Question

If $f$ is defined in $[1,3]$ by $f\left(x\right)=x^3+b{x}^2+ax$, such that $f\left(1\right)-f\left(3\right)=0$ and $f^{\prime }\left(c\right)=0$ where $c=2+\frac{1}{\sqrt{3}}$, then $(a,b)=$

• A. $(-6,11)$
• B. $(2-\frac{1}{\sqrt{3}},2+\frac{1}{\sqrt{3}})$
• C. $(11,-6)$
• D. $(6,11)$

Answer 1

The correct option is C $(11,-6)$

$f^{{}^{\prime }}\left(x\right)=3x^2+2bx+a$
$f^{{}^{\prime }}\left(c\right)=0$
$3c^2+2bc+a=0$
Irrational roots occur in conjugate pairs.
So, $c=2+\frac{1}{\sqrt{3}}$ and $c=2-\frac{1}{\sqrt{3}}$
Using formula for sum and product of roots , we get
$\Rightarrow a=11,b=-6$