Differentiation (Part-38) | S N De

Differentiation (Part-32) | S N Dey

9. In the mean value theorem, \( f(a+h) = f(a) + h f'(a+\theta h) \) \( (0 < \theta < 1) \), find \( \theta \) when \( f(x) = \sqrt{x} \), \( a = 1 \), and \( h = 3 \).

Solution.

We have:

$$ f(x) = \sqrt{x} = x^{1/2} \quad \rightarrow (1) $$

$$ \therefore f'(x) = \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2\sqrt{x}} \quad \rightarrow (2) $$

Given the mean value relation:

$$ f(a+h) = f(a) + h f'(a+\theta h) $$

Substituting (1) and (2):

$$ \sqrt{a+h} = \sqrt{a} + h \cdot \frac{1}{2\sqrt{a+\theta h}} $$

Now substitute \( a = 1 \) and \( h = 3 \):

$$ \sqrt{1+3} = \sqrt{1} + 3 \cdot \frac{1}{2\sqrt{1+\theta \times 3}} $$

$$ \implies \sqrt{4} = 1 + \frac{3}{2\sqrt{1+3\theta}} $$

$$ \implies 2 - 1 = \frac{3}{2\sqrt{1+3\theta}} \implies 1 = \frac{3}{2\sqrt{1+3\theta}} $$

$$ \implies \frac{2}{3} = \frac{1}{\sqrt{1+3\theta}} $$

Squaring both sides:

$$ \left(\frac{2}{3}\right)^2 = \left(\frac{1}{\sqrt{1+3\theta}}\right)^2 \implies \frac{4}{9} = \frac{1}{1+3\theta} $$

$$ \implies 4(1+3\theta) = 9 \implies 4 + 12\theta = 9 $$

$$ \implies 12\theta = 5 \implies \theta = \frac{5}{12} $$

\( \therefore \theta = \frac{5}{12} \).

10. In the mean value theorem, \( f(x+h) = f(x) + h f'(x+\theta h) \) \( (0 < \theta < 1) \), if \( f(x) = e^x \), then show that the value of \( \theta \) is independent of \( x \).

Solution.

We have:

$$ f(x) = e^x \quad \rightarrow (1) $$

$$ \therefore f'(x) = e^x \quad \rightarrow (2) $$

Applying the mean value relation:

$$ f(x+h) = f(x) + h f'(x+\theta h) $$

$$ \implies e^{x+h} = e^x + h \cdot e^{x+\theta h} $$

$$ \implies e^x \cdot e^h - e^x = h \cdot e^x \cdot e^{\theta h} $$

$$ \implies e^x (e^h - 1) = e^x \cdot h e^{\theta h} $$

Dividing both sides by \( e^x \):

$$ e^h - 1 = h e^{\theta h} \implies e^{\theta h} = \frac{e^h - 1}{h} $$

Taking the natural logarithm (\(\log_e\)) on both sides:

$$ \log_e(e^{\theta h}) = \log_e \left(\frac{e^h - 1}{h}\right) $$

$$ \implies \theta h = \log_e \left(\frac{e^h - 1}{h}\right) $$

$$ \implies \theta = \frac{1}{h} \log_e \left(\frac{e^h - 1}{h}\right) \quad \rightarrow (3) $$

Conclusion: From equation (3), \( \theta \) depends only on \( h \) and is independent of \( x \).

11 (i). Using Lagrange's mean value theorem, prove that \( e^x > 1 + x \).

Solution.

Let \( f(x) = \log_e(1+x) \quad \rightarrow (1) \)

$$ \therefore f'(x) = \frac{1}{1+x} \quad \rightarrow (2) $$

According to Lagrange's mean value theorem:

$$ f(a+h) = f(a) + h f'(a+\theta h) \quad \text{where } 0 < \theta < 1 \quad \rightarrow (3) $$

Substitute \( a = 0 \) and \( h = x \) into equation (3):

$$ f(0+x) = f(0) + x f'(0+\theta x) \implies f(x) = f(0) + x f'(\theta x) $$

$$ \implies \log_e(1+x) = \log_e(1+0) + x \cdot \frac{1}{1+\theta x} $$

$$ \implies \log_e(1+x) = 0 + \frac{x}{1+\theta x} $$

Since \( x > 0 \) and \( 0 < \theta < 1 \), we have \( 1+\theta x > 1 \implies \frac{1}{1+\theta x} < 1 \):

$$ \therefore \log_e(1+x) < x \cdot 1 = x $$

$$ \implies 1+x < e^x \implies e^x > 1+x $$

Hence proved.

11 (ii). Using Lagrange's mean value theorem, prove that \( \log(1+x) < x \) (when \( x > 0 \)).

Solution.

Let \( f(x) = \log_e(1+x) \quad \rightarrow (1) \)

$$ \therefore f'(x) = \frac{1}{1+x} \quad \rightarrow (2) $$

The function \( f(x) \) is continuous in \( [0, x] \) and differentiable in \( (0, x) \). By Lagrange's mean value theorem, there exists at least one value \( c \in (0, x) \) such that:

$$ f'(c) = \frac{f(x) - f(0)}{x - 0} $$

$$ \implies f'(c) = \frac{f(x) - 0}{x} \quad [\because f(0) = \log_e(1) = 0] $$

$$ \implies \frac{1}{1+c} = \frac{\log_e(1+x)}{x} \quad \rightarrow (3) $$

Since \( 0 < c < x \implies c > 0 \):

$$ \implies 1 + c > 1 \implies \frac{1}{1+c} < 1 \quad \rightarrow (4) $$

Combining (3) and (4):

$$ \frac{\log_e(1+x)}{x} < 1 \implies \log_e(1+x) < x \quad (\text{when } x > 0) $$

Hence proved.

11 (iii). Using Lagrange's mean value theorem, prove that \( b^n - a^n < n b^{n-1}(b - a) \) when \( b > a \).

Solution.

Let \( f(x) = x^n \) defined on \( [a, b] \quad \rightarrow (1) \)

$$ \therefore f'(x) = n x^{n-1} \quad \rightarrow (2) $$

The function \( f(x) \) is continuous in \( [a, b] \) and differentiable in \( (a, b) \). By Lagrange's mean value theorem, there exists at least one value \( c \in (a, b) \) such that:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

$$ \implies n c^{n-1} = \frac{b^n - a^n}{b - a} $$

$$ \implies b^n - a^n = n c^{n-1}(b - a) \quad \rightarrow (3) $$

Since \( a < c < b \implies c < b \):

$$ \implies c^{n-1} < b^{n-1} \quad \rightarrow (4) $$

Combining (3) and (4):

$$ b^n - a^n < n b^{n-1}(b - a) \quad (\text{when } b > a) $$

Hence proved.

11 (iv). Using Lagrange's mean value theorem, prove that \( (b-a)\sec^2 a < \tan b - \tan a < (b-a)\sec^2 b \) when \( 0 < a < b < \frac{\pi}{2} \).

Solution.

Let \( f(x) = \tan x \) for \( x \in \left[a, b\right] \subset \left(0, \frac{\pi}{2}\right) \quad \rightarrow (1) \)

$$ \therefore f'(x) = \sec^2 x \quad \rightarrow (2) $$

Since \( f(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \), by Lagrange's mean value theorem, there exists at least one point \( c \in (a, b) \) such that:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

$$ \implies \sec^2 c = \frac{\tan b - \tan a}{b - a} \quad \rightarrow (3) $$

Since \( 0 < a < c < b < \frac{\pi}{2} \), and the function \( \sec^2 x \) is strictly increasing in \( \left(0, \frac{\pi}{2}\right) \):

$$ \sec^2 a < \sec^2 c < \sec^2 b \quad \rightarrow (4) $$

Substituting equation (3) into (4):

$$ \sec^2 a < \frac{\tan b - \tan a}{b - a} < \sec^2 b $$

Multiplying by \( (b - a) \) (since \( b - a > 0 \)):

$$ (b - a)\sec^2 a < \tan b - \tan a < (b - a)\sec^2 b \quad \left(\text{when } 0 < a < b < \frac{\pi}{2}\right) $$

Hence proved.