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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Verify the hypothesis and conclusion of Lagrange's mean value theorem for the function
$\text{f}(\text{x})=\frac{1}{4\text{x}-1},1\leq\text{x}\leq4.$
✓

Answer

The given function $\text{f}(\text{x})=\frac{1}{4\text{x}-1}.$

Clearly, f(x) is does not exist for x = 0

Since for each $\text{x}\in[1,4],$ the function attains a unique definite value, f(x) is continuous on [1, 4].

Also, $\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$ exists for all $\text{x}\in[1,4],$

Thus, both the conditions of Lagrange's mean value theorem are verified.

Concequently, there exists some $\text{c}\in[1,4]$ such that

$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$

$=\frac{\text{f}(4)-\text{f}(1)}{3}$

Now,

$\text{f}(\text{x})=\frac{1}{4\text{x}-1}\Rightarrow\text{f}'(\text{x})=\frac{-4}{(4\text{x}-1)^2}$

$\text{f}(4)=\frac{1}{15},\text{f}(1)=\frac{1}{3}$

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$

$\Rightarrow\text{f}'(\text{x})=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}==\frac{-4}{45}$

$\Rightarrow\frac{-4}{(4\text{x}-1)^2}=\frac{-4}{45}$

$\Rightarrow(4\text{x}-1)^2=45$

$\Rightarrow16\text{x}^2-8\text{x}-44=0$

$\Rightarrow4\text{x}^2-2\text{x}-11=0$

$\Rightarrow\text{x}=\frac{1}{4}\big(1+3\sqrt{5}\big)$

Thus, $\text{c}=\frac{1}{4}\big(1+3\sqrt{5}\big)\in(1,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}.$

Hence, Lagrange's theorem is verified.

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