#### explain solution RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (viii) maths

Answer:   $4 \pm \frac{2}{\sqrt{3}}$

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

Given: $f(x)=(x-1)(x-2)(x-3) \text { on }[0,4]$

Solution:

\begin{aligned} &f(x)=(x-1)(x-2)(x-3) \\ &f(x)=x^{3}-6 x^{2}+11 x-6 \end{aligned}

$f(x)$  is a polynomial function.

It is continuous in [0, 4]

$f^{\prime}(x)=3 x^{2}-12 x+11$

(Which is defined in [0, 4])

$\therefore f(x)$ is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(0)}{4-0} \end{aligned}

\begin{aligned} &3 c^{2}-12 c+11=\frac{(4-1)(4-2)(4-3)-(0-1)(0-2)(0-3)}{4} \\ &3 c^{2}-12 c+11=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4} \end{aligned}

\begin{aligned} &3 c^{2}-12 c+11=\frac{6+6}{4} \\ &3 c^{2}-12 c+11=\frac{12}{4} \end{aligned}

\begin{aligned} &3 c^{2}-12 c+11=3 \\ &3 c^{2}-12 c+11-3=0 \\ &3 c^{2}-12 c+8=0 \end{aligned}

\begin{aligned} &c=\frac{12 \pm \sqrt{144-96}}{6} \\ &c=\frac{12 \pm \sqrt{48}}{6} \end{aligned}

\begin{aligned} &c=\frac{12 \pm 4 \sqrt{3}}{6} \\ &c=4 \pm \frac{2 \sqrt{3}}{3} \\ &c=4 \pm \frac{2}{\sqrt{3}} \end{aligned}

Both values lie between [0, 4]