Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[t]$, gives the greatest integer less than or equal to $t$. A key property is that $t - 1 < [t] \le t$, which implies $t - \{t\} = [t]$, where $\{t\}$ is the fractional part of $t$ and $0 \le \{t\} < 1$.
• Limit of a Function: For a function $f(x)$ to have a limit $L$ as $x \to c$, the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal to $L$.
• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if:
  1. $f(c)$ is defined.
  2. $\mathop {\lim }\limits_{x \to c} f(x)$ exists.
  3. $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$.
• Discontinuity: A function is discontinuous at a point if any of the conditions for continuity are not met. For functions involving the greatest integer function or trigonometric functions, discontinuities often occur at integer values of arguments or where the argument of the greatest integer function becomes an integer.

Step-by-Step Solution

Step 1: Evaluate the limit A. We are given $A = \mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$. To evaluate this limit, we can use the property $[t] = t - \{t\}$, where $\{t\}$ is the fractional part of $t$. So, $\left[ {{4 \over x}} \right] = {{4 \over x}} - \left\{ {{4 \over x}} \right\}$. Substituting this into the limit expression: $A = \mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$ $A = \mathop {\lim }\limits_{x \to 0} \left( x \cdot {4 \over x} - x \cdot \left\{ {{4 \over x}} \right\} \right)$ $A = \mathop {\lim }\limits_{x \to 0} \left( 4 - x\left\{ {{4 \over x}} \right\} \right)$ Now, we need to evaluate $\mathop {\lim }\limits_{x \to 0} x\left\{ {{4 \over x}} \right\}$. We know that $0 \le \left\{ {{4 \over x}} \right\} < 1$. Multiplying by $|x|$ (since $x \to 0$, $x$ can be positive or negative), we get $0 \le |x|\left\{ {{4 \over x}} \right\} < |x|$. As $x \to 0$, $|x| \to 0$. By the Squeeze Theorem, $\mathop {\lim }\limits_{x \to 0} |x|\left\{ {{4 \over x}} \right\} = 0$. Therefore, $\mathop {\lim }\limits_{x \to 0} x\left\{ {{4 \over x}} \right\} = 0$. So, the limit $A$ becomes: $A = 4 - 0 = 4$

Step 2: Determine the value of $\sqrt{A+1}$. From Step 1, we found $A=4$. We need to evaluate the potential points of discontinuity given in the options, which are of the form $\sqrt{A+k}$. Let's calculate $\sqrt{A+1}$: $\sqrt{A+1} = \sqrt{4+1} = \sqrt{5}$

Step 3: Analyze the continuity of $f(x) = [x^2]\sin(\pi x)$. A function of the form $[g(x)]h(x)$ is generally discontinuous when $g(x)$ is an integer or when $h(x)$ is zero (if the limit of $[g(x)]$ is finite). In our case, $f(x) = [x^2]\sin(\pi x)$. The term $\sin(\pi x)$ is zero when $\pi x = n\pi$, which means $x=n$ for any integer $n$. At these integer points, $f(x) = [n^2]\sin(n\pi) = [n^2] \cdot 0 = 0$. We need to check if the limit as $x \to n$ is also 0. The term $[x^2]$ changes its value when $x^2$ is an integer. This happens when $x = \pm \sqrt{k}$ for some integer $k \ge 0$.

We are looking for a point of discontinuity. The options are of the form $\sqrt{A+k}$. We found $A=4$, so the options are $\sqrt{5}, \sqrt{9}=3, \sqrt{25}=5, \sqrt{4}=2$. The function $f(x) = [x^2]\sin(\pi x)$ can be discontinuous at:

1. Points where $x^2$ is an integer (i.e., $x = \pm \sqrt{k}$, where $k$ is a non-negative integer).
2. Points where $\sin(\pi x) = 0$ (i.e., $x = n$, where $n$ is an integer).

Let's check the point $x = \sqrt{A+1} = \sqrt{5}$. At $x = \sqrt{5}$: $x^2 = 5$. So, $[x^2] = [5] = 5$. $f(\sqrt{5}) = [(\sqrt{5})^2]\sin(\pi \sqrt{5}) = [5]\sin(\pi \sqrt{5}) = 5\sin(\pi \sqrt{5})$.

Now let's check the limit as $x \to \sqrt{5}$. We need to consider the left-hand limit (LHL) and the right-hand limit (RHL). As $x \to \sqrt{5}^-$, $x < \sqrt{5}$. Since $x$ is close to $\sqrt{5}$, $x$ is positive. Then $x^2 < 5$. So, $[x^2]$ will be $4$ for values of $x$ slightly less than $\sqrt{5}$ such that $x^2 \ge 4$ (i.e., $x \ge 2$). Since $\sqrt{5} \approx 2.23$, this condition is met. For $x$ very close to $\sqrt{5}$ from the left, $x^2$ is slightly less than 5, so $[x^2]$ will be $4$. LHL = $\mathop {\lim }\limits_{x \to \sqrt{5}^-} [x^2]\sin(\pi x) = 4 \sin(\pi \sqrt{5})$.

As $x \to \sqrt{5}^+$, $x > \sqrt{5}$. Then $x^2 > 5$. So, $[x^2]$ will be $5$ for values of $x$ slightly greater than $\sqrt{5}$. RHL = $\mathop {\lim }\limits_{x \to \sqrt{5}^+} [x^2]\sin(\pi x) = 5 \sin(\pi \sqrt{5})$.

Since LHL ($4\sin(\pi \sqrt{5})$) $\ne$ RHL ($5\sin(\pi \sqrt{5})$) (unless $\sin(\pi \sqrt{5}) = 0$, which is not generally true for $\sqrt{5}$), the limit $\mathop {\lim }\limits_{x \to \sqrt{5}} f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $x = \sqrt{5}$.

Step 4: Verify continuity at other potential points. The problem statement mentions that "at $x = 2, 3$ and $5$, $f(x)$ is continuous." Let's quickly check why this might be true.

• At $x=2$: $f(2) = [2^2]\sin(2\pi) = [4] \cdot 0 = 0$. LHL: $\mathop {\lim }\limits_{x \to 2^-} [x^2]\sin(\pi x)$. For $x$ slightly less than 2, $x^2$ is slightly less than 4, so $[x^2]=3$. Thus, LHL $= 3 \sin(2\pi) = 0$. RHL: $\mathop {\lim }\limits_{x \to 2^+} [x^2]\sin(\pi x)$. For $x$ slightly greater than 2, $x^2$ is slightly greater than 4, so $[x^2]=4$. Thus, RHL $= 4 \sin(2\pi) = 0$. Since LHL = RHL = $f(2) = 0$, $f(x)$ is continuous at $x=2$.

• At $x=3$: $f(3) = [3^2]\sin(3\pi) = [9] \cdot 0 = 0$. LHL: $\mathop {\lim }\limits_{x \to 3^-} [x^2]\sin(\pi x)$. For $x$ slightly less than 3, $x^2$ is slightly less than 9, so $[x^2]=8$. Thus, LHL $= 8 \sin(3\pi) = 0$. RHL: $\mathop {\lim }\limits_{x \to 3^+} [x^2]\sin(\pi x)$. For $x$ slightly greater than 3, $x^2$ is slightly greater than 9, so $[x^2]=9$. Thus, RHL $= 9 \sin(3\pi) = 0$. Since LHL = RHL = $f(3) = 0$, $f(x)$ is continuous at $x=3$.

• At $x=5$: $f(5) = [5^2]\sin(5\pi) = [25] \cdot 0 = 0$. LHL: $\mathop {\lim }\limits_{x \to 5^-} [x^2]\sin(\pi x)$. For $x$ slightly less than 5, $x^2$ is slightly less than 25, so $[x^2]=24$. Thus, LHL $= 24 \sin(5\pi) = 0$. RHL: $\mathop {\lim }\limits_{x \to 5^+} [x^2]\sin(\pi x)$. For $x$ slightly greater than 5, $x^2$ is slightly greater than 25, so $[x^2]=25$. Thus, RHL $= 25 \sin(5\pi) = 0$. Since LHL = RHL = $f(5) = 0$, $f(x)$ is continuous at $x=5$.

The points $x=2, 3, 5$ are integers. At integer points $n$, $\sin(\pi n) = 0$, so $f(n)=0$. The continuity at integer points depends on whether the limit of $[x^2]\sin(\pi x)$ as $x \to n$ is 0. If $x \to n$, then $\sin(\pi x) \to 0$. If $[x^2]$ does not behave pathologically near $n$ (e.g., if $n^2$ is not an integer, which is true for non-integer $n$, or if $n^2$ is an integer, the jump in $[x^2]$ is multiplied by $\sin(\pi x) \to 0$), the function is often continuous at integers.

Step 5: Connect the result to the options. We found that $f(x)$ is discontinuous at $x = \sqrt{5}$. We calculated $\sqrt{A+1} = \sqrt{5}$. This matches option (A).

Common Mistakes & Tips

• Misinterpreting the Greatest Integer Function: Remember that $[t] \le t$. When evaluating limits, the behavior of $[t]$ as $t$ approaches an integer is crucial. For example, as $y \to 5^-$, $[y]=4$, and as $y \to 5^+$, $[y]=5$.
• Ignoring the Fractional Part: The term $x\{4/x\}$ in the limit evaluation is important. Forgetting to show that its limit is 0 can lead to an incorrect value for A.
• Assuming Continuity: Don't assume a function is continuous at points where the argument of the greatest integer function becomes an integer or where trigonometric functions become zero. Always check the LHL, RHL, and function value.
• Checking Only Integer Points: The question asks for points of discontinuity. While integer points are common places for discontinuity with $\sin(\pi x)$, the presence of $[x^2]$ means that points where $x^2$ is an integer (i.e., $x = \pm \sqrt{k}$) are also critical.

Summary

The problem requires us to first evaluate the limit $A$, which we found to be 4. Then, we need to identify the point of discontinuity for the function $f(x) = [x^2]\sin(\pi x)$. By analyzing the behavior of the greatest integer function and the sine function, we determined that discontinuities can arise when $x^2$ is an integer or when $\sin(\pi x) = 0$. We tested the point $x = \sqrt{A+1} = \sqrt{5}$. At this point, $x^2 = 5$, which is an integer. We calculated the left-hand and right-hand limits of $f(x)$ as $x \to \sqrt{5}$ and found them to be unequal, indicating a discontinuity. The other options were examined and found to be points of continuity.

The final answer is $\boxed{\sqrt {A + 1} }$.