JEE Advanced 2026 — Paper 2 — MATH-II solution

JEE Advanced2026Paper 2MATH-II

Let $y = f (x)$ be the real valued function defined on the interval $(0, \infty)$ , satisfying $y (1) = 0$ and the differential equation

$x \frac{d y}{d x} = y - x^{3}$

Then which of the following statements is (are) TRUE?

(A) The function $f$ has a local minimum at $x = \frac{1}{3}$

(B) The function $f$ has a local maximum at $x = \frac{1}{3}$

(D) If $g (x) = 4 x^{3} - 5 x^{2} + \frac{3}{2} x$ for $x > 0$ , then the number of elements in the set ${x \in (0, \infty) : f (x) = g (x)}$ is $2$

A
The function $f$ has a local minimum at $x = \frac{1}{3}$
B
The function $f$ has a local maximum at $x = \frac{1}{3}$
C
The function $f$ is increasing in the interval $(1, 2)$
D
If $g (x) = 4 x^{3} - 5 x^{2} + \frac{3}{2} x$ for $x > 0$ , then the number of elements in the set ${x \in (0, \infty) : f (x) = g (x)}$ is $2$

Solution

Rearrange: $x d y - y d x = - x^{3} d x$ . Divide both sides by $x^{2}$ :

$\frac{x d y - y d x}{x ^{2}} = - x d x \Rightarrow d (\frac{y}{x}) = - x d x$ .

Integrating: $\frac{y}{x} = - \frac{x ^{2}}{2} + C$ , so $y = - \frac{x ^{3}}{2} + C x$ .

The condition $y (1) = 0$ gives $C = \frac{1}{2}$ , hence $f (x) = - \frac{x ^{3}}{2} + \frac{x}{2}$ on $(0, \infty)$ .

$f^{'} (x) = - \frac{3 x ^{2}}{2} + \frac{1}{2} = - \frac{3}{2} (x - \frac{1}{3}) (x + \frac{1}{3})$ .

On $(0, \infty)$ : $f^{'} > 0$ for $x < \frac{1}{3}$ and $f^{'} < 0$ for $x > \frac{1}{3}$ , so $f$ has a local maximum at $x = \frac{1}{3}$ . Option (B) is correct; (A) is wrong.

On $(1, 2)$ , $f^{'} (x) < 0$ , so $f$ is decreasing. Option (C) is wrong.

(D): $f (x) = g (x)$ gives $- \frac{x ^{3}}{2} + \frac{x}{2} = 4 x^{3} - 5 x^{2} + \frac{3 x}{2}$ , i.e. $9 x^{3} - 10 x^{2} + 2 x = 0$ , or $x (9 x^{2} - 10 x + 2) = 0$ . Discarding $x = 0$ , the two positive roots are $x = \frac{5 \pm 7}{9}$ , both in $(0, \infty)$ . So there are exactly $2$ solutions. Option (D) is correct.

$∴$ Correct options: (B), (D).

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