Let f(x) = x cdot left[ frac{x}{2} right] for | vedclass

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(A) The function is defined as $f(x) = x \cdot \left[ \frac{x}{2} \right]$ for $x \in (-10, 10)$.The greatest integer function $[t]$ is discontinuous

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$8$

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(A) The function is defined as $f(x) = x \cdot \left[ \frac{x}{2} ight]$ for $x \in (-10, 10)$.The greatest integer function $[t]$ is discontinuous at all integer values of $t$.Here,$t = \frac{x}{2}$. Thus,$f(x)$ is potentially discontinuous when $\frac{x}{2} = k$,where $k \in \mathbb{Z}$.Given $-10 The possible integer values for $k = \frac{x}{2}$ are $\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.Let us check the continuity at these points:$1$. For $k 
eq 0$ (i.e.,$x 
eq 0$),the function $f(x) = x \cdot \left[ \frac{x}{2} ight]$ is discontinuous because the greatest integer function $\left[ \frac{x}{2} ight]$ jumps at these points,and $x$ is non-zero.$2$. For $k = 0$,$x = 0$. We check the continuity at $x = 0$:$f(0) = 0 \cdot [0] = 0$.$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} x \cdot [0] = 0$.$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} x \cdot [-1] = 0$.Since $f(0) = \lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = 0$,the function is continuous at $x = 0$.Therefore,the points of discontinuity are $\frac{x}{2} \in \{-4, -3, -2, -1, 1, 2, 3, 4\}$,which corresponds to $x \in \{-8, -6, -4, -2, 2, 4, 6, 8\}$.There are $8$ such points.

Let $f(x) = x \cdot \left[ \frac{x}{2} \right]$ for $-10 < x < 10$,where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

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