Question

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

• A. $-3,2$
• B. $2,-4$
• C. $1,6$
• D. none of these

Answer 1

The correct option is D none of these

$f\left(x\right)=a{x}^3+b{x}^2+11x-6$
$f^{\prime }\left(x\right)=3a{x}^2+2bx+11$
$f\left(1\right)=f\left(3\right)$
$\Rightarrow 13a+4b+11=0$ .....(1)
Also,$f^{\prime }(2+\frac{1}{\sqrt{3}})=0$
$\Rightarrow 3a(6+\sqrt{3})+2b(6+\sqrt{3})+33=0$ .....(2)
Solving (1) and (2), we get
$a=1,b=-6$