Official solution
A = \[\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]\]
= \[\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)\]
= \[\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)\]
= 4
Now, when x = \[\sqrt {A + 1} \] = \[\sqrt 5 \], f(x) = [x2]sin(\[\pi \]x) is discontinuous at this non integer point.
But at x = 2, 3 and 5, f(x) is continuous.