Question

If the function $f\left(x\right)=x^3+b{x}^2+ax+5$ on $[1,3]$ satisfies the conditions of Rolle's Theorem with $c=2+\frac{1}{\sqrt{3}}$, then find $a$ + $b$.

Answer 1

$f\left(x\right)=x^3+b{x}^2+ax+5$

$f\left(1\right)=f\left(3\right)$

$1+b+a+5=27+9b+3a+5$

$a+4b=-13$

$f^{\prime }\left(x\right)=3x^2-2bx+a$

$f^{\prime }\left(c\right)=0$

Now $c=2+\frac{1}{\sqrt{3}}$

$3(2+\frac{1}{\sqrt{3}}{)}^2+b(2+\frac{1}{\sqrt{3}})+a=0$

$13+4\sqrt{3}+2b\frac{(6+\sqrt{3})}{3}+a=0$

solving the above equation

$x=11$

$y=-6$

$x+y=5$