JEE Main 2024 — Mathematics Calculus

Given: $(x\mathrm{cos}(\frac{y}{x}))\frac{dy}{dx}=y\mathrm{cos}(\frac{y}{x})+x$(xcos(xy​))dxdy​=ycos(xy​)+x

$$\Rightarrow \mathrm{cos}(\frac{y}{x})\frac{dy}{dx}=\frac{y}{x}\mathrm{cos}(\frac{y}{x})+1$$⇒cos(xy​)dxdy​=xy​cos(xy​)+1

Putting, $y=vx$y=vx

$$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$⇒dxdy​=v+xdxdv​

$$\Rightarrow \mathrm{cos}v(v+x\frac{dv}{dx})=v\mathrm{cos}v+1$$⇒cosv(v+xdxdv​)=vcosv+1

$$\Rightarrow v+x\frac{dv}{dx}=v+\frac{1}{\mathrm{cos}v}$$⇒v+xdxdv​=v+cosv1​

$$\Rightarrow x\frac{dv}{dx}=\frac{1}{\mathrm{cos}v}$$⇒xdxdv​=cosv1​

$$\Rightarrow (\mathrm{cos}v)dv=\frac{dx}{x}$$⇒(cosv)dv=xdx​

$$\Rightarrow \int (\mathrm{cos}v)dv=\int \frac{dx}{x}$$⇒∫(cosv)dv=∫xdx​

$$\Rightarrow \mathrm{sin}v=\mathrm{log}|x|+c$$⇒sinv=log∣x∣+c

$$\Rightarrow \mathrm{sin}(\frac{y}{x})=\mathrm{log}|x|+c$$⇒sin(xy​)=log∣x∣+c

Now, $f(1)=\frac{\pi }{3}$f(1)=3π​

$$\Rightarrow \mathrm{sin}(\frac{\pi }{3})=\mathrm{log}|1|+c$$⇒sin(3π​)=log∣1∣+c

$$\Rightarrow \frac{\sqrt{3}}{2}=c$$⇒23​​=c

$$\Rightarrow \mathrm{sin}(\frac{y}{x})=\mathrm{log}|x|+\frac{\sqrt{3}}{2}$$⇒sin(xy​)=log∣x∣+23​​

So, on comparing we get,

$$\Rightarrow \alpha =\sqrt{3}$$⇒α=3​

$$\Rightarrow {\alpha }^{2}=3$$⇒α2=3