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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Differentiation

JEE Advanced · Mathematics · Differentiation

If f(x)=(x-1)^ 100 (x-2)^ 2(99) (x-3)^ 3(98) . .(x-n)^ 100 , let k=...

If f(x)=(x−1)100(x−2)2(99)(x−3)3(98)…..(x−n)100f(x)=(x-1)^{100}(x-2)^{2(99)}(x-3)^{3(98)} \ldots . .(x-n)^{100}, let k=f′(101)f(101)k=\frac{f^{\prime}(101)}{f(101)}, then (k50−97)\left(\frac{k}{50}-97\right) is ______

Correct answer

4

Step-by-step solution

f(x)=∏i=1100(x−i)i(101−i)f(x)=\prod_{i=1}^{100}(x-i)^{i(101-i)} ln⁡(f(x))=∑i=1100i(101−i)ln⁡(x−i)\ln (f(x))=\sum_{i=1}^{100} i(101-i) \ln (x-i) ⇒1f(x)⋅f′(x)=∑i=1100i(101−i)(x−i)\Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)=\sum_{i=1}^{100} \frac{i(101-i)}{(x-i)}
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