Question

The value of the function $(x–1){(x–2)}^2$ at its maxima is

• A. $1$
• B. $2$
• C. $0$
• D. $\frac{4}{27}$

Answer 1

The correct option is D

$\frac{4}{27}$

Explanation for the correct option:

Given function:$(x–1){(x–2)}^2$

Let $f\left(x\right)=(x–1){(x–2)}^2$

$$\begin{gathered} \Rightarrow f\left(x\right)=(x–1)(x^2+4–4x) \\ \Rightarrow f\left(x\right)=x^3–5x^2+8x–4 \end{gathered}$$

On differentiating both sides with respect to $x$ we get,

$f\text{'}\left(x\right)=3x^2–10x+8$

For maximum or minimum $f’\left(x\right)=0$

$$\begin{gathered} \therefore 3x^2–10x+8=0 \\ \Rightarrow (3x–4)(x–2)=0 \\ \Rightarrow x=\frac{4}{3}\mathrm{or}x=2 \end{gathered}$$

Again differentiating with respect to $x$ we get

$f\text{'}\text{'}\left(x\right)=6x–10$

Now,

$$\begin{gathered} f\text{'}\text{'}\left(\frac{4}{3}\right)=6\times \left(\frac{4}{3}\right)–10 \\ =-2<0 \end{gathered}$$

$$\begin{gathered} f\text{'}\text{'}\left(2\right)=6\times \left(2\right)–10 \\ =2>0 \end{gathered}$$

Therefore, at $x=\frac{4}{3}$, the function will attain its maximum value.

$$\begin{gathered} \therefore f\left(\frac{4}{3}\right)=\left(\frac{4}{3}–1\right){\left(\frac{4}{3}–2\right)}^2 \\ =\frac{1}{3}\times \frac{4}{9} \\ =\frac{4}{27} \end{gathered}$$

Hence, option D is the correct option.