JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Continuity and Differentiability

$f(x)=[x^3]\log (1+{\sin }^2\pi x)$ where $x\in (-3,3)$ so $x^3\in (-27,27)$. At integers $x=0,\pm 1,\pm 2$, ${\sin }^2\pi x=0$ so $\log (1+{\sin }^2\pi x)=0$ and $f$ is continuous. At all other integers in $(-27,27)$, $[x^3]$ jumps and $f$ is discontinuous. Total integers in $(-27,27)$: 53. Excluding the 5 where $f$ is continuous: $|A|=48$. For $g(x)=x^3{\sin }^2(\pi \log (1+\{x\}))$, discontinuities occur at integers except 0. In $(-3,3)$: $x=\pm 1,\pm 2$, so $|B|=4$. Since $A\cap B$ consists of integers where both are discontinuous (excluding 0): $|A\cap B|=0$. Thus $|A|+2|B|-|A\cap B|=48+8-0=56$.