#### Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 11 sub question (i)

$b=-5, c=8$

Hint:

$f(x)=x^{3}+b x^{2}+c x, x \in[1,2]$

Given:

$f(x)=x^{3}+b x^{2}+c x, x \in[1,2], x=\frac{4}{3}$

Explanation:

It is given that the Rolle’s Theorem holds for the function

$f(x)=x^{3}+b x^{2}+c x, x \in[1,2]$

At the point,$\begin{array}{l} x=\frac{4}{3} \end{array}$

We need to find the value of b and c

$f(x)=x^{3}+b x^{2}+c x$

Since it satisfies the Rolle’s Theorem, we have

\begin{aligned} & f(1)=f(2) \\\\ \Rightarrow &(1)^{3}+b(1)^{2}+c(1)=(2)^{3}+b(2)^{2}+c(2) \\\\ \Rightarrow \quad & 1+b+c=8+4 b+2 c \end{aligned}

$\Rightarrow \quad 3 b+c=-7$                                    … (i)

Differentiating the given function, we have

$\begin{array}{l} f^{\prime}(x)=3 x^{2}+2 b x+c \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]\\\\ \Rightarrow f^{\prime}\left(\frac{4}{3}\right)=3\left(\frac{4}{3}\right)^{2}+2 b\left(\frac{4}{3}\right)+c \\\\ \Rightarrow \quad 0=\frac{16}{3}+\frac{8 b}{3}+c \end{array}$          … (ii)

Solving the equation (i) and (ii), we have

$b=-5, c=8$