Question

Let f(x)=[x²]sinπx, x>0. Then

• A. f is discontinuous everywhere.
• B. f is continuous everywhere.
• C. f is continuous at only these points which are perfect squares.
• D. f is continuous at only these points which are not perfect squares.

Answer 1

The Correct Option is C

Let \(f(x) = [x^2] \sin(\pi x)\), where \(x > 0\). We need to analyze the continuity of this function.

Understanding the Components: 

* \([x^2]\) is the greatest integer function of \(x^2\). It is discontinuous at points where \(x^2\) is an integer, which means \(x = \sqrt{n}\), where n is a non-negative integer.

* \(\sin(\pi x)\) is a continuous function for all real numbers.

Analyzing Discontinuities:

The function \(f(x)\) can only be discontinuous at points where \([x^2]\) is discontinuous, which are \(x = \sqrt{n}\), where n is a non-negative integer.

Now, we need to check the continuity at these points.

Let \(x = \sqrt{n}\), where n is an integer.

For f(x) to be continuous at \(x = \sqrt{n}\), we need to have:

\(\lim_{x \to \sqrt{n}^-} f(x) = \lim_{x \to \sqrt{n}^+} f(x) = f(\sqrt{n})\)

\(f(\sqrt{n}) = [\sqrt{n}^2] \sin(\pi \sqrt{n}) = n \sin(\pi \sqrt{n})\)

Case 1: n is a perfect square

If n is a perfect square, then \(\sqrt{n}\) is an integer, so \(\sin(\pi \sqrt{n}) = 0\).

Therefore, \(f(\sqrt{n}) = n \cdot 0 = 0\). Now, we need to check the limits. Since \(\sin(\pi x)\) is continuous and equal to zero when x is an integer, then when x equals the square root of an integer we expect continuity at the points which are perfect squares. We already have 0 for these points: \(\sin(n\pi)=0\).

Case 2: n is not a perfect square

Now let's analyze the limit at a point where \(x=\sqrt{n}\), where n is an integer: \([x^2] \sin(\pi x)\).

Since \(\sin(\pi x)\) is continuous, \(\lim_{x\to \sqrt{n}} sin(\pi x)=sin(\pi \sqrt{n})\)

If we choose n as a non perfect square integer, the main factor defining continuity is whether \([x^2]\) is continuous. But by our earlier reasoning this function does have discrete values. However, \(\lim_{x \to \sqrt{n}^-} [x^2]\) is equal to \((n-1)\), and \(\lim_{x \to \sqrt{n}^+} [x^2]=n\).

Conclusion:

The greatest integer function is not continuous, thus, at points which are not perfect integers, there exists a discrete step. Since the non integer points go to \(\sin(\pi x)\) that means they are equal to 0. This also can never happen due to the domain condition (\(x>0\)). Hence, this only holds true for perfect squares, thus the function can only be continuous at perfect square points. f is continuous at only those points which are perfect squares.