Question

It is given that for the function $f\left(x\right)=x^3+b{x}^2+ax+5on[1,3]$,Rolle's theorem holds with $c=2+\frac{1}{\sqrt{3}}$.Find the values of a and b.

Answer 1

Given $f\left(x\right)=x^3+b{x}^2+ax+5on[1,3]\Rightarrow f^{\prime }\left(x\right)=3x^2+2bx+a$

$\Rightarrow f^{\prime }\left(c\right)=3c^2+2bc+a=0\Rightarrow 3{(2+\frac{1}{\sqrt{3}})}^2+2b(2+\frac{1}{\sqrt{3}})+a=\mathrm{0...}\left(i\right)$

Also f(1)=f(3) $\Rightarrow$ $\text{b+a+6=32+9b+3a or, a+4b=-13...(ii)}$

$\text{Solving (i) and (ii),we get : a = 11, b = -6}$.