JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let for a differentiable function \(f:(0, \infty) \rightarrow R\), \(f(x)-f(y) \geq \log _e\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, \infty) \text {. }\) Then \(\sum_{\mathrm{n}=1}^{20} \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{n}^2}\right)\) is equal to
- A \(8569\)
- B \(2890\)
- C \(1256\)
- D \(3564\)
Answer & Solution
Correct Answer
(B) \(2890\)
Step-by-step Solution
Detailed explanation
\( f(x)-f(y) \geq \ln x-\ln y+x-y \) \( \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1 \) Let \(x>y \) \(\lim _{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1 \quad \ldots\) \( ........(1) \) Let \(x>y\)…
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