उत्तर (2890)

समाधान

$f(x)-f(y) \geq \ln x-\ln y+x-y$

$\dfrac{f(x)-f(y)}{x-y} \geq \dfrac{\ln x-\ln y}{x-y}+1 \ldots (1)$

मान लीजिए $x>y$

$\lim _{y \rightarrow x} \dfrac{f(x)-f(y)}{x-y} \geq \lim _{y \rightarrow x} \dfrac{\ln x-\ln y}{x-y} +1 $

$\Rightarrow f^{\prime}\left(x^{-}\right) \geq \dfrac{1}{x}+1 $

मान लीजिए $x<y$

$\lim _{y \rightarrow x} \dfrac{f(x)-f(y)}{x-y} \leq\lim _{y \rightarrow x} \dfrac{\ln x-\ln y}{x-y}+1 $

$\Rightarrow f^{\prime}\left(x^{+}\right) \leq \dfrac{1}{x}+1 \ldots (2)$

समीकरण (1) और (2) से

$f^{\prime}\left(x^{-}\right)=f^{\prime}\left(x^{+}\right)$

$f^{\prime}(x)=\dfrac{1}{x}+1$

$f^{\prime}\left(\dfrac{1}{x^{2}}\right)=x^{2}+1$

$\sum _{x=1}^{20}\left(x^{2}+1\right)=\sum _{x=1}^{20} x^{2}+20$

$=\dfrac{20 \times 21 \times 41}{6}+20$

$=2890$