Question

The minimum value of $f\left(x\right)={\sin }^4\left(x\right)+{\cos }^4\left(x\right)$ , $0\le x\le \frac{\mathrm{\pi }}{2}$ is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

Find the minimum value of the given function

Given,

$\begin{array}{rcl}f\left(x\right)& =& {\sin }^4\left(x\right)+{\cos }^4\left(x\right)\\ & =& {\left({\sin }^2\left(x\right)\right)}^2+{\left({\cos }^2\left(x\right)\right)}^2\\ & =& {\left[{\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right]}^2-2{\sin }^2\left(x\right){\cos }^2\left(x\right)\\ & =& 1-2{\sin }^2\left(x\right){\cos }^2\left(x\right)\left[2\sin \left(x\right)\cos \left(x\right)=\sin \left(2x\right)\right]\\ & =& 1-\left[\frac{{\sin }^2\left(2x\right)}{2}\right]\end{array}$

We know that the range of $\sin \left(x\right)$ is $\left[-1,1\right]$, whereas the range of ${\sin }^2\left(2x\right)$ is $\left[0,1\right]$.

$0\le {\sin }^2\left(2x\right)\le 1$

This can be written as,

$0\ge -{\sin }^2\left(2x\right)\ge -1$

Divide by $2$,

$0\ge \frac{-{\sin }^2\left(2x\right)}{2}\ge \frac{-1}{2}$

Add $1$ on both side,

$1\ge 1-\frac{{\sin }^2\left(2x\right)}{2}\ge \frac{1}{2}$

Therefore, the minimum value of the function is $\frac{1}{2}$.

Hence option $\left(D\right)$ is the correct answer.