Question

The difference between the greatest and least values of the function $f\left(x\right)=\sin 2x-x$ on $[-\pi /2,\pi /2]is?.$

• A. $\pi .$
• B. $\pi /2.$
• C. $2\pi .$
• D. $2/\pi .$

Answer 1

The correct option is A $\pi .$

$f\left(\frac{\pi }{2}\right)=-\frac{\pi }{2},f(-\frac{\pi }{2})=\frac{\pi }{2}$

$\frac{dy}{dx}=2\cos 2x-1=0\therefore \cos 2x=\frac{1}{2}$

$\therefore 2x=-\frac{\pi }{3},\frac{\pi }{3}$ or $x=-\frac{\pi }{6},\frac{\pi }{6}$ in $[-\frac{\pi }{2},\frac{\pi }{2}]$

$\frac{d^{2}y}{d{x}^2}=-4\sin 2x==+ive$ for $-\frac{\pi }{6}$

$\therefore$ Min. at $x=-\frac{\pi }{6}$ and MIn.value $=-\frac{\sqrt{3}}{2}+\frac{\pi }{6}$

$\frac{d^{2}y}{d{x}^2}=-ive$ for $x=\frac{\pi }{6}$

$\therefore$ Max. at $x=\frac{\pi }{6}$ and Max. value $=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

$x=-\frac{\pi }{2},-\frac{\pi }{6},\frac{\pi }{6},\frac{\pi }{2}$

value $=\frac{\pi }{2},(-\frac{\sqrt{3}}{2}+\frac{\pi }{2}),(\frac{\sqrt{3}}{2}-\frac{\pi }{2}),-\frac{\pi }{2}$

$\frac{\pi }{2}$ is greatest and $-\frac{\pi }{2}$ is least value and their difference is $\pi .$

Ans: A