JEE Main 2024 — Mathematics Calculus

Plotting the line $Y-y={Y}^{'}(x)(X-x)$Y−y=Y′(x)(X−x) we get,

Now, from the above diagram,

The area of the triangle formed will be,

$$A=\frac{1}{2}(\frac{-y}{{Y}^{'}(x)}+x)(y-x{Y}^{'}(x))$$A=21​(Y′(x)−y​+x)(y−xY′(x))

Now, according to the given condition we get,

$$\frac{1}{2}(\frac{-y}{{Y}^{'}(x)}+x)(y-x{Y}^{'}(x))=\frac{-{y}^{2}}{2{Y}^{'}(x)}+1$$21​(Y′(x)−y​+x)(y−xY′(x))=2Y′(x)−y2​+1

$$\Rightarrow (-y+x{Y}^{'}(x))(y-x{Y}^{'}(x))=-{y}^{2}+2{Y}^{'}(x)$$⇒(−y+xY′(x))(y−xY′(x))=−y2+2Y′(x)

$$\Rightarrow -{y}^{2}+xy{Y}^{'}(x)+xy{Y}^{'}(x)-{x}^{2}{[{Y}^{'}(x)]}^{2}=-{y}^{2}+2{Y}^{'}(x)$$⇒−y2+xyY′(x)+xyY′(x)−x2[Y′(x)]2=−y2+2Y′(x)

$$\Rightarrow 2xy-{x}^{2}{Y}^{'}(x)=2$$⇒2xy−x2Y′(x)=2

$$\Rightarrow \frac{dy}{dx}=\frac{2xy-2}{{x}^{2}}$$⇒dxdy​=x22xy−2​

$$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=\frac{-2}{{x}^{2}}$$⇒dxdy​−x2​y=x2−2​

It is a linear differential equation,

Now, finding $\mathrm{IF}={e}^{-2\mathrm{ln}x}=\frac{1}{{x}^{2}}$IF=e−2lnx=x21​

Now, the solution is given by,

$$y\cdot \frac{1}{{x}^{2}}=\frac{2}{3}{x}^{-3}+c$$y⋅x21​=32​x−3+c

Given at $x=1,y=1$x=1,y=1

$$\Rightarrow 1=\frac{2}{3}+c\Rightarrow c=\frac{1}{3}$$⇒1=32​+c⇒c=31​

Hence, the equation of the curve will be,

$$Y=\frac{2}{3}\cdot \frac{1}{X}+\frac{1}{3}{X}^{2}$$Y=32​⋅X1​+31​X2

$$\Rightarrow 12Y(2)=\frac{5}{3}\times 12=20$$⇒12Y(2)=35​×12=20