Question

If the function $f\left(x\right)=a{x}^3+b{x}^2+11x-6$ satisfies the condition of Rolle's theorem in $\left[1,3\right]$ and $f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=0$, then the values of $a$ and $b$ respectively are:

• A. $-1,6$
• B. $-2,1$
• C. $1,-6$
• D. $-1,\frac{1}{2}$

Answer 1

The correct option is C

$1,-6$

Explanation for correct option

Given function, $f\left(x\right)=a{x}^3+b{x}^2+11x-6$

The function $f\left(x\right)$ satisfies the Rolle's theorem in $\left[1,3\right]$

So, $f\left(1\right)=f\left(3\right)$

$$\begin{gathered} \to f\left(1\right)=a{\left(1\right)}^3+b{\left(1\right)}^2+11\left(1\right)-6 \\ =a+b+5 \end{gathered}$$

$$\begin{gathered} \to f\left(3\right)=a{\left(3\right)}^3+b{\left(3\right)}^2+11\left(3\right)-6 \\ =27a+9b+27 \end{gathered}$$

Equating the two results,

$a+b+5=27a+9b+27$

$\Rightarrow 26a+8b+22=0$

$\Rightarrow 13a+4b+11=0$ $...\left(1\right)$

Now, differentiate the $f\left(x\right)$, we get $f\text{'}\left(x\right)=3a{x}^2+2bx+11$

Evaluate, $f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)$,

$$\begin{gathered} f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=3a{\left(2+\frac{1}{\sqrt{3}}\right)}^2+2b\left(2+\frac{1}{\sqrt{3}}\right)+11 \\ =3a\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)+4b+\frac{2b}{\sqrt{3}}+11 \\ =13a+4\sqrt{3}a+4b+\frac{2b}{\sqrt{3}}+11[\because 13\mathrm{a}+4\mathrm{b}+11=0] \\ =4\sqrt{3}a+\frac{2b}{\sqrt{3}} \\ =\frac{2\left(6a+b\right)}{\sqrt{3}} \end{gathered}$$

So, $6a+b=0$ $...\left(2\right)$

Solve $\left(1\right)$ and $\left(2\right)$,

$a=1$ and $b=-6$

Therefore, option (C) i.e. $1,-6$ is the correct answer.