Question

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

Answer 1

Hint: Greatest integer of a number is the integer that just comes before and is less than the real number. The real number is written between the square brackets. A function is discontinuous at a point x if the limit of function at that point are not equal for the points \[{x^ - }\] and \[{x^ + }\].

Formula Used: Intervals of a variable can be changed by dividing or multiplying the variable, if \[X \in \left( {a,b} \right)\] then \[\dfrac{X}{2} \in \left( {\dfrac{a}{2},\dfrac{b}{2}} \right)\]. Greatest integer near zero are,
\[
  \left[ {{0^ + }} \right] = 0 \\
  \left[ 0 \right] = 0 \\
  \left[ {{0^ - }} \right] = - 1
 \]


Complete step by step answer:
We are given an equation that is \[f(x) = x\left[ {\dfrac{x}{2}} \right]\]
Now we find the value of \[\dfrac{x}{2}\] by dividing \[2\] in the given equation, we get
\[
  X \in \left( { - 10,10} \right) \\
  \dfrac{x}{2} \in \left( {\dfrac{{ - 10}}{2},\dfrac{{10}}{2}} \right) \\
  \dfrac{x}{2} \in \left( { - 5,5} \right)
 \]
Now, we check the continuity at x = 0.
For x = 0,
\[
  f\left( x \right) = x\left[ {\dfrac{x}{2}} \right] \\
  f\left( 0 \right) = \left( 0 \right)\left[ {\dfrac{0}{2}} \right] \\
  f\left( 0 \right) = \left( 0 \right)\left[ 0 \right] \\
  f\left( 0 \right) = 0
 \]

For \[x = {0^ + }\],
\[
  f\left( x \right) = x\left[ {\dfrac{x}{2}} \right] \\
  f\left( {{0^ + }} \right) = \left( {{0^ + }} \right)\left[ {\dfrac{{{0^ + }}}{2}} \right] \\
  f\left( {{0^ + }} \right) = \left( 0 \right)\left[ 0 \right] \\
  f\left( {{0^ + }} \right) = 0
 \]

For \[x = {0^ - }\],
\[
  f\left( x \right) = x\left[ {\dfrac{x}{2}} \right] \\
  f\left( {{0^ - }} \right) = \left( {{0^ - }} \right)\left[ {\dfrac{{{0^ - }}}{2}} \right] \\
  f\left( {{0^ - }} \right) = \left( 0 \right)\left( { - 1} \right) \\
  f\left( {{0^ - }} \right) = 0
 \]

Since the function is continuous at \[x = 0\]. The function will be discontinuous when
\[\dfrac{x}{2} = \pm 4, \pm 3, \pm 2, \pm 1\]
The \[8\] points of discontinuity are \[ - 4, - 3, - 2, - 1,0,1,2,3,4\].
Therefore, the number of points of discontinuity of \[f\] is equal to \[8\].



Hence, the number of points of discontinuity of \[f\] is equal to \[8\].

Additional information: A Limitation with a denominator equal to zero is a point of discontinuity because it breaks the graph at that point. A point of discontinuity is the point at which a mathematical function goes away to be continuous. This is also known as the point at which the function is undefined. In mathematics, a continuous function is one that has no discontinuities, implying no unexpected changes in value. A function is continuous if we can ensure arbitrarily small changes in its input by restricting enough minor changes. The given function is said to be discontinuous if it is not continuous.


Note: Many students made miscalculations while writing the values of \[x\] in continuity. Also, they do not calculate the exact values of the function. They are confused in between the negative and positive signs.