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Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

117. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

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alok kumar singh | Contributor-Level 10

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117. Solution :
Mean Value Theorem states that for a function f[a,b] →R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that $f' (c) = \frac{f (b) - f (a)}{b - a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5,

...more

117. Solution :
Mean Value Theorem states that for a function f[a,b] →R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that $f' (c) = \frac{f (b) - f (a)}{b - a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5, 9).

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - 1 - n}{h} = \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$ .

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

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

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

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f (x) = x^{2} - 1$ for $x \in [1, 2]$

It can be proved as follows.

$\begin{array}{l} f (1) = 1^{2} - 1 = 0, f (2) = 2^{2} - 1 = 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (2) - f (1)}{2 - 1} = \frac{3 - 0}{1} = 3 \\ f' (x) = 2 x \\ ∴ f' (c) = 3 \\ \Rightarrow 2 c = 3 \\ \Rightarrow c = \frac{3}{2} = 1.5, w h e r e, 1.5 \in (1, 2) \end{array}$

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117. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

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