Question

If \[f:\mathbb{R} \to \mathbb{R}\] is defined by \[f(x) = x - \left[ x \right] - \dfrac{1}{2}\] for \[x \in \mathbb{R}\], where \[\left[ x \right]\] is the greatest integer not exceeding x, then find the interval for which\[\left\{ {x \in \mathbb{R}:f(x) = \dfrac{1}{2}} \right\}\].
A. Z, the set of all integers,
B. N, the set of all natural numbers,
C. \[\phi \]
D. R, set of all real numbers.

Answer 1

Hints: First write x as the sum of integer part and the fractional part then obtain the value of box x and substitute in the given function. Then substitute \[\dfrac{1}{2}\] for f(x) in the obtained equation and solve for x. Now, the fractional part of x always satisfies the inequality \[0 \le x < 1\] but we conclude that \[x = 1\] , which is a contradiction.

Complete step by step solution
The given equation is \[f(x) = x - \left[ x \right] - \dfrac{1}{2}\].
Now, we know that any number is the sum of its integer part and fractional part, therefore,
\[x = \left[ x \right] + \left\{ x \right\}\], where \[\left[ x \right]\] is the integer part and \[\left\{ x \right\}\] is the fractional part.
So, \[x - \left[ x \right] = \left\{ x \right\}\]
Substitute \[x - \left[ x \right] = \left\{ x \right\}\]in the given equation we have,
\[f(x) = \left\{ x \right\} - \dfrac{1}{2}\]
Now, it is given that \[f(x) = \dfrac{1}{2}\] .
So,
\[\dfrac{1}{2} = \left\{ x \right\} - \dfrac{1}{2}\]
\[\therefore \left\{ x \right\} = 1\]
But the fractional part always satisfies the range \[0 \le x < 1\], and can not equal to 1.
Therefore,
\[\left\{ x \right\} = 1\], is a contradiction.
Hence, the required interval is \[\phi \].

Note Sometime students directly substitute \[\dfrac{1}{2}\] for \[f\left(x\right)\] in the given equation and calculate to obtain the value of the interval, but in this way, we cannot conclude the result. First, we need to apply the concept that any number is the sum of its integer part and the fractional part then have to calculate the rest.