JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Differential Equations

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

Answer: (B), (D)

Solution

From $xdy-ydx=-x^{3}dx$, we get $d(\frac{y}{x})=-xdx$, so $\frac{y}{x}=-\frac{x^2}{2}+c$. Using $y(1)=0$: $c=\frac{1}{2}$. Thus $y=\frac{-x^3}{2}+\frac{x}{2}$. Then $\frac{dy}{dx}=\frac{-3x^2}{2}+\frac{1}{2}=\frac{-3}{2}(x-\frac{1}{\sqrt{3}})(x+\frac{1}{\sqrt{3}})$. For $x>0$: $f$ increases on $(0,\frac{1}{\sqrt{3}})$ and decreases on $(\frac{1}{\sqrt{3}},\infty )$, so local maximum at $x=\frac{1}{\sqrt{3}}$ (option B). In $(1,2)$, $f$ is decreasing (option C false). For $f(x)=g(x)$: $9x^3-10x^2+2x=0\Rightarrow x(9x^2-10x+2)=0$. Roots are $x=0$ and $x=\frac{5\pm \sqrt{7}}{9}$ (both positive), giving 2 positive solutions (option D).

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · One or more correct options