JEE Advanced 2026 · Paper 2, Mathematics question 8

Solve. $xy'-y=-x^3$ gives $\left(\frac yx\right)'=-x$, so $\frac yx=-\frac{x^2}2+C$. From $y(1)=0$, $C=\frac12$:
$$f(x)=\frac{x-x^3}{2}.$$

• $f'(x)=\frac{1-3x^2}2$ vanishes at $x=\frac1{\sqrt3}$, and $f''=-3x<0$ there: a local maximum. (B) true, (A) false.
• On $(1,2)$, $f'<0$, so $f$ is decreasing. (C) false.
• (D) $f=g$ means $\frac x2-\frac{x^3}2=4x^3-5x^2+\frac32x$, i.e. $x(9x^2-10x+2)=0$. For $x>0$: $9x^2-10x+2=0$ has discriminant $28>0$, and both roots $\frac{10\pm\sqrt{28}}{18}$ are positive. Two solutions. True.