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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 Marks
Question
If the value of c prescribed in Roll's theorem for the function$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$ on the interval $\big[0,2\sqrt3\big]$ is $\frac{3}{4},$ write the value of n (a positive integers).
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Answer

We have,
$\text{f}(\text{x})=2\text{x}(\text{x}-3)^{\text{n}}$
Differentiating the given function with respect to x, we get
$\text{f}'(\text{x})=2\big[\text{xn}(\text{x}-3)^{\text{n}-1}+(\text{x}-3)^{\text{n}}\big]$
$\Rightarrow\text{f}'(\text{x})=2(\text{x}-3)^{\text{n}}\Big[\frac{\text{xn}}{(\text{x}-3)}+1\Big]$
$\Rightarrow\text{f}'(\text{c})=2(\text{c}-3)^{\text{n}}\Big[\frac{\text{cn}}{(\text{c}-3)}+1\Big]$
Given:
$\text{f}'\Big(\frac{3}{4}\Big)=0$
$\therefore\ 2-\Big(\frac{9}{4}\Big)^{\text{n}}\Bigg[\frac{\frac{3}{4}\text{n}}{\big(\frac{-9}{4}\big)}+1\Bigg]=0$
$\Rightarrow2-\Big(\frac{9}{4}\Big)^{\text{n}}\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow\Big[\frac{-\text{n}}{3}+1\Big]=0$
$\Rightarrow-\text{n}+3=0$
$\Rightarrow\text{n}=3$

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