JEE Main 2024 — Mathematics Calculus

Given: $(1+{y}^{2})(1+{\mathrm{log}}_{e}x)dx+xdy=0$(1+y2)(1+loge​x)dx+xdy=0

$$\Rightarrow \frac{(1+{\mathrm{log}}_{e}x)}{x}dx=\frac{-1}{(1+{y}^{2})}dy$$⇒x(1+loge​x)​dx=(1+y2)−1​dy

$$\Rightarrow \int (\frac{1}{x}+\frac{\mathrm{log}x}{x})dx+\int \frac{dy}{1+{y}^{2}}=0$$⇒∫(x1​+xlogx​)dx+∫1+y2dy​=0

$$\Rightarrow \mathrm{log}x+\frac{(\mathrm{log}x{)}^{2}}{2}+{\mathrm{tan}}^{-1}y=C$$⇒logx+2(logx)2​+tan−1y=C

This curve passes through the point $(1,1)$(1,1).

$$\Rightarrow \mathrm{log}(1)+\frac{(\mathrm{log}(1){)}^{2}}{2}+{\mathrm{tan}}^{-1}(1)=C$$⇒log(1)+2(log(1))2​+tan−1(1)=C

$$\Rightarrow C=\frac{\pi }{4}$$⇒C=4π​

$$\Rightarrow \mathrm{log}x+\frac{(\mathrm{log}x{)}^{2}}{2}+{\mathrm{tan}}^{-1}y=\frac{\pi }{4}$$⇒logx+2(logx)2​+tan−1y=4π​

Now, putting $x=e$x=e

$$\Rightarrow \mathrm{log}e+\frac{(\mathrm{log}e{)}^{2}}{2}+{\mathrm{tan}}^{-1}y=\frac{\pi }{4}$$⇒loge+2(loge)2​+tan−1y=4π​

$$\Rightarrow {\mathrm{tan}}^{-1}y=\frac{\pi }{4}-\frac{3}{2}$$⇒tan−1y=4π​−23​

$$\Rightarrow y=\mathrm{tan}(\frac{\pi }{4}-\frac{3}{2})$$⇒y=tan(4π​−23​)

$$\Rightarrow y=\frac{\mathrm{tan}\frac{\pi }{4}-\mathrm{tan}(\frac{3}{2})}{1+\mathrm{tan}\frac{\pi }{4}\mathrm{tan}(\frac{3}{2})}$$⇒y=1+tan4π​tan(23​)tan4π​−tan(23​)​

$$\Rightarrow y=\frac{1-\mathrm{tan}(\frac{3}{2})}{1+\mathrm{tan}(\frac{3}{2})}$$⇒y=1+tan(23​)1−tan(23​)​

Hence, on comparing we get,

$$\Rightarrow \alpha =\beta =1$$⇒α=β=1

$$\Rightarrow \alpha +2\beta =3$$⇒α+2β=3