Question

The value of ${\int }_{-\pi /4}^{\pi /4}f\left(\theta \right)d\theta$ is equal to?

• A. $\frac{3\pi }{2}-\sqrt{2}$
• B. $\frac{\pi }{2}+\sqrt{2}$
• C. $\frac{3\pi }{2}+\sqrt{2}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3\pi }{2}+\sqrt{2}$

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

$=2+\cos \theta +1$ ....... $[\because \cos 2\theta =2{\cos }^2\theta -1]$

$=3+\cos \theta$
Now, $f(-\theta )=3+\cos (-\theta )=3+\cos \theta =f\left(\theta \right)$
Hence, $f$ is an even function.
So, ${\int }_{-\pi /4}^{\pi /4}f\left(\theta \right)d\theta$

$=2{\int }_0^{\pi /4}(3+\cos \theta )d\theta$ ..... $\left[f\right(\theta \left)\text{is even}\right]$
$=2{[3\theta +\sin \theta ]}_0^{\pi /4}$
$=2[\frac{3\pi }{4}+\sin \frac{\pi }{4}-0]$
$=\frac{3\pi }{2}+\frac{2}{\sqrt{2}}=\frac{3\pi }{2}+\sqrt{2}$