Question

$f(x)$ be a real values function defined on the interval [-2, 2] as follows

$\qquad \begin{cases} \qquad 1 \qquad \text{if} \quad -2 \le x \le -1 \\ f(x)= \begin{cases} \qquad x+2 \qquad \text{if} \quad -1 < x < 1 \\ \qquad 1-x \qquad \text{if} \quad 1 \le x \le 2 \end{cases} \end{cases}$

then number of solutions of the equation $\{f(x)\} = \frac{1}{2}$ is, (where $\{f(x)\} \rightarrow$ denotes fractional part of $f(x)$)

Answer 1

3

Answer 2

The equation to solve is $\{f(x)\} = \frac{1}{2}$, where $\{y\}$ denotes the fractional part of $y$. The expression $\{y\} = \frac{1}{2}$ implies that $y$ must be of the form $n + \frac{1}{2}$ for some integer $n$. So, we need to find the values of $x$ for which $f(x) = n + \frac{1}{2}$ for some integer $n$.

We analyze the function $f(x)$ in each of its defined intervals:

Case 1: $-2 \le x \le -1$ In this interval, $f(x) = 1$. We need $\{f(x)\} = \{1\} = 0$. Since $0 \ne \frac{1}{2}$, there are no solutions in this interval.

Case 2: $-1 < x < 1$ In this interval, $f(x) = x+2$. As $x$ ranges from $-1$ (exclusive) to $1$ (exclusive): When $x \to -1^+$, $f(x) \to (-1+2)^+ = 1^+$. When $x \to 1^-$, $f(x) \to (1+2)^- = 3^-$. So, the range of $f(x)$ in this interval is $(1, 3)$.

We need $f(x) = n + \frac{1}{2}$ where $n$ is an integer and $1 < f(x) < 3$. Possible values for $n$:

• If $n=1$, $f(x) = 1 + \frac{1}{2} = \frac{3}{2}$. This value lies in $(1, 3)$. Setting $x+2 = \frac{3}{2} \implies x = \frac{3}{2} - 2 = -\frac{1}{2}$. This value $x = -\frac{1}{2}$ is within the interval $(-1, 1)$. So, $x = -\frac{1}{2}$ is a solution.
• If $n=2$, $f(x) = 2 + \frac{1}{2} = \frac{5}{2}$. This value lies in $(1, 3)$. Setting $x+2 = \frac{5}{2} \implies x = \frac{5}{2} - 2 = \frac{1}{2}$. This value $x = \frac{1}{2}$ is within the interval $(-1, 1)$. So, $x = \frac{1}{2}$ is a solution.
• For any other integer $n$ (e.g., $n=0 \implies f(x)=1/2$, or $n=3 \implies f(x)=7/2$), $n+\frac{1}{2}$ would not fall within the range $(1, 3)$.

Thus, there are 2 solutions in the interval $(-1, 1)$.

Case 3: $1 \le x \le 2$ In this interval, $f(x) = 1-x$. As $x$ ranges from $1$ (inclusive) to $2$ (inclusive): When $x=1$, $f(x) = 1-1 = 0$. When $x=2$, $f(x) = 1-2 = -1$. So, the range of $f(x)$ in this interval is $[-1, 0]$.

We need $f(x) = n + \frac{1}{2}$ where $n$ is an integer and $-1 \le f(x) \le 0$. Possible values for $n$:

• If $n=-1$, $f(x) = -1 + \frac{1}{2} = -\frac{1}{2}$. This value lies in $[-1, 0]$. Setting $1-x = -\frac{1}{2} \implies x = 1 + \frac{1}{2} = \frac{3}{2}$. This value $x = \frac{3}{2}$ is within the interval $[1, 2]$. So, $x = \frac{3}{2}$ is a solution.
• For any other integer $n$ (e.g., $n=0 \implies f(x)=1/2$, or $n=-2 \implies f(x)=-3/2$), $n+\frac{1}{2}$ would not fall within the range $[-1, 0]$.

Thus, there is 1 solution in the interval $[1, 2]$.

Total Number of Solutions: Summing the solutions from all intervals: $0 + 2 + 1 = 3$. The solutions are $x = -\frac{1}{2}$, $x = \frac{1}{2}$, and $x = \frac{3}{2}$.