1. Find $A$ from the limit

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

Use the property of greatest integer function: $y-1<[y]\le y.$ Putting $y=\frac{4}{x}$, $\frac{4}{x}-1<\left[\frac{4}{x}\right]\le \frac{4}{x}.$

Now multiply by $x$.

• For $x>0$: $4-x<x\left[\frac{4}{x}\right]\le 4.$ So as $x\to 0^+$, $x\left[\frac{4}{x}\right]\to 4.$

• For $x<0$, multiplying reverses inequality: $4\le x\left[\frac{4}{x}\right]<4-x.$ So as $x\to 0^-$, $x\left[\frac{4}{x}\right]\to 4.$

Hence, $A=4.$

2. Interpret the function

The given function is $f(x)=[x^2]\sin(\pi x).$ Here $[x^2]$ means the greatest integer of $x^2$.

So, $f(x)=\lfloor x^2\rfloor\sin(\pi x).$

We must find where this function is discontinuous among the given options.

Since $\sin(\pi x)$ is continuous everywhere, discontinuity can only come from $\lfloor x^2\rfloor$, which changes value when $x^2$ crosses an integer.

Thus possible discontinuity points are when $x^2=n,\quad n\in \mathbb{Z}_{\ge 0},$ that is, $x=\pm\sqrt n.$

But not every such point makes the product discontinuous, because if $\sin(\pi x)=0,$ then the jump may disappear. This happens when $x$ is an integer.

So we test the given options.

3. Evaluate the options using $A=4$

• Option A: $\sqrt{A+1}=\sqrt{5}$
• Option B: $\sqrt{A+5}=\sqrt{9}=3$
• Option C: $\sqrt{A+21}=\sqrt{25}=5$
• Option D: $\sqrt{A}=\sqrt{4}=2$

So the candidate points are: $\sqrt5,\ 3,\ 5,\ 2.$

4. Check continuity at each point

At $x=\sqrt5$

Here $x^2=5,$ so $\lfloor x^2\rfloor$ jumps from $4$ to $5$. Also, $\sin(\pi\sqrt5)\ne 0$ since $\sqrt5$ is not an integer.

Thus left and right limits are: $\lim_{x\to \sqrt5^-} f(x)=4\sin(\pi\sqrt5),$ $\lim_{x\to \sqrt5^+} f(x)=5\sin(\pi\sqrt5).$ These are unequal, so $f$ is discontinuous at $x=\sqrt5$.

At $x=3$

Here $x^2=9$, so $\lfloor x^2\rfloor$ jumps, but $\sin(3\pi)=0.$ Near $x=3$, $f(x)=\lfloor x^2\rfloor\sin(\pi x),$ with bounded $\lfloor x^2\rfloor$ and continuous $\sin(\pi x)\to 0$. Thus both side limits are $0=f(3)$, so it is continuous.

At $x=5$

Similarly, $\sin(5\pi)=0,$ so $f$ is continuous at $x=5$.

At $x=2$

Similarly, $\sin(2\pi)=0,$ so $f$ is continuous at $x=2$.

5. Conclusion

The function is discontinuous only at $x=\sqrt5=\sqrt{A+1}.$

So the correct option is A.