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Mean Value Theorems — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\log(\text{x}^2+2)-\log3\text{ on }[-1,1]$
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Answer

The given function is $\text{f}(\text{x})=\log(\text{x}^2+2)-\log3,$ which can be rewritten as$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
We know that, logarithmic function is differentiable and so continuous in its domain $\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$ is continuous is $[-1,1]$ and differentiable is $(-1,1).$
Also,
f(1) = f(-1) = 0
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have show that there must exists $\text{c}\in(-1,1)$ such that f'(c) = 0.
We have,
$\text{f}(\text{x})=\log\Big(\frac{\text{x}^2+2}{3}\Big)$
$\text{f}'(\text{x})=\frac{3(2\text{x})}{\text{x}^2+2}=\frac{6\text{x}}{\text{x}^2+2}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{6\text{x}}{\text{x}^2+2}=0$
$\Rightarrow\text{x}=0$
Thus, $\text{c}=0\in(-1,1)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.

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