JEE Advanced 2026 Paper 2Question 1
Let $y=f(x)$ be the real valued function defined on the interval $(0,\infty)$, satisfying $y(1)=0$ and the differential equation $$x\frac{dy}{dx} = y - x^3.$$ Then which of the following statements is (are) TRUE ?
Differential Equation
Solution
Concept. $xy'-y=-x^3$ is $\left(\frac yx\right)'=-x$; then study $f$ with derivatives. Formulas used. (1) $\left(\dfrac yx\right)'=\dfrac{xy'-y}{x^2}$(2) $f'(a)=0$, $f''(a)<0$ $\Rightarrow$ local maximum at $a$Step 1 - Solve. By (1), $\left(\frac yx\right)'=-x$, so $\frac yx=-\frac{x^2}{2}+C$. $y(1)=0$ gives $C=\frac12$: $$f(x)=\frac{x-x^3}{2}.$$ Step 2 - (A), (B). $f'(x)=\frac{1-3x^2}{2}=0$ at $x=\frac1{\sqrt3}$, and $f''(x)=-3x<0$ there. By (2) it is a local maximum: (B) true, (A) false. Step 3 - (C). On $(1,2)$, $1-3x^2<0$, so $f$ is decreasing: (C) false. Step 4 - (D). $f=g$ means $x-x^3=8x^3-10x^2+3x$, i.e. $x(9x^2-10x+2)=0$. For $x>0$: $9x^2-10x+2=0$ has discriminant $100-72=28>0$, and both roots $\frac{10\pm\sqrt{28}}{18}$ are positive. So $2$ elements: (D) true. Answer: (B), (D)