Question

Find the values of x for which A $si{n}^{2}x$ is minimum and Maximum. A is some positive constant.

• A. Maxima : $x=N\pi$
• B. Maxima: $x=(2N+1)\frac{\pi }{2}$
• C. Minima: $x=(2N+1)\pi$
• D. none of these

Answer 1

The correct option is B

Maxima: $x=(2N+1)\frac{\pi }{2}$

$y=Asi{n}^{2}x$
For minima For maxima
$\frac{dy}{dx}=0\frac{dy}{dx}=0$
$\frac{d^{2}y}{d{x}^2}>0\frac{d^{2}y}{d{x}^2}<0$
$\frac{dy}{dx}=2Asinxcosx=Asin2x\left(Chainrule\right)$
$\Rightarrow Asin2x=0$
$\Rightarrow 2x=n\pi$
$\Rightarrow x=\frac{n\pi }{2}$ where n=0,1,2,.......
So minima and maxima occur at these points $0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi .....$
$\frac{d^{2}y}{d{x}^2}>0$ for minima
2Acos2x $>$ 0 putting value of x
$cos2\left(\frac{n\pi }{2}\right)>0$
$cosn\pi >0$
only possible when n is even or n=2N
so at $x=\frac{2N\pi }{2}=N\pi$ where N=0,1,2......
we will have maxima
$\frac{d^{2}y}{d{x}^2}<0$ for maxima
$\Rightarrow 2Acos2x<0$
Putting value of x
$cos\frac{n\pi }{2}.2<0$
only possible if n is odd or n=(2N+1)
so values of x at which function is x = $(2N+1)\frac{\pi }{2}$