Question

If the function $f\left(x\right)=a{x}^3-\frac{11x^2}{b}+11x-6$ satisfies conditions of Rolle's theorem in $[1,3]$ and $f^{\prime }\left(2\right)=0$, then find $a+b$.

• A. $\frac{17}{6}$
• B. $\frac{11}{6}$
• C. $\frac{7}{6}$
• D. $4$

Answer 1

The correct option is D $4$

If given function $f\left(x\right)$ satisfies Rolle's theorem on interval $[1,3]$, then it will satisfy following conditions:
1) $f\left(x\right)$ will be continuous on closed interval $[1,3]$
2) $f\left(x\right)$ will be differentiable on open interval $(1,3)$
3) $f\left(1\right)=f\left(3\right)$
And there will be some point $c$ in $(1,3)$ such that $f^{\prime }\left(c\right)=0$.
Using $3^{rd}$ condition we will get,
$a-\frac{11}{b}+11-6=27a-\frac{99}{b}+33-6⟹26a-\frac{88}{b}+22=0\to$(i)
As given $f^{\prime }\left(2\right)=0⟹12a-\frac{44}{b}+11-6=0\to$ (ii)
Solving equation (i) & (ii), we will get
$a=0$ & $b=4$
$\therefore a+b=0+4=4$