Let for a differentiable function f:(0, ) R, f(x)-f(y) _e (x y )+… - Vidyadip

MCQ

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
$8569$
✓
$2890$
C
$1256$
D
$3564$

✓

Answer

Correct option: B.

$2890$

b
$ 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 $x>y$

$ \lim _{y \rightarrow \mathbb{x}} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \quad \ldots $ $..........(2)$

$ f^1\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$

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