Question

lf ${\int }_a^b|\sin x|dx=8$ and ${\int }_a^{a+b}|\cos x|dx=\frac{9}{2}$ then $a$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{4}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\pi }{4}$

${\int }_a^b|\sin x|dx=8\Rightarrow b-a=4\pi$
$\because$ period of $|\sin x|$ is $\pi$ and ${\int }_{\pi }^0|\sin x|dx=2$.
Also $|\cos x|$ is periodic with period $\pi$ and ${\int }_{\frac{\pi }{2}}^0|\cos x|=1$
So, ${\int }_a^{a+b}|\cos x|dx=\frac{9}{2}\Rightarrow a+b=\frac{9\pi }{2}$
$\Rightarrow b=\frac{17\pi }{4}$ and $a=\frac{\pi }{4}$