Question

If $f\left(x\right)=x^3+b{x}^2+ax$ satisfies the conditions of Rolles theorem on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$ then $(a,b)=$

• A. $(11,6)$
• B. $(11,-6)$
• C. $(-6,11)$
• D. $(6,11)$

Answer 1

Answer: (B)

$(11,-6)$

For Rolle's to be applicable
$f\left(x\right)$ should be continous
$f^{\prime }\left(x\right)$ should exist
$f\left(1\right)=f\left(3\right)$
$1+b+a=27+9b+3a$
$26+8b+2a=0$
$a+4b+13=0$
$f^{\prime }\left(c\right)=0$ (Rolle's theorem)
$3c^2+2bc+a=0$
$3(4+\frac{1}{3}+\frac{4}{\sqrt{3}})+2b(2+\frac{1}{\sqrt{3}})+a=0$
$13+\frac{12}{\sqrt{3}}+4b+\frac{2b}{\sqrt{3}}+a=0$
$\frac{12}{\sqrt{3}}=\frac{-2b}{\sqrt{3}}$
$b=-6$ so $a=11$