Question

if $f\left(\theta \right)=\left|\begin{array}{cccc}1& cos& \theta & 1\\ -sin\theta & 1& & -cos\theta \\ -1& sin& \theta & 1\end{array}\right|$ 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)

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