Question

if $f \left(\theta\right)=\begin{vmatrix}1&cos&\theta&1\\ -sin\,\theta&1&&-cos\,\theta\\ -1&sin&\theta&1\end{vmatrix}$ and $A$ and $B$ are respectively the maximum and the minimum values of $f \left(\theta\right)$, then $(A, B)$ is equal to :

• A. $(3, -1)$
• B. $\left(4,2-\sqrt{2}\right)$
• C. $\left(2+\sqrt{2},2-\sqrt{2}\right)$
• D. $\left(2+\sqrt{2},-1\right)$

Answer 1

Answer: (C)

$f \left(\theta\right)=\begin{vmatrix}1&cos&\theta&1\\ -sin\,\theta&1&&-cos\,\theta\\ -1&sin&\theta&1\end{vmatrix}$
$f (\theta)=1(1+\sin \theta \cos \theta)+\cos \theta(\sin \theta+\cos \theta)+$
$1\left(1-\sin ^{2} \theta\right)$
$=1+2 \sin \theta \cos \theta+2 \cos ^{2} \theta$
$=1+\sin 2 \theta+(1+\cos 2 \theta)$
$=2+\sin 2 \theta+\cos 2 \theta$
Now, $\sin 2 \theta+\cos 2 \theta$ lies between $-\sqrt{2}$ to $\sqrt{2}$
$\left[\sqrt{2}\left(\sin \left(\frac{\pi}{4}+2 \theta\right)\right) \rightarrow \pm \sqrt{2}\right]$
$\therefore A =2+\sqrt{2} ; \quad B =2-\sqrt{2}$