JEE Mains · Maths · STD 12 - 9. differential equations
Let \(f(x)\) be a real differentiable function such that \(f(0)=1\) and \(f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)\) for all \(x, y \in \mathbf{R}\). Then \(\sum_{\mathrm{n}=1}^{100} \log _{\mathrm{e}} f(\mathrm{n})\) is equal to :
- A \(2525\)
- B \(5220\)
- C \(2384\)
- D \(2406\)
Answer & Solution
Correct Answer
(A) \(2525\)
Step-by-step Solution
Detailed explanation
\(\because f(x+y)=f(x) \cdot f(y)+f(x) \cdot f(y), \forall x, y \in R\) ....(i) And \(f(0)=1\) ....(ii) Now replace \(x\) by zero and \(y\) by zero we get…
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