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Question
If the value of c prescribed in Rolle's theorem for the function f (x) = 2x (x − 3)n on the interval \[[0, 2\sqrt{3}] \text { is } \frac{3}{4},\] write the value of n (a positive integer) ?
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Solution
We have
Differentiating the given function with respect to x, we get
\[f'\left( x \right) = 2\left[ xn \left( x - 3 \right)^{n - 1} + \left( x - 3 \right)^n \right]\]
\[ \Rightarrow f'\left( x \right) = 2 \left( x - 3 \right)^n \left[ \frac{xn}{\left( x - 3 \right)} + 1 \right]\]
\[ \Rightarrow f'\left( c \right) = 2 \left( c - 3 \right)^n \left[ \frac{cn}{\left( c - 3 \right)} + 1 \right]\]
Given:
\[\therefore 2 \left( \frac{- 9}{4} \right)^n \left[ \frac{\frac{3}{4}n}{\left( \frac{- 9}{4} \right)} + 1 \right] = 0\]
\[ \Rightarrow 2 \left( \frac{- 9}{4} \right)^n \left[ \frac{- n}{3} + 1 \right] = 0\]
\[ \Rightarrow \left[ \frac{- n}{3} + 1 \right] = 0\]
\[ \Rightarrow - n + 3 = 0\]
\[ \Rightarrow n = 3\]
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