-4{\sin }^2\left(\theta \right)$

$\Rightarrow$ $f\left(\theta \right)=4\cos \left(2\theta \right)$

$f:\left[\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right]\to R$, and $m$ is the minimum value and $M$ is the maximum value, thus

$$\begin{gathered} m=4\cos \left(2\times \frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow m=-4 \end{gathered}$$

$$\begin{gathered} M=4\cos \left(2\times \frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow M=4\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow M=0 \end{gathered}$$

Therefore, $\left(m,M\right)=\left(-4,0\right)$

Hence option $\left(B\right)$ is the correct option.