Question

Rolle's theorem hold for the function $f\left(x\right)=x^3+b{x}^2+cx,1\le x\le 2$ at the point $4/3$, the values of b and c are

• A. $b=8,c=-5$
• B. $b=-5,c=8$
• C. $b=5,c=-8$
• D. $b=-5,c=-8$

Answer 1

The correct option is B $b=-5,c=8$

$f\left(x\right)=x^3+b{x}^2+cx,1\le x\le 2$
$f^{\prime }\left(x\right)=3x^2+2bx+c$
Since, Rolle's theorem holds for $c=\frac{4}{3}$
$\Rightarrow f^{\prime }\left(\frac{3}{4}\right)=0$
$\Rightarrow 8b+3c=-16$ .....(i)
Also, $f\left(1\right)=f\left(2\right)$
$\Rightarrow 3b+c=-7$ ......(ii)
$\Rightarrow b=-5,c=8$