Let f(x) = [x^2] sin(pi x), for x 0. Then: | vedclass

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(D) The function $f(x) = [x^2] \sin(\pi x)$ is a product of two functions: $g(x) = [x^2]$ and $h(x) = \sin(\pi x)$.$g(x) = [x^2]$ is discontinuous at

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None of these

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(D) The function $f(x) = [x^2] \sin(\pi x)$ is a product of two functions: $g(x) = [x^2]$ and $h(x) = \sin(\pi x)$.$g(x) = [x^2]$ is discontinuous at all points where $x^2$ is an integer, i.e.,$x = \sqrt{n}$ for $n \in \{1, 2, 3, \dots\}$.For the product $f(x) = g(x)h(x)$ to be continuous at a point $x_0$ where $g(x)$ is discontinuous, $h(x_0)$ must be $0$.Here, $h(x) = \sin(\pi x) = 0$ when $x$ is an integer.If $x^2 = n$ (where $n$ is an integer) and $x$ is an integer, then $x = \sqrt{n}$ must be an integer, which means $n$ is a perfect square.If $x^2 = n$ but $x$ is not an integer, then $\sin(\pi x) 
eq 0$, so the function $f(x)$ remains discontinuous at these points.Therefore, $f(x)$ is continuous only at points where $x^2$ is an integer and $\sin(\pi x) = 0$, which occurs when $x$ is an integer.Since the options provided do not correctly describe the set of points of continuity, the correct choice is $D$.

Let $f(x) = [x^2] \sin(\pi x)$, for $x > 0$. Then:

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