Differentiation Question 2
Question 2 - 2024 (27 Jan Shift 1)
Let for a differentiable function $f:(0, \infty) \rightarrow R, \mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y}) \geq \log {\mathrm{e}}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{x}-\mathrm{y}, \forall \mathrm{x}, \mathrm{y} \in(0, \infty)$. Then $\sum{\mathrm{n}=1}^{20} \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{n}^{2}}\right)$ is equal to
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Answer (2890)
Solution
$f(x)-f(y) \geq \ln x-\ln y+x-y$
$\frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y})}{\mathrm{x}-\mathrm{y}} \geq \frac{\ln \mathrm{x}-\ln \mathrm{y}}{\mathrm{x}-\mathrm{y}}+1$
Let $x>y$
$\lim _{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1 \ldots(1)$
Let $\mathrm{x}<\mathrm{y}$
$\lim _{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1$..
$\mathrm{f}^{\mathrm{l}}\left(\mathrm{x}^{-}\right)=\mathrm{f}^{1}\left(\mathrm{x}^{+}\right)$
$\mathrm{f}^{1}(\mathrm{x})=\frac{1}{\mathrm{x}}+1$
$\mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}^{2}}\right)=\mathrm{x}^{2}+1$
$\sum_{\mathrm{x}=1}^{20}\left(\mathrm{x}^{2}+1\right)=\sum_{\mathrm{x}=1}^{20} \mathrm{x}^{2}+20$
$=\frac{20 \times 21 \times 41}{6}+20$
$=2890$
