Question

Let $f(x)=\underset{x}{\overset{x+\frac{\pi }{3}}{\int }}|\sin \theta |d\theta ,0\le x\le \pi$. If $m$ and $M$ are minimum and maximum values of $f(x)$ and $m+M=\sqrt{p}-\sqrt{q}$ where $p,q\in N$, then find the value of $(p+q)$.

Answer 1

Answer: 12

$f(x)=\underset{x}{\overset{x+\frac{\pi }{3}}{\int }}|\sin \theta |d\theta ;x\in [0,\pi ]$
for maximum $/$ minimum $f^{\prime }(x)=0$
$f^{\prime }(x)=\left|\sin \left(x+\frac{\pi }{3}\right)\right|-|\sin x|=0$
$\therefore {\sin }^2\left(x+\frac{\pi }{3}\right)={\sin }^{2}x$
${\sin }^2\left(x+\frac{\pi }{3}\right)-{\sin }^{2}x=0$
$\sin \left(2x+\frac{\pi }{3}\right)\cdot \sin \frac{\pi }{3}=0$
$\therefore 2x+\frac{\pi }{3}=n\pi \Rightarrow x=\frac{n\pi }{2}-\frac{\pi }{6};n\in I$
$\therefore \text{ in }[0,\pi ]x=\frac{\pi }{3}\text{ or }\frac{5\pi }{6}$

Now${f(0)=\underset{0}{\overset{\pi /3}{\int }}\sin \theta d\theta =\cos \theta ]}_{z/3}^0=1-\frac{1}{2}=\frac{1}{2}$
$f(\pi )=\underset{\pi }{\overset{\pi +\frac{\pi }{3}}{\int }}|\sin \theta |d\theta =\underset{\pi }{\overset{\frac{4\pi }{3}}{\int }}-\sin \theta d\theta$
$\therefore f(\pi )=\cos \theta {]}_{\pi }^{4\pi /3}=\frac{-1}{2}+1=\frac{1}{2}$
$f\left(\frac{\pi }{3}\right){=\underset{\pi /3}{\overset{2\pi /3}{\int }}\sin \theta d\theta =\cos \theta ]}_{2\pi /3}^{\pi /3}=\frac{1}{2}-\left(\frac{-1}{2}\right)=1$
$f\left(\frac{5\pi }{6}\right)=\underset{5\pi /6}{\overset{7\pi /6}{\int }}|\sin \theta |d\theta =\underset{5\pi /6}{\overset{\pi }{\int }}\sin \theta d\theta +\underset{\pi }{\overset{7\pi /6}{\int }}-\sin \theta d\theta =2-\sqrt{3}$
Hence minimum value $m=2-\sqrt{3}$ and maximum value $M=1$
$\therefore m+M=3-\sqrt{3}=\sqrt{9}-\sqrt{3}\equiv \sqrt{p}-\sqrt{q}$
$\therefore p=9;q=3\Rightarrow p+q=12$