Question

Let $f\left(b\right)$ be the minimum value of the expression $y=x^{2}-2x+\left(b^{3} - 3 b^{2} + 4\right)\forall x\in R.$ Then, the maximum value of $f\left(b\right)$ as $b$ varies from $0$ to $4$ is

• A. 20
• B. 19
• C. 63
• D. 64

Answer 1

Answer: (B)

$y=\left(x - 1\right)^{2}+b^{3}-3b^{2}+3$
$\therefore y_{\min}=f\left(b\right)=b^{3}-3b^{2}+3$
Now, $f^{'} \left(b\right) = 3 b^{2} - 6 b$
$=3b\left(b - 2\right)$

Also, $f\left(0\right)=3$
$f\left(2\right)=-1$
$f\left(4\right)=19$
$\therefore \max f \left(b\right)=19$