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