#### provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (ii)

Answer:    $\frac{1}{3}$

Hint:You must know the formula of Lagrange’s Mean Value Theorem.

Given:  $f(x)=x^{3}-2 x^{2}-x+3 \text { on }[0,1]$

Solution:

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

$f(x)$  is a polynomial function.

It is continuous in [0, 1]

$f^{\prime}(x)=3 x^{2}-4 x-1$

(Which is defined in [0, 1])

$\therefore f(x)$ is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \end{aligned}

$3 c^{2}-4 c-1=\frac{\left((1)^{3}-2(1)^{2}-1+3\right)-\left((0)^{3}-2(0)^{2}-0+3\right)}{1}$

\begin{aligned} &3 c^{2}-4 c-1=(1-2-1+3)-(3) \\ &3 c^{2}-4 c-1=1-3 \\ &3 c^{2}-4 c-1=-2 \\ &3 c^{2}-4 c-1+2=0 \\ &3 c^{2}-4 c+1=0 \end{aligned}

\begin{aligned} &3 c^{2}-3 c-c+1=0 \\ &3 c(c-1)-(c-1)=0 \\ &(3 c-1)(c-1)=0 \end{aligned}

Therefore, $c=\frac{1}{3} \text { and } \mathrm{c}=1$