Question

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

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

Answer 1

Answer: (A)

$1,-6$

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

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

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

Therefore, $a+b+11-6=a\times 27+b\times 9+11\times 3-6$

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

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

$\Rightarrow 13a+4b=-11$ ........(1)

Given, $f{}^{\prime }(2+\frac{1}{3})=0$

Thus $f^{\prime }\left(x\right)=3a{x}^2+2bx+11$

$\Rightarrow 3a\times {\left(\frac{7}{3}\right)}^2+2b\times \frac{7}{3}+11=0$

$\Rightarrow \frac{49a}{3}+\frac{14b}{3}+11=0$
$\Rightarrow 49a+14b=-33$ .......(2)

Solving equations (1) and (2), we get

$a=1,b=-6$