Question

Let $f(x)=\int\limits_x^{x+\frac{\pi}{3}}|\sin \theta| d \theta, 0 \leq x \leq \pi$. 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

$ f ( x )=\int\limits_{ x }^{ x +\frac{\pi}{3}}|\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(2 x+\frac{\pi}{3}\right) \cdot \sin \frac{\pi}{3}=0$
$\therefore 2 x+\frac{\pi}{3}=n \pi \Rightarrow x=\frac{n \pi}{2}-\frac{\pi}{6} ; n \in I$
$\therefore \text { in }[0, \pi] x=\frac{\pi}{3} \text { or } \frac{5 \pi}{6}$

Now$\left.f(0)=\int\limits_0^{\pi / 3} \sin \theta d \theta=\cos \theta\right]_{z / 3}^0=1-\frac{1}{2}=\frac{1}{2}$
$f (\pi)=\int\limits_\pi^{\pi+\frac{\pi}{3}}|\sin \theta| d \theta=\int\limits_\pi^{\frac{4 \pi}{3}}-\sin \theta d \theta $
$\therefore f (\pi)=\cos \theta]_\pi^{4 \pi / 3}=\frac{-1}{2}+1=\frac{1}{2} $
$f \left(\frac{\pi}{3}\right)\left.=\int\limits_{\pi / 3}^{2 \pi / 3} \sin \theta d \theta=\cos \theta\right]_{2 \pi / 3}^{\pi / 3}=\frac{1}{2}-\left(\frac{-1}{2}\right)=1$
$f \left(\frac{5 \pi}{6}\right)=\int\limits_{5 \pi / 6}^{7 \pi / 6}|\sin \theta| d \theta=\int\limits_{5 \pi / 6}^\pi \sin \theta d \theta+\int\limits_\pi^{7 \pi / 6}-\sin \theta d \theta=2-\sqrt{3}$
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 } $
$\therefore p =9 ; q =3 \Rightarrow p + q =12$