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

$c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2]$, Hence Rolle’s Theorem is verified.

Hint:

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

Given:

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

Explanation:

We have,

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

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

$\\f^{\prime}(x) =\left(x^{2}-1\right) \frac{d}{d x}(x-2)+(x-2) \frac{d}{d x}\left(x^{2}-1\right) \\\\ =\left(x^{2}-1\right)(1)+(x-2)(2 x) \\\\ =x^{2}-1+2 x^{2}-4 x \\\\ =3 x^{2}-4 x-1 \\\\$

3.

$\\ f(-1)=\left((-1)^{2}-1\right)(-1-2)=(1-1)(-1-2)=0\\\\ f(2)=\left(2^{2}-1\right)(2-2)=(4-1)(0)=0 \\\\ \therefore f(-1)=f(2)$

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

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

Using discrimination method,

$\begin{array}{l} \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\\\ c=\frac{4 \pm \sqrt{16+12}}{6}=\frac{4 \pm \sqrt{28}}{6} \\\\ c=\frac{4 \pm 2 \sqrt{7}}{6} \\\\ c=\frac{2(2 \pm \sqrt{7})}{6} \\\\ c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2] \end{array}$

Hence, Rolle’s Theorem is verified.