#### Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 2 sub question (iii) maths textbook solution

$c = \frac{4}{3} \in (1,2)$

Hint:

Using discrimination method,$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given:

$f(x)=(x-1)(x-2)^{2}$on [1,2]

Explanation:

We have,

$f(x)=(x-1)(x-2)^{2}$

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

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

Which exists in (1.2)

$\therefore f(x)$is derivable in (1,2)

3.

$\\ f(1)=(1-1)(1-2)^{2}=0 \\\\ f(2)=(2-1)(2-2)^{2}=0\\\\ \therefore f(1)=f(2)$

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

There exists at least one$c \in(1,2)$ such that$f^{\prime}(c)=0$

Now, $f^{\prime}(c)=0$

\begin{aligned} &=2(c-1)(c-2)+(c-2)^{2} \\ &=2\left(c^{2}-2 c-c+2\right)+c^{2}+4-4 c \\ &=2\left(c^{2}-3 c+2\right)+c^{2}+4-4 c \\ &=2 c^{2}-6 c+2+c^{2}+4-4 c \\ &=3 c^{2}-10 c+8 \end{aligned}

Using discrimination method,

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

$\\c=\frac{-(-10) \pm \sqrt{(-10)^{2}-4 \times 3 \times 8}}{2 \times 3}\\\\ c=\frac{10 \pm \sqrt{100-96}}{6}\\\\ c=\frac{10 \pm \sqrt{4}}{6}=\frac{10 \pm 2}{6}\\\\ =\frac{8}{6}, \frac{12}{6}\\\\ c=\frac{4}{3}, 2\\\\ c=\frac{4}{3} \in(1,2)$

[Neglecting the value 2 ]

Hence, Rolle’s Theorem is verified.