Question

Let $f(\theta)=\begin{vmatrix} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{vmatrix}$
Suppose $A$ and $B$ are respectively maximum and minimum value of $f(\theta)$. Then $(A, B)$ is equal to:

• A. $(2,1)$
• B. $(2,0)$
• C. $(\sqrt{2}, 1)$
• D. $\left(2, \frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B)

Using $C_1 \rightarrow C_1+C_3$, we get
$f(\theta)=\begin{vmatrix}0 & \cos \theta & -1 \\-(\sin \theta+\cos \theta) & 1 & -\cos \theta \\0 & \sin \theta & 1\end{vmatrix}$
Evaluating along $C_1$, we get
$f(\theta) =(\sin \theta+\cos \theta)\begin{vmatrix}\cos \theta & -1 \\\sin \theta & 1\end{vmatrix}$
$ =(\sin \theta+\cos \theta)^2 $
$ =\left[\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta\right)\right]^2 $
$ =2 \sin ^2\left(\frac{\pi}{4}+\theta\right) $
As $0 \leq \sin ^2\left(\frac{\pi}{4}+\theta\right) \leq 1, \text { we get } $
$0 \leq f(\theta) \leq 2$
$A =2 \text { for } \theta=\frac{\pi}{4} $
$B =0 \text { for } \theta=-\frac{\pi}{4}$
As $0 \leq \sin ^2\left(\frac{\pi}{4}+\theta\right) \leq 1$, we get
and $B=0$ for $\theta=-\frac{\pi}{4}$
Thus, $(A, B)=(2,0)$