Question

Let $f\left(b\right)$ be the minimum value of the expression $y=x^2-2x+\left(b^3-3b^2+4\right)\forall x\in R.$ 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, $f^{{}^{\prime }}\left(b\right)=3b^2-6b$
$=3b\left(b-2\right)$

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