#### Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 1 sub question 1 maths textbook solution

Rolle’s Theorem is not applicable

Hint:

$f (x)$is continuous for al lx and hence continuous in$[1,3]$.

Given:

$f (x) = 3 + ( x -2 ) ^ \frac{2}{3}$on$[1,3]$

Explanation:

We have

$f (x) = 3 + ( x -2 ) ^ \frac{2}{3}$                                                                                                                         …(i)

1. Being polynomial$f (x)$is continuous for all x and hence continuous in$[1,3]$.

2.

$\\f{}' (x) = \frac{2}{3} ( x -2) ^{\frac{2}{3}-1} \\\\ = \frac{2}{3} (x-2 ) ^{\frac{-1}{3}}$

Which exists in $(1,3)$

$f (x)$is derivable in$(1,3)$

3.

$\\f (1) = 3 + ( 1-2) ^{\frac{2}{3}} = 3 + (-1)\frac{2}{3}\\\\ = 3-1 = 2 \\\\ f (3) = 3 + ( 3-2)^\frac{2}{3} = 3 + ( 1) ^\frac{2}{3} \\\\ = 3+1 = 4 \\\\ f (1) \neq f (3 )$

Thus, third condition of Rolle’s Theorem is not satisfied.

Hence, Rolle’s Theorem is not applicable.