Let $f (x) = x \cdot [\frac{x}{2}],$ 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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Solution

$f (x) = x [\frac{x}{2}]$
$x \in (- 10, 10)$
$\frac{x}{2} \in (- 5, 5) \to 9$ integers

Check continuity at $x = 0$
$\begin{matrix} f (0) = 0 f (0^{+}) = 0 f (0^{-}) = 0 \end{matrix} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Continous at x = 0$

Function will be discontinuous when
$\Rightarrow \frac{x}{2} = \pm 4, \pm 3, \pm 2, \pm 1$
Points are $- 4, - 3, - 2, - 1, 0, 1, 2, 3, 4$
$\Rightarrow 8$ points of discontinuity.

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Let f(x)=x⋅[x2], for −10<x<10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to

f(x)=x[x2] x∈(−10,10) x2∈(−5,5)→9 integers Check continuity at x=0 f(0)=0f(0+)=0f(0−)=0⎫⎪ ⎪⎬⎪ ⎪⎭Continous at x=0 Function will be discontinuous when ⇒x2=±4,±3,±2,±1 Points are −4,−3,−2,−1,0,1,2,3,4 ⇒8 points of discontinuity.

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