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

$c=\left(1, \frac{1}{3}\right) \in(0,1)$, hence Rolle’s Theorem is verified.

Hint:

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

Given:

$f(x)=x(x-1)^{2}$on  [0,1]

Explanation:

We have,

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

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

2.

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

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

3.

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

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

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

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

$3 c^{2}-4 c+1$

Using discrimination method,

\\ \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\\\\\ \begin{aligned} c &=\frac{-(-4) \pm \sqrt{16-12}}{6} \\\\ c &=\frac{4 \pm \sqrt{4}}{6} \\\\ c &=\frac{4 \pm 2}{6} \\\\ &=\frac{6}{6}, \frac{2}{6} \\\\ c &=1, \frac{1}{3} \\\\ c &=\left(1, \frac{1}{3}\right) \in(0,1) \end{aligned}

Hence, Rolle’s Theorem is verified.