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Given n is a positive integer for the function ${⨍ (x)=5x(x-2)}^{n},x∈ [0, 2],$ by Rolle’s
Theorem, the value of c is $\frac{1}{5}$ then n is equal to

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Detailed Solution

${⨍ (x)=5x(x-2)}^{n}$

For Rolle's Theorem in $[0, 2]$

$⨍^{'} (\frac{1}{5}) =0$

$\Rightarrow {[{(x - 2)}^{n} + nx {(x - 2)}^{n - 1}]}_{x = \frac{1}{5}} = 0 \Rightarrow {[(x - 2) + nx]}_{x = \frac{1}{5}} = 0 \Rightarrow (\frac{1}{5} - 2) + \frac{1}{5} n = 0 \Rightarrow n = 9$

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