JEE Main 2023 — Mathematics Calculus

Given,

$${x}^{2}f(x)-x=4{\int }_{0}^{x}tf(t)dt$$x2f(x)−x=4∫0x​tf(t)dt

Now Differentiating both side w.r.t. $x$x we get,

$${x}^{2}{f}^{'}(x)+2xf(x)-1=4xf(x)$$x2f′(x)+2xf(x)−1=4xf(x)

$$\Rightarrow {x}^{2}{f}^{'}(x)-1=2xf(x)$$⇒x2f′(x)−1=2xf(x)

$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=\frac{1}{{x}^{2}}$⇒dxdy​−x2​y=x21​ $(\mathrm{Let}y=f(x)\frac{dy}{dx}={f}^{'}(x))$(Lety=f(x)dxdy​=f′(x))

Which is a linear differential equation,

So, $\displaystyle I.F={e}^{\int -\frac{2}{x}dx}={e}^{-2\mathrm{ln}x}=\frac{1}{{x}^{2}}$I.F=e∫−x2​dx=e−2lnx=x21​

Now solution of the differential equation is given by,

$$\frac{y}{{x}^{2}}=\int \frac{1}{{x}^{4}}dx+C$$x2y​=∫x41​dx+C

$$\Rightarrow \frac{y}{{x}^{2}}=-\frac{1}{3{x}^{3}}+C$$⇒x2y​=−3x31​+C

Now given, $f(1)=\frac{2}{3}$f(1)=32​

So, $\frac{2}{3}=-\frac{1}{3}+C$32​=−31​+C

$$\Rightarrow C=1$$⇒C=1

Hence, the function will be,

$$f(x)=-\frac{1}{3x}+{x}^{2}$$f(x)=−3x1​+x2

So, the required value is given by,

$$18f(3)=18[-\frac{1}{9}+9]=160$$18f(3)=18[−91​+9]=160