Question

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

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

Hint:

Apply \(C_2\to C_2-C_1\) and then \(R_1\to R_1-R_2\) to reduce the determinant to a multiple of \(\cos 2\theta\).

Answer 1

The Correct Option is B

Apply \(C_2\to C_2-C_1\): the second column becomes \((-1,\,-1,\,-2)^T\).
Now \(f(\theta)=\begin{vmatrix} -\sin^2\theta & -1 & 1 \\ -\cos^2\theta & -1 & 1 \\ 12 & -2 & -2 \end{vmatrix}\).
Apply \(R_1\to R_1-R_2\): first row becomes \((\cos^2\theta-\sin^2\theta,\,0,\,0)=(\cos2\theta,0,0)\).
Expand along row 1: \(f(\theta)=\cos2\theta\begin{vmatrix}-1&1\\-2&-2\end{vmatrix}=\cos2\theta\,(2+2)=4\cos2\theta\).
For \(\theta\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]\), \(2\theta\in\left[\frac{\pi}{2},\pi\right]\) so \(\cos2\theta\) decreases from \(0\) to \(-1\).
Hence \(f\) ranges over \([-4,0]\): \(m=-4\), \(M=0\).
Answer: (2) \((-4, 0)\)