Question

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

• A. $-3,2$
• B. $2,-4$
• C. $1,6$
• D. $1,-6$

Answer 1

The correct option is D $1,-6$

Given : $f\left(x\right)=a{x}^3+b{x}^2+11x-6$ is continuous and differentiable in $[1,3].$
For Rolle's theorem to be applicable :
$\Rightarrow f\left(1\right)=f\left(3\right)$
$\Rightarrow a+b+11-6=27a+9b+33-6\Rightarrow 13a+4b=-11\cdots \left(i\right)$
and $f^{\prime }\left(x\right)=3a{x}^2+2bx+11$
$\Rightarrow f^{\prime }\left(c\right)=0$
$\Rightarrow f^{\prime }(2+\frac{1}{\sqrt{3}})=3a{(2+\frac{1}{\sqrt{3}})}^2+2b(2+\frac{1}{\sqrt{3}})+11=0$
$\Rightarrow 3a(4+\frac{1}{3}+\frac{4}{\sqrt{3}})+2b(2+\frac{1}{\sqrt{3}})+11=0\cdots \left(ii\right)$
Solving equations $\left(i\right)$ and $\left(ii\right)$ :
we get, $a=1,b=-6$