Question

If $f\left(x\right)=x^3+b{x}^2+ax$ satisfies the condition of Rolle's theorem on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}.$ Then $(a+b)=$

• A. $0$
• B. $3$
• C. $-4$
• D. $5$

Answer 1

The correct option is D $5$

$f\left(x\right)=x^3+b{x}^2+ax$
$\Rightarrow f\left(1\right)=a+b+1\cdots \left(i\right)$
$f\left(3\right)=3a+9b+27\cdots \left(ii\right)$
$f\left(1\right)=f\left(3\right)\Rightarrow a+4b+13=0\cdots \left(iii\right)$
$\Rightarrow f^{\prime }\left(c\right)=0,$
$\Rightarrow 3c^2+2bc+a=0$
putting the value of $c$
$\Rightarrow 3(4+\frac{1}{3}+\frac{4}{\sqrt{3}})+2b(2+\frac{1}{\sqrt{3}})+a=0$
$\Rightarrow 13+4\sqrt{3}+4b+\frac{2b}{\sqrt{3}}+a=0$
Comparing with $\left(iii\right)$, we get
$4\sqrt{3}+\frac{2b}{\sqrt{3}}=0$
$\Rightarrow b=-6$
Putting $b$ in$\left(iii\right)$
$\Rightarrow a=11$
$\therefore a+b=5$