Explanation

(b)
Given, $y = x^{40} - x^{20}$
$\therefore\ \ \ \ \ \frac{dy}{dx} = 40x^{39} - 20x^{19}$
Now, put $\frac{dy}{dx} = 0\ ie,\ 20x^{19}\ \left( 2x^{20} - 1 \right) = 0$
$\Rightarrow \ \ \ \ \ x = 0\ \ \ \ or\ \ \ \ x^{20} = \frac{1}{2}$
When $x = 0,\ \ \ y = 0$
And $x^{20} = \frac{1}{2}$
$y = \left( \frac{1}{2} \right)^{2} - \frac{1}{2} = - \frac{1}{4}$
Also, $x = 1,\ \ \ \ y = 0$
Hence, absolute maximum value of $y$ is 0