Question

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

• A. $-\frac{1}{4}$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

$0$

Explanation for the correct option :

Step-1 : Finding the critical points of $y$

To find the critical points of $y$, we have to differentiate $y$ and then equate it to zero.

Differentiating $y$ with respect to $x$ , we get

$$\begin{gathered} y\text{'}=40x^{39}–20x^{19} \\ =20x^{19}(2x^{20}–1) \end{gathered}$$

Now, the critical points of $y$ will be the solutions of $y\text{'}=0$. Then

$$\begin{gathered} y\text{'}=0 \\ \Rightarrow 20x^{19}(2x^{20}–1)=0 \\ \Rightarrow x=0\text{or}{x}^{20}=\frac{1}{2} \end{gathered}$$

Step-3 : Fining the maximum value of $y$

Substituting $x=0$, we get

$$\begin{gathered} y=x^{40}-x^{20} \\ =0^{40}-0^{20} \\ =0 \end{gathered}$$

Substituting $x^{20}=\frac{1}{2}$, we get

$$\begin{gathered} \Rightarrow y=\frac{1}{4}–\frac{1}{2} \\ =-\frac{1}{4} \end{gathered}$$

Thus, the absolute maximum value is $y=0$.

Hence, option (B) is the correct answer.