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(D) The function is $f(x) = x \cos(\pi(x + [x]))$. $A$ function is discontinuous at points where the left-hand limit $(LHL)$ is not equal to the right

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$A, C, D$

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(D) The function is $f(x) = x \cos(\pi(x + [x]))$. $A$ function is discontinuous at points where the left-hand limit $(LHL)$ is not equal to the right-hand limit $(RHL)$ or the function value.For any integer $n$,in the interval $[n, n+1)$,$[x] = n$. Thus,$f(x) = x \cos(\pi(x + n))$.$1$. At $x = -1$:$LHL = \lim_{x 	o -1^-} x \cos(\pi(x - 2)) = -1 \cos(-3\pi) = -1(-1) = 1$.$RHL = \lim_{x 	o -1^+} x \cos(\pi(x - 1)) = -1 \cos(-2\pi) = -1(1) = -1$.Since $LHL 
eq RHL$,$f(x)$ is discontinuous at $x = -1$.$2$. At $x = 0$:$LHL = \lim_{x 	o 0^-} x \cos(\pi(x - 1)) = 0 \cos(-\pi) = 0$.$RHL = \lim_{x 	o 0^+} x \cos(\pi x) = 0 \cos(0) = 0$.Since $LHL = RHL = f(0) = 0$,$f(x)$ is continuous at $x = 0$.$3$. At $x = 1$:$LHL = \lim_{x 	o 1^-} x \cos(\pi(x)) = 1 \cos(\pi) = -1$.$RHL = \lim_{x 	o 1^+} x \cos(\pi(x + 1)) = 1 \cos(2\pi) = 1$.Since $LHL 
eq RHL$,$f(x)$ is discontinuous at $x = 1$.$4$. At $x = 2$:$LHL = \lim_{x 	o 2^-} x \cos(\pi(x + 1)) = 2 \cos(3\pi) = -2$.$RHL = \lim_{x 	o 2^+} x \cos(\pi(x + 2)) = 2 \cos(4\pi) = 2$.Since $LHL 
eq RHL$,$f(x)$ is discontinuous at $x = 2$.Thus,the function is discontinuous at $x = -1, 1, 2$. The correct option is $D$.

Let $[x]$ be the greatest integer less than or equal to $x$. At which of the following point$(s)$ is the function $f(x) = x \cos(\pi(x + [x]))$ discontinuous?
$[A]$ $x = -1$
$[B]$ $x = 0$
$[C]$ $x = 2$
$[D]$ $x = 1$

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Let $[x]$ be the greatest integer less than or equal to $x$. At which of the following point$(s)$ is the function $f(x) = x \cos(\pi(x + [x]))$ discontinuous?$[A]$ $x = -1$$[B]$ $x = 0$$[C]$ $x = 2$$[D]$ $x = 1$