JEE Main 2024 — Mathematics Calculus

Given,

$$f(1)=1,f(a)=0$$f(1)=1,f(a)=0

And $f(x)=\sqrt{\underset{r\rightarrow x}{\mathrm{lim}}{\frac{2{r}^{2}[(f(r){)}^{2}-f(x)f(r)]}{{r}^{2}-{x}^{2}}-{r}^{3}{e}^{\frac{f(r)}{r}}}}$f(x)=r→xlim​r2−x22r2[(f(r))2−f(x)f(r)]​−r3erf(r)​​

$$\Rightarrow {f}^{2}(x)=\underset{r\rightarrow x}{\mathrm{lim}}(\frac{2{r}^{2}({f}^{2}(r)-f(x)f(r))}{{r}^{2}-{x}^{2}}-{r}^{3}{e}^{\frac{f(r)}{r}})$$⇒f2(x)=r→xlim​(r2−x22r2(f2(r)−f(x)f(r))​−r3erf(r)​)

$$\Rightarrow {f}^{2}(x)=\underset{r\rightarrow x}{\mathrm{lim}}(\frac{2{r}^{2}f(r)}{r+x}\frac{(f(r)-f(x))}{r-x}-{r}^{3}{e}^{\frac{f(r)}{r}})$$⇒f2(x)=r→xlim​(r+x2r2f(r)​r−x(f(r)−f(x))​−r3erf(r)​)

$$\Rightarrow {f}^{2}(x)=\frac{2{x}^{2}f(x)}{x+x}(\underset{r\rightarrow x}{\mathrm{lim}}\frac{(f(r)-f(x))}{r-x})-{x}^{3}{e}^{\frac{f(x)}{x}}$$⇒f2(x)=x+x2x2f(x)​(r→xlim​r−x(f(r)−f(x))​)−x3exf(x)​

$$\Rightarrow {f}^{2}(x)=\frac{2{x}^{2}f(x)}{2x}{f}^{'}(x)-{x}^{3}{e}^{\frac{f(x)}{x}}$$⇒f2(x)=2x2x2f(x)​f′(x)−x3exf(x)​

Now, taking $f(x)=y$f(x)=y we get,

$$\Rightarrow {y}^{2}=xy\frac{dy}{dx}-{x}^{3}{e}^{\frac{y}{x}}$$⇒y2=xydxdy​−x3exy​

$$\Rightarrow \frac{y}{x}=\frac{dy}{dx}-\frac{{x}^{2}}{y}{e}^{\frac{y}{x}}$$⇒xy​=dxdy​−yx2​exy​

Now, let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$y=vx⇒dxdy​=v+xdxdv​

$$\Rightarrow v=v+x\frac{dv}{dx}-\frac{x}{v}{e}^{v}$$⇒v=v+xdxdv​−vx​ev

$$\Rightarrow \frac{dv}{dx}=\frac{{e}^{v}}{v}$$⇒dxdv​=vev​

$$\Rightarrow {e}^{-v}vdv=dx$$⇒e−vvdv=dx

Integrating both side

$${e}^{v}(x+c)+1+v=0$$ev(x+c)+1+v=0

Now using $f(1)=1\Rightarrow x=1,y=1$f(1)=1⇒x=1,y=1 we get,

$$\Rightarrow c=-1-\frac{2}{e}$$⇒c=−1−e2​

Hence, ${e}^{v}(-1-\frac{2}{e}+x)+1+v=0$ev(−1−e2​+x)+1+v=0

$$\Rightarrow {e}^{\frac{y}{x}}(-1-\frac{2}{e}+x)+1+\frac{y}{x}=0$$⇒exy​(−1−e2​+x)+1+xy​=0

Now taking, x=a&y=0 we get,

$${e}^{0}(-1-\frac{2}{e}+a)+1+0=0$$e0(−1−e2​+a)+1+0=0

$$\Rightarrow a=\frac{2}{e}$$⇒a=e2​

$$\Rightarrow ae=2$$⇒ae=2