It is given that the Rolle's theorem holds for the function

Solution:

As, the Rolle's theorem holds for the function $f(x)=x^{3}+b x^{2}+c x, x \in[1,2]$ at the point $x=\frac{4}{3}$

So, $f(1)=f(2)$

$\Rightarrow(1)^{3}+b(1)^{2}+c(1)=(2)^{3}+b(2)^{2}+c(2)$

$\Rightarrow 1+b+c=8+4 b+2 c$

$\Rightarrow 3 b+c+7=0$           ......(1)

And $f^{\prime}\left(\frac{4}{3}\right)=0$

$\Rightarrow 3\left(\frac{4}{3}\right)^{2}+2 b\left(\frac{4}{3}\right)+c=0$                   $\left[\right.$ As, $\left.f^{\prime}(x)=3 x^{2}+2 b x+c\right]$

$\Rightarrow \frac{16}{3}+\frac{8 b}{3}+c=0$

$\Rightarrow 8 b+3 c+16=0$                      .......(2)

$($ ii $)-($ i $) \times 3$, we ge

$8 b-9 b+16-21=0$

$\Rightarrow-b-5=0$

$\Rightarrow b=-5$

Substituting $b=-5$ in (i), we get

$3(-5)+c+7=0$

$\Rightarrow-15+c+7=0$

$\Rightarrow c=8$