JEE Main 2026 — Mathematics Calculus

Given the limit $\displaystyle \lim_{t \to x} \frac{t^2 y(x) - x^2 y(t)}{x - t} = 3$t→xlim​x−tt2y(x)−x2y(t)​=3.

The limit is in $\frac{0}{0}$00​ form. Applying L'Hopital's Rule with respect to $t$t:

$$\lim_{t \to x} \frac{\frac{d}{dt}(t^2 y(x) - x^2 y(t))}{\frac{d}{dt}(x - t)} = 3$$t→xlim​dtd​(x−t)dtd​(t2y(x)−x2y(t))​=3

$$\lim_{t \to x} \frac{2t y(x) - x^2 y'(t)}{-1} = 3$$t→xlim​−12ty(x)−x2y′(t)​=3

Substituting $t = x$t=x:

$$\frac{2x y(x) - x^2 y'(x)}{-1} = 3$$−12xy(x)−x2y′(x)​=3

$$x^2 y'(x) - 2x y(x) = 3$$x2y′(x)−2xy(x)=3

Dividing by $x^4$x4 to make it a linear differential equation or recognizing the quotient rule form:

$$\frac{x^2 y'(x) - 2x y(x)}{x^4} = \frac{3}{x^4}$$x4x2y′(x)−2xy(x)​=x43​

$$\frac{d}{dx} \left( \frac{y(x)}{x^2} \right) = \frac{3}{x^4}$$dxd​(x2y(x)​)=x43​

Integrating both sides with respect to $x$x:

$$\frac{y(x)}{x^2} = \int 3x^{-4} dx = \frac{3x^{-3}}{-3} + C = -\frac{1}{x^3} + C$$x2y(x)​=∫3x−4dx=−33x−3​+C=−x31​+C

Given $y(1) = 2$y(1)=2:

$$\frac{2}{1^2} = -\frac{1}{1^3} + C \implies 2 = -1 + C \implies C = 3$$122​=−131​+C⟹2=−1+C⟹C=3

So, $\frac{y(x)}{x^2} = -\frac{1}{x^3} + 3$x2y(x)​=−x31​+3

$$y(x) = -\frac{1}{x} + 3x^2$$y(x)=−x1​+3x2

We need to find $2y(2)$2y(2):

$$y(2) = -\frac{1}{2} + 3(2^2) = -\frac{1}{2} + 12 = \frac{23}{2}$$y(2)=−21​+3(22)=−21​+12=223​

$$2y(2) = 2 \times \frac{23}{2} = 23$$2y(2)=2×223​=23