Question

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

• A. 1
• B. 2
• C. $\frac{4}{27}$

Answer 1

The correct option is C $\frac{4}{27}$

Given $f\left(x\right)=(x-1)(x-2{)}^2$
$f\left(x\right)=(x-1)(x^2+4-4x);f\left(x\right)=(x^3-5x^2+8x-4)$
Now $f\left(x\right)=3x^2-10x+8.f\left(x\right)=0$
$\Rightarrow 3x^2-10x+8=0\Rightarrow (3x-4)(x-2)=0\Rightarrow x=\frac{4}{3},2$
Now $f^{\prime }\left(x\right)=6x-10$
$f^{\prime }\left(\frac{4}{3}\right)=6\times \frac{4}{3}-10<0f^{\prime }\left(2\right)=12-10>0$
Hence at $x=\frac{4}{3}$ the function will occupy maximum value.
$\therefore$ Maximum value =$f\left(\frac{4}{3}\right)=\frac{4}{27}$.