If $f(x)=\sin x+\cos x$ and $g(x)=\left\{\begin{array}{c}\frac{|x|}{x}, x \neq 0 \\ 2, x=0\end{array}\right.$ then the value of $\int\limits_{-\pi / 4}^{2 \pi} gof(x) d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$f(x)=\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$
and, $g(x)=\left\{\begin{array}{cl}\frac{|x|}{x}, & x \neq 0 \\ 2, & x=0\end{array}\right.$
$\Rightarrow g(x)=\left\{\begin{aligned} 1, & x>0 \\ 2 & , x=0 \\ -1 & , x<0\end{aligned}\right.$
∴ $gof(x)=\left\{\begin{aligned} 1, & \text { if } x \in(-\pi / 4,3 \pi / 4) \in(7 \pi / 4,2 \pi) \\ 2, & \text { if } x=-\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{7 \pi}{4} \\ -1, & \text { if } x \in(3 \pi / 4,7 \pi / 4)\end{aligned}\right. $
∴ $\int\limits_{-\pi / 4}^{2 \pi} gof(x) d x=\int\limits_{-\pi / 4}^{3 \pi / 4} 1 d x+\int\limits_{3 \pi / 4}^{7 \pi / 4}-1 d x+\int\limits_{7 \pi / 4}^{2 \pi} 1 d x$
$\Rightarrow \int\limits_{-\pi / 4}^{2 \pi} gof(x) d x=1\left(\frac{3 \pi}{4}+\frac{\pi}{4}\right)-\left(\frac{7 \pi}{4}-\frac{3 \pi}{4}\right)+\left(2 \pi-\frac{7 \pi}{4}\right)$
$\Rightarrow \int\limits_{-\pi / 4}^{2 \pi} gof(x) d x=\pi-\pi+\frac{\pi}{4}=\frac{\pi}{4}$