JEE Main 2024 — Mathematics Calculus

Given: $\frac{dy}{dx}=\frac{x+y-2}{x-y}$dxdy​=x−yx+y−2​

Now, finding the intersection point of $x+y-2=0$x+y−2=0 and $x-y=0$x−y=0.

$$\Rightarrow x+x-2=0,x=y$$⇒x+x−2=0,x=y

$$\Rightarrow x=1,y=1$$⇒x=1,y=1

$$\Rightarrow (x,y)\equiv (1,1)$$⇒(x,y)≡(1,1)

Let, $x=X+1,y=Y+1$x=X+1,y=Y+1

$$\Rightarrow \frac{dY}{dX}=\frac{X+Y}{X-Y}$$⇒dXdY​=X−YX+Y​

Putting, $Y=tX$Y=tX

$$\Rightarrow \frac{dY}{dX}=t+X\frac{dt}{dX}$$⇒dXdY​=t+XdXdt​

$$\Rightarrow \frac{1+t}{1-t}=t+X\frac{dt}{dX}$$⇒1−t1+t​=t+XdXdt​

$$\Rightarrow \frac{1+t-t+{t}^{2}}{1-t}=X\frac{dt}{dX}$$⇒1−t1+t−t+t2​=XdXdt​

$$\Rightarrow \frac{1}{X}dX=\frac{1-t}{1+{t}^{2}}dt$$⇒X1​dX=1+t21−t​dt

$$\Rightarrow \int \frac{1}{X}dX=\int \frac{1-t}{1+{t}^{2}}dt$$⇒∫X1​dX=∫1+t21−t​dt

$$\Rightarrow \mathrm{log}|X|={\mathrm{tan}}^{-1}t-\frac{1}{2}\mathrm{log}|1+{t}^{2}|+C$$⇒log∣X∣=tan−1t−21​log∣1+t2∣+C

$$\Rightarrow \mathrm{log}|x-1|={\mathrm{tan}}^{-1}(\frac{y-1}{x-1})-\frac{1}{2}\mathrm{log}|1+{(\frac{y-1}{x-1})}^{2}|+C$$⇒log∣x−1∣=tan−1(x−1y−1​)−21​log∣1+(x−1y−1​)2∣+C

It is given that the solution curve passes through $(2,1)$(2,1).

$$\Rightarrow \mathrm{log}|2-1|={\mathrm{tan}}^{-1}(\frac{1-1}{2-1})-\frac{1}{2}\mathrm{log}|1+{(\frac{1-1}{2-1})}^{2}|+C$$⇒log∣2−1∣=tan−1(2−11−1​)−21​log∣1+(2−11−1​)2∣+C

$$\Rightarrow 0={\mathrm{tan}}^{-1}(0)-\frac{1}{2}\mathrm{log}|1|+C$$⇒0=tan−1(0)−21​log∣1∣+C

$$\Rightarrow C=0$$⇒C=0

$$\Rightarrow \alpha =1,\beta =2$$⇒α=1,β=2

$\Rightarrow 5\beta +\alpha =11$⇒5β+α=11.