JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ and $t \to x lim (\frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

A
18
B
23
C
27
D
12

Solution

The limit is $\frac{0}{0}$ form as $t \to x$ . Apply L'Hopital with respect to $t$ :

$t \to x lim \frac{2 t y ( x ) - x ^{2} y ^{'} ( t )}{- 1} = 3 \Rightarrow - (2 x y (x) - x^{2} y^{'} (x)) = 3 \Rightarrow x^{2} y^{'} (x) - 2 x y (x) = 3$ .

This is linear: $y^{'} (x) - \frac{2}{x} y (x) = \frac{3}{x ^{2}}$ . Integrating factor $= e^{- \int \frac{2}{x} d x} = \frac{1}{x ^{2}}$ .

$\frac{d}{d x} (\frac{y}{x ^{2}}) = \frac{3}{x ^{4}} \Rightarrow \frac{y}{x ^{2}} = - \frac{1}{x ^{3}} + C \Rightarrow y = C x^{2} - \frac{1}{x}$ .

Use $y (1) = 2$ : $C - 1 = 2 \Rightarrow C = 3$ . So $y (x) = 3 x^{2} - \frac{1}{x}$ .

$y (2) = 12 - \frac{1}{2} = \frac{23}{2}$ . Therefore $2 y (2) = 23$ .

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