JEE Main 2022 — Mathematics Calculus

Given,

$$\frac{dy}{dx}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$$dxdy​=x−yx+y−2​=(x−1)−(y−1)(x−1)+(y−1)​

Now let $x-1=X,y-1=Y$x−1=X,y−1=Y

So, $\frac{dy}{dx}=\frac{X+Y}{X-Y}........(1)$dxdy​=X−YX+Y​........(1)

Now let $Y=VX\frac{dY}{dX}=V+X\frac{dV}{dX}$Y=VXdXdY​=V+XdXdV​

Putting the value in equation $(1)$(1) we get,

$$V+X\frac{dV}{dX}=\frac{1+V}{1-V}$$V+XdXdV​=1−V1+V​

$$\Rightarrow X\frac{dV}{dX}=\frac{{V}^{2}+1}{1-V}$$⇒XdXdV​=1−VV2+1​

$$\Rightarrow \int \frac{1-V}{1+{V}^{2}}dV=\int \frac{dX}{X}$$⇒∫1+V21−V​dV=∫XdX​

$$\Rightarrow \int \frac{dV}{1+{V}^{2}}-\frac{1}{2}\int \frac{2VdV}{1+{V}^{2}}=\int \frac{dX}{X}$$⇒∫1+V2dV​−21​∫1+V22VdV​=∫XdX​

$$\Rightarrow {\mathrm{tan}}^{-1}V-\frac{1}{2}\mathrm{ln}(1+{V}^{2})=\mathrm{ln}X+c$$⇒tan−1V−21​ln(1+V2)=lnX+c

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{Y}{X})-\frac{1}{2}\mathrm{ln}(1+\frac{{Y}^{2}}{{X}^{2}})=\mathrm{ln}(X)+c$$⇒tan−1(XY​)−21​ln(1+X2Y2​)=ln(X)+c

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{y-1}{x-1})-\frac{1}{2}\mathrm{ln}(1+\frac{{(y-1)}^{2}}{{(x-1)}^{2}})=\mathrm{ln}(x-1)+c$$⇒tan−1(x−1y−1​)−21​ln(1+(x−1)2(y−1)2​)=ln(x−1)+c

Now given curve passes through $(2,1)$(2,1)

So, $0-\frac{1}{2}\mathrm{ln}1=\mathrm{ln}1+c\Rightarrow c=0$0−21​ln1=ln1+c⇒c=0

Now given curve also passes through $(k+1,2)$(k+1,2)

So, ${\mathrm{tan}}^{-1}(\frac{1}{k})-\frac{1}{2}In(1+\frac{1}{{k}^{2}})=\mathrm{ln}k$tan−1(k1​)−21​In(1+k21​)=lnk

$$\Rightarrow 2{\mathrm{tan}}^{-1}(\frac{1}{k})=In(\frac{1+{k}^{2}}{{k}^{2}})+2\mathrm{ln}k$$⇒2tan−1(k1​)=In(k21+k2​)+2lnk

$$\Rightarrow 2{\mathrm{tan}}^{-1}(\frac{1}{k})=In(1+{k}^{2})$$⇒2tan−1(k1​)=In(1+k2)