JEE Main 2024 — Mathematics Calculus

Given: $f(x)-f(y)\geq {\mathrm{log}}_{e}(\frac{x}{y})+x-y$f(x)−f(y)≥loge​(yx​)+x−y

Now, taking x=x+h&y=x we get,

$$f(x+h)-f(x)\geq {\mathrm{log}}_{e}(\frac{x+h}{x})+x+h-x$$f(x+h)−f(x)≥loge​(xx+h​)+x+h−x

$$f(x+h)-f(x)\geq {\mathrm{log}}_{e}(\frac{x+h}{x})+h$$f(x+h)−f(x)≥loge​(xx+h​)+h

Now, using ${f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(x+h)-f(x)}{h}$f′(x)=h→0lim​hf(x+h)−f(x)​

$$\Rightarrow {f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{log}}_{e}(\frac{x+h}{x})+h}{h}$$⇒f′(x)=h→0lim​hloge​(xx+h​)+h​

$$\Rightarrow {f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{log}}_{e}(1+\frac{h}{x})}{x\cdot (\frac{h}{x})}+1$$⇒f′(x)=h→0lim​x⋅(xh​)loge​(1+xh​)​+1

$$\Rightarrow {f}^{'}(x)=\frac{1}{x}+1$$⇒f′(x)=x1​+1

$$\Rightarrow {f}^{'}(\frac{1}{{x}^{2}})={x}^{2}+1$$⇒f′(x21​)=x2+1

$$\Rightarrow \sum _{x=1}^{20}{f}^{'}(\frac{1}{{x}^{2}})=\sum _{x=1}^{20}({x}^{2}+1)$$⇒x=1∑20​f′(x21​)=x=1∑20​(x2+1)

$$\Rightarrow \sum _{x=1}^{20}{f}^{'}(\frac{1}{{x}^{2}})=\sum _{x=1}^{20}{x}^{2}+20$$⇒x=1∑20​f′(x21​)=x=1∑20​x2+20

$$\Rightarrow \sum _{x=1}^{20}{f}^{'}(\frac{1}{{x}^{2}})=\frac{20\times 21\times 41}{6}+20$$⇒x=1∑20​f′(x21​)=620×21×41​+20

$$\Rightarrow \sum _{x=1}^{20}{f}^{'}(\frac{1}{{x}^{2}})=2870+20$$⇒x=1∑20​f′(x21​)=2870+20

$$\Rightarrow \sum _{x=1}^{20}{f}^{'}(\frac{1}{{x}^{2}})=2890$$⇒x=1∑20​f′(x21​)=2890