Question

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

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

Answer 1

The correct option is A

$1,-6$

Explanation for the correct option

Step 1: Simplify by applying Rolle's theorem

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

Rolle's theorem states that if a function f is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ such that $f\left(a\right)=f\left(b\right)$, then $f\prime \left(x\right)=0$ for some $x$ with $a\le x\le b$.

As the give function satisfies Rolle's theorem in the interval $[1,3]$.

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

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

$\Rightarrow$ $f\left(1\right)=a+b+5$

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

$\Rightarrow$ $f\left(3\right)=27a+9b+27$

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

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

$\Rightarrow$$13a+4b=-11$ $...\left(i\right)$

Step 2: Solve for the required values

Differentiating the given function with respect to $x$ we get

$\Rightarrow$$f\text{'}\left(x\right)=3\times a{x}^2+2\times bx+11\times 1-0$

$\Rightarrow$$f\text{'}\left(x\right)=3a{x}^2+2bx+11$

It is given that $f\prime \left(2+\frac{1}{\sqrt{3}}\right)=0$

$\Rightarrow$$3a{\left(2+\frac{1}{\sqrt{3}}\right)}^2+2b\left(2+\frac{1}{\sqrt{3}}\right)+11=0$

$\Rightarrow$ $3a\left(4+\frac{4}{\sqrt{3}}+\frac{1}{3}\right)+2b\left(2+\frac{1}{\sqrt{3}}\right)=-11$

$\Rightarrow$ $12a+a+4\sqrt{3}a+4b+\frac{2b}{\sqrt{3}}=-11$

$\Rightarrow$ $13a+4b+4\sqrt{3}a+\frac{2b}{\sqrt{3}}=-11$

$\Rightarrow$ $-11+4\sqrt{3}a+\frac{2b}{\sqrt{3}}=-11$ …[From $\left(i\right)$]

$\Rightarrow$ $\frac{2b}{\sqrt{3}}=-4\sqrt{3}a$

$\Rightarrow$ $b=-6a$ $...\left(ii\right)$

Solving $\left(i\right),\left(ii\right)$ simultaneously we get

$\Rightarrow$$13a+4(-6a)=-11$

$\Rightarrow$ $-11a=-11$

$\Rightarrow$ $a=1$

$\Rightarrow$ $b=-6$

Thus the values of $a$ and $b$ are $1,-6$ respectively.

Hence option(A) is the correct answer.