Continuity and Differentiability — Class 12 Maths Solution

(i)f(x) $=$ [x] for x $\in$ [5, 9]
f(x) $=$ [x] in the interval [5, 9] is neither continuous, nor differentiable.

therefore, Mean value theorem is not applicable.

(ii) f(x) $= [x]$ for x $\in [ - 2$, 2]

Again,f(x)$=$ [x] in the interval [$-$2, 2] is neither continuous, nor differentiable.

Hence, mean value theorem is not applicable.

(iii) $f(x) = {x^2} - 1$ for $x \in$ [1, 2]

It is a polynomial. Therefore, it is continuous in the interval [1, 2] and differentiable in the interval (1, 2).

So all conditions of mean value theorem are satisfied. Therefore, there exists at least one $c \in (1,2)$ such that

$f'(c) = \cfrac{{f(2) - f(1)}}{{2 - 1}} \Rightarrow 2c = \cfrac{{3 - 0}}{{2 - 1}} = \cfrac{3}{1}$

As $c = \cfrac{3}{2} \in (1,2)$ ,so mean value theorem is verified.

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