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  • If <img alt="f(x)=(x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \ldots(x-100)^{100}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%28x-1%29%5E%7B100%7D%28x-2%29%5E%7B2%2899%29%7D%28

If f(x)=(x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \ldots(x-100)^{100} then the value of \frac{f^{\prime}(101)}{f(101)} is 

 

Option: 1

5050


Option: 2

2575


Option: 3

3030


Option: 4

1250


Answers (1)

best_answer

f(x)=\prod_{i=1}^{100}(x-i)^{i(101-i)}

Taking log both sides, we get

\log (f(x))=\sum_{i=1}^{100} i(101-i) \log (x-i)

Differentiating w.r.t. x, we get

\begin{gathered} \frac{1}{f(x)} \cdot f^{\prime}(x)=\sum_{i=1}^{100} \frac{i(101-i)}{(x-i)} \\ \\\Rightarrow \quad \frac{f^{\prime}(101)}{f(101)}=\sum_{i=1}^{100} i\left(\frac{101-i}{101-i}\right)=5050 \end{gathered}

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Divya Prakash Singh

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