Question

The absolute maximum of $x^{40}-x^{20}$ on the interval $[0,1]$ is

• A. $-1/4$
• B. $0$
• C. $1/4$
• D. $1/2$

Answer 1

Answer: (B)

Given, $y=x^{40}-x^{20}$
$\therefore$ $\frac{dy}{dx}=40x^{39}-20x^{19}$
$\frac{dy}{dx}=20x^{19}(2x^{20}-1)$
Now, put $\frac{dy}{dx}=0$ ie,
$20x^{19}(2x^{20}-1)=0$
$\Rightarrow$ $x=0$ or $x^{20}=1/2$
$y_{x=0}=0$ and $y_{x=1}=0$
Now, $y=x^{40}-x^{20}={(x^{20})}^2-x^{20}$
$y_{x^{20}=\frac{1}{2}}={\left(\frac{1}{2}\right)}^2-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$
$\therefore$ Absolute maximum value of y is 0.