Practicing Success
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 |
$b=8, c=-5$ $b=-5, c=8$ $b=5, c=-8$ $b=-5, c=-8$ |
$b=-5, c=8$ |
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 $8 b+3 c+16=0$ $\Rightarrow b=-5, c=8$ |