JEE Advanced · 2006 · 09 Apr 2006 · Application of Derivatives

Let $f(x)=a{x}^3+b{x}^2+cx+d$. Then $f^{\prime }(x)=3a{x}^2+2bx+c$.

Given conditions:

1. $f^{\prime }(-1)=0$ (local maximum at $x=-1$)
2. $f^{\prime \prime }(0)=0$ (since $f^{\prime }(x)$ has local minimum at $x=0$)
3. $f(1)=-1$ and $f(2)=18$

From $f^{\prime \prime }(x)=6ax+2b$ and $f^{\prime \prime }(0)=0$: $b=0$

From $f^{\prime }(-1)=0$: $3a-c=0$, so $c=3a$

From $f(1)=-1$: $a+c+d=-1$, so $a+3a+d=-1$, giving $4a+d=-1$

From $f(2)=18$: $8a+2c+d=18$, so $8a+6a+d=18$, giving $14a+d=18$

Solving: $10a=19$, so $a=\frac{19}{4}$, and $d=-1-4a=-1-19=-20$...

The source gives $f(x)=\frac{1}{4}(19x^3-57x+34)$.

This has local maximum at $x=-1$ and local minimum at $x=1$.

For $x\in [1,2\sqrt{5}]$, since the local minimum is at $x=1$, $f(x)$ is increasing on this interval.

Therefore, options (B) and (C) are correct.