Question

If $f\left(x\right)=\sin x+\cos x$ and $g\left(x\right)=\{\begin{array}{cc}\frac{|x|}{x},& x\ne 0\\ 2,& x=0\end{array}$ then the value of ${\int }_{-\pi /4}^{2\pi }gof\left(x\right)dx$ is equal to

• A. $3\pi /4$
• B. $\pi /4$
• C. $\pi$
• D. None of the above

Answer 1

Answer: (B)

$\pi /4$

We have
$f\left(x\right)=\sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})$
and, $g\left(x\right)=\{\begin{array}{cc}\frac{|x|}{x},& x\ne 0\\ 2,& x=0\end{array}$
$\Rightarrow g\left(x\right)=\{\begin{array}{cc}1,& x>0\\ 2,& x=0\\ -1,& x<0\end{array}$
$\therefore gof\left(x\right)=\{\begin{array}{cc}1,& \text{if}x\in (-\pi /4,3\pi /4)\cup (7\pi /4,2\pi )\\ 2,& \text{if}x=-\frac{\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4}\\ -1,& \text{if}x\in (3\pi /4,7\pi /4)\end{array}$
$\therefore {\int }_{-\pi /4}^{2\pi }\text{gof}\left(x\right)dx={\int }_{-\pi /4}^{3\pi /4}1dx+{\int }_{3\pi /4}^{7\pi /4}-1dx+{\int }_{7\pi /4}^{2\pi }1dx$
$=1(\frac{3\pi }{4}+\frac{\pi }{4})-(\frac{7\pi }{4}-\frac{3\pi }{4})+(2\pi -\frac{7\pi }{4})$ $=\pi -\pi +\frac{\pi }{4}=\frac{\pi }{4}$