JEE Main 2016 — Mathematics Calculus

Let $L= \underset{t\rightarrow x}{\mathrm{lim}}⁡\frac{{t}^{2}f(x)-{x}^{2}f(t)}{t-x}=1$L=t→xlim​⁡t−xt2f(x)−x2f(t)​=1

Applying L'Hospital Rule

$$L= \underset{t\rightarrow x}{\mathrm{lim}}⁡\frac{2t f(x)-{x}^{2}{f}^{'}(t)}{1}=1$$L=t→xlim​⁡12tf(x)−x2f′(t)​=1

Or $2x f(x)-{x}^{2}{f}^{'}(x)=1$2xf(x)−x2f′(x)=1 $\Rightarrow \frac{dy}{dx}+(-\frac{2}{x})y=-\frac{1}{{x}^{2}}$⇒dxdy​+(−x2​)y=−x21​..........$(1)$(1)

Solving the above linear differential equation we get

Integrating factor $\displaystyle ={e}^{\int -\frac{2}{x}dx}={e}^{-2\mathrm{ln}x}=\frac{1}{{x}^{2}}$=e∫−x2​dx=e−2lnx=x21​

After multiplying the equation $(1)$(1) by $\frac{1}{{x}^{2}}$x21​, and simplifying the equation $(1)$(1) becomes,

$$\frac{d}{dx}(\frac{y}{{x}^{2}})=-\frac{1}{{x}^{4}}$$dxd​(x2y​)=−x41​

$$\Rightarrow \frac{y}{{x}^{2}}=\int -\frac{1}{{x}^{4}}dx=\frac{1}{3{x}^{3}}+c$$⇒x2y​=∫−x41​dx=3x31​+c

$$\Rightarrow y=\frac{1}{3x}+c{x}^{2}$$⇒y=3x1​+cx2

Putting $x=1,y=1$x=1,y=1, we get

$c=\frac{2}{3}$c=32​.

$$\therefore f(x)=\frac{2}{3}{x}^{2}+\frac{1}{3x}$$∴f(x)=32​x2+3x1​

Put $x=\frac{3}{2}$x=23​

$$f(\frac{3}{2})=\frac{2}{3} \times {(\frac{3}{2})}^{2}+\frac{2}{3\times 3}$$f(23​)=32​×(23​)2+3×32​

$=\frac{3}{2}+\frac{2}{9}=\frac{27+4}{18}=\frac{31}{18}$=23​+92​=1827+4​=1831​.