JEE Mathematics 2026 Question with Solution

Given: $y = y(x)$ is differentiable on $(0,\infty)$, $y(1) = 2$, and

for each $x > 0$.

Find: $2y(2)$.

Rewrite the given limit as

This is the derivative with respect to $t$ of $x^2y(t)-t^2y(x)$ evaluated at $t = x$.

Therefore,

Given that the limit equals $3$, we get

Now divide by $x^2$ to obtain the differential equation

Its integrating factor is

Multiplying throughout by the integrating factor,

Integrating,

Hence,

Using the condition $y(1)=2$,

So,

Now evaluate at $x=2$:

Thus,

Therefore, the correct option is A.

Given: the limit expression involves two variables, $t$ and $x$, with $x$ treated as fixed while $t \to x$.

Find: convert the limit into a differential equation for $y$.

Consider

Here, for fixed $x$, the quantity $y(x)$ is a constant with respect to $t$. Then

is precisely $F'(x)$, because

So the given limit becomes

Differentiating with respect to $t$,

Putting $t=x$,

Since the limit equals $3$,

From here, solving the linear differential equation gives $y(x)=3x^2-\frac{1}{x}$ and hence $2y(2)=23$.