JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Continuity and Differentiability

Writing $\left[\frac{4}{x}\right]=\frac{4}{x}-\left\{\frac{4}{x}\right\}$, the product $x\left[\frac{4}{x}\right]=4-x\left\{\frac{4}{x}\right\}\to 4$ as $x\to 0$, since the fractional part stays bounded, so $A=4$. The function $f(x)=[x^2]\sin (\pi x)$ breaks only where $[x^2]$ jumps, that is where $x^2$ is an integer while $x$ itself is not, because at an integer $x$ the factor $\sin (\pi x)$ vanishes and smooths the jump. Testing $\sqrt{A+1}=\sqrt{5}$: here $x^2=5$ is an integer but $x=\sqrt{5}$ is not, so $f$ is discontinuous at $x=\sqrt{5}$.