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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 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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