#### Need solution for RD Sharma math class class 12 chapter Mean value theorem exercise 14.1 question 2 sub question (vi)

$c=\left(4, \frac{4}{3}\right)\in[0,4]$, hence Rolle’s Theorem is verified.

Hint:

Using discrimination method,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given:

$f(x)=x(x-4)^{2}$on [0,4]

Explanation:

We have,

$f(x)=x(x-4)^{2}$

1. Being polynomial f(x) is continuous for all x and hence continuous in [0,4]
2.

$\\f^{\prime}(x)=x \frac{d}{d x}(x-4)^{2}+(x-4)^{2} \frac{d}{d x} x \\\\ x \cdot 2 (x-4)+ ( x-4)^2 (1)$

$\\ = 2 x^{2}-8 x+x^{2}+16-8 x \\ \quad=3 x^{2}-16 x+16 \\$

$\therefore f(x)$ is derivable in (0,4)

1.

$\\(0)=0(0-4)^{2}=0\\\\ f(4)=4(4-4)^{2}=0\\\\ \therefore f(0)=f(4)$

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least one$c \in (0,4)$such that $f' (c ) = 0$

Now, $f' (c ) = 0$

$3 c^{2}-16 c+16$

Using discrimination method,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\\\\\ \begin{array}{l} c=\frac{16 \pm \sqrt{256-192}}{6}=\frac{16 \pm \sqrt{64}}{6} \\\\ c=\frac{16 \pm 8}{6} \\\\ c=\frac{24}{6}, \frac{8}{6} \\\\ c=4, \frac{4}{3} \\\\ c=\left(4, \frac{4}{3}\right) \in[0,4] \end{array}$

Hence, Rolle’s Theorem is verified.