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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify mean value theorem for the function:
$\text{f(x)}=\frac{1}{4\text{x}-1}\text{ in }[1,4].$
✓

Answer

We have, $\text{f(x)}=\frac{1}{4\text{x}-1}\text{ in }[1,4]$
f(x) is continuous in [1, 4].
Also, at $\text{x}=\frac{1}{4}.$ f(x) is discontinuous.
Hence, f(x) is continuous in [1, 4].
$\text{f}'(\text{x})=-\frac{4}{(4\text{x}-1)^2},$ which exists in (1, 4).
Since, conditions of mean value theorem are satisfied.
Hence, there exists a real number $\text{c}\in[1,4]$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(1)}{4-1}$
$\Rightarrow\ \frac{-4}{(4\text{c}-1)^2}=\frac{\frac{1}{16-1}-\frac{1}{4-1}}{4-1}=\frac{\frac{1}{15}-\frac{1}{3}}{3}$
$\Rightarrow\ \frac{-4}{(4\text{c}-1)^2}=\frac{1-5}{45}=\frac{-4}{45}$
$\Rightarrow\ (4\text{c}-1)^2=45$
$\Rightarrow\ 4\text{c}-1=\pm3\sqrt{5}$
$\Rightarrow\ \text{c}=\frac{3\sqrt{5}+1}{4}\in(1,4)$ [neglecting (-ve) value]
Hence, mean value theorem has been verified.

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