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

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Continuity and Differentiability

Let

y = y (x)

be a differentiable function in the interval

(0, \infty)

such that

y (1) = 2

.
and

lim_{t \to x} (\frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t}) = 3

for each

x > 0

.
Then

2 y (2)

is equal to

Answer: (B)

Both numerator and denominator vanish as

t \to x

, so applying L'Hopital in

t

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

. At

t = x

this reads

- (2 x y - x^{2} y^{'}) = 3

, i.e.

x^{2} y^{'} - 2 x y = 3

, so

y^{'} - \frac{2}{x} y = \frac{3}{x ^{2}}

. The integrating factor is

x^{- 2}

, giving

\frac{d}{d x} (y x^{- 2}) = 3 x^{- 4}

and

y x^{- 2} = - x^{- 3} + c

, hence

y = c x^{2} - \frac{1}{x}

. From

y (1) = 2

c = 3

, so

y (x) = 3 x^{2} - \frac{1}{x}

and

y (2) = \frac{23}{2}

. Therefore

2 y (2) = 23

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option