Question

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

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

Answer 1

The correct option is B $(-4,0)$

$f\left(\theta \right)=\left|\begin{array}{ccc}-{\sin }^2\theta & -1-{\sin }^2\theta & 1\\ -{\cos }^2\theta & -1-{\cos }^2\theta & 1\\ 12& 10& -2\end{array}\right|$
$C_1\to C_1-C_2,C_3\to C_3+C_2$
$=\left|\begin{array}{ccc}1& -1-{\sin }^2\theta & -{\sin }^2\theta \\ 1& -1-{\cos }^2\theta & -{\cos }^2\theta \\ 2& 10& 8\end{array}\right|C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}1& -1& -{\sin }^2\theta \\ 1& -1& -{\cos }^2\theta \\ 2& 2& 8\end{array}\right|$
$=1(2{\cos }^2\theta -8)+(8+2{\cos }^2\theta )-4{\sin }^2\theta$
$f\left(\theta \right)=4{\cos }^2\theta -4{\sin }^2\theta =4\cos 2\theta$
Given, $\theta \in [\frac{\pi }{4},\frac{\pi }{2}]$
$2\theta \in [\frac{\pi }{2},\pi ]$
$\therefore f\left(\theta \right)\in [-4,0]$
$(m,M)=(-4,0)$