Let $$[t]$$ denote the greatest integer $$\le t$$ and $$\lim_{x \to 0} x\left[\frac{4}{x}\right] = A$$. Then the function, $$f(x) = [x^2]\sin(\pi x)$$ is discontinuous, when $$x$$ is equal to:

First we evaluate the constant $$A$$ that appears in the statement. The limit to be found is

$$A=\lim_{x\to 0}\;x\left[\frac{4}{x}\right].$$

To recognise how the greatest-integer (floor) function behaves, recall the identity

$$t=\,[t]+\{t\},\quad\text{where }0\le\{t\}<1.$$

Applying this to $$t=\dfrac4x$$ gives

$$\left[\dfrac4x\right]=\dfrac4x-\left\{\dfrac4x\right\}.$$

Hence

$$x\left[\dfrac4x\right]=x\left(\dfrac4x-\left\{\dfrac4x\right\}\right)=4-x\left\{\dfrac4x\right\}.$$

The fractional part $$\displaystyle\left\{\dfrac4x\right\}$$ is always bounded between $$0$$ and $$1$$, so that the product $$x\left\{\dfrac4x\right\}$$ is squeezed between $$-\,|x|$$ and $$|x|$$. Therefore

$$\lim_{x\to 0}x\left\{\dfrac4x\right\}=0,$$

and we immediately get

$$\lim_{x\to 0}x\left[\dfrac4x\right]=\lim_{x\to 0}\left(4-x\left\{\dfrac4x\right\}\right)=4-0=4.$$

So we have established

$$A=4.$$

Now we turn to the function whose points of discontinuity are to be located:

$$f(x)=\,[x^{2}]\,\sin(\pi x).$$

The greatest-integer function $$[x^{2}]$$ can change its value only when $$x^{2}$$ itself passes through an integer. Hence potential trouble points satisfy

$$x^{2}=n,\quad n\in\{0,1,2,3,\dots\}\quad\Longrightarrow\quad x=\pm\sqrt n.$$

At such points $$[x^{2}]$$ has a jump. However, the overall product may still be continuous if the factor $$\sin(\pi x)$$ happens simultaneously to be zero, because

$$\sin(\pi x)=0\quad\Longleftrightarrow\quad x\in\mathbb Z.$$

Putting the two observations together:

• If $$x=\sqrt n$$ but $$\sqrt n\notin\mathbb Z,$$ then $$[x^{2}]$$ jumps while $$\sin(\pi x)\neq 0$$, so $$f(x)$$ is discontinuous.
• If $$x=\sqrt n$$ and in addition $$\sqrt n\in\mathbb Z$$ (namely $$x=0,\,\pm1,\,\pm2,\,\pm3,\dots$$), then $$\sin(\pi x)=0$$ on both sides and the product tends to $$0$$ from either side, making $$f(x)$$ continuous.

Thus the function is discontinuous precisely at all non-integral square roots of natural numbers.

With $$A=4$$ already found, the option list yields

$$\sqrt{A+1}=\sqrt5,\qquad \sqrt{A+5}=\sqrt9=3,\qquad \sqrt{A+21}=\sqrt{25}=5,\qquad \sqrt A=\sqrt4=2.$$

Among these, $$\sqrt5$$ is not an integer, so it satisfies the discontinuity condition, whereas $$2,3,5$$ are integers and correspond to points where $$f(x)$$ remains continuous.

Hence, the correct answer is Option A.