Rolle's theorem hold for the function $f(x)=x^3+b x^2+c x, 1 \leq x \leq 2$ at the point $4 / 3$, the values of b and c are
Answer & explanation
Correct answer: option 2
It is given that the Rolle's theorem holds for $f(x)=x^3+b x^2+c x$ on $[1,2]$
∴ $f(1)=f(2)$ and $f'\left(\frac{4}{3}\right)=0$
$\Rightarrow 1+b+c=8+4 b+2 c$
and, $3\left(\frac{4}{3}\right)^2+2 b\left(\frac{4}{3}\right)+c=0$ $\left[∵ f'(x)=3 x^2+2 b x+c\right]$
$\Rightarrow 3 b+c+7=0$ and $8 b+3 c+16=0$
$\Rightarrow b=-5, c=8$