Question

Find the local minimum value of the function $f\left(x\right)=si{n}^{4}x+co{s}^{4}x,0<x<\frac{\pi }{2}$

• A. $1/\sqrt{2}$
• B. $1/2$
• C. $\sqrt{3}/2$
• D. $0$

Answer 1

Answer: (B)

We have, $y=f\left(x\right)=si{n}^{4}x+co{s}^{4}x$
$\Rightarrow \frac{dy}{dx}=f\left(x\right)=4si{n}^{3}xcosx-4co{s}^{3}xsinx$
$\Rightarrow \frac{dy}{dx}=-4cosxsinx\left(co{s}^{2}x-si{n}^{2}x\right)$
$\Rightarrow \frac{dy}{dx}=-2sin2xcos2x=-sin4x$
For a local maximum or a local minimum, we have
$\frac{dy}{dx}=0$
$\Rightarrow -sin4x=0$
$\Rightarrow sin4x=0$
$\Rightarrow 4x=\pi$
$\Rightarrow x=\frac{\pi }{4}$
$\left[\because 0<x<\frac{\pi }{2}\therefore 0<4x<2\pi \right]$
$\therefore \frac{d^{2}y}{d{x}^2}=-4cos4x$
$\Rightarrow {\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=\pi /4}$
So, $x=\frac{\pi }{4}$ is a point of local minimum.
The local minimum value of $f\left(x\right)$ at $x=\frac{\pi }{4}$ is
$f\left(\frac{\pi }{4}\right)={\left(sin\frac{\pi }{4}\right)}^4+{\left(cos\frac{\pi }{4}\right)}^4$
$=\frac{1}{2}$.