Question

Let $f\left(\theta \right)=\left|\begin{array}{ccc}\cos \frac{\theta }{2}& 1& 1\\ 1& \cos \frac{\theta }{2}& -\cos \frac{\theta }{2}\\ -\cos \frac{\theta }{2}& 1& -1\end{array}\right|$$f\left(\pi \right)+f(-\pi )$ is equal to

• A. The maximum value of $f\left(\theta \right)$
• B. The minimum value of $f\left(\theta \right)$
• C. Average value of the range of $f\left(\theta \right)$
• D. None of these

Answer 1

The correct option is A The maximum value of $f\left(\theta \right)$

$f\left(\theta \right)=\left|\begin{array}{ccc}\cos \frac{\theta }{2}& 1& 1\\ 1& \cos \frac{\theta }{2}& -\cos \frac{\theta }{2}\\ -\cos \frac{\theta }{2}& 1& -1\end{array}\right|$
$f\left(\theta \right)=2+2{\cos }^2\frac{\theta }{2}=3+\cos \theta$

Now,

$\because -1\le \cos \theta \le 1$

$\Rightarrow 2\le 3+\cos \theta \le 4$

Maximum value of $f\left(\theta \right)$ is 4.

Now, $f\left(\pi \right)=3-1=2$

$f(-\pi )=3-1=2$

Hence, $f\left(\pi \right)+f(-\pi )=4$(Maximum value of $f\left(\theta \right)$)