Mathematics - Functions Question with Solution | TestHub

Since $\{x+1\}=\{x\}$, we have $f(x+1)=\max (\{x+1\},1-\{x+1\})=\max (\{x\},1-\{x\})=f(x)$. Thus, $f(x)$ is periodic with period 1 (A is correct).

Let $y=\{x\}$. Then $f(x)=\max (y,1-y)$. If $0\le y\le \frac{1}{2}$, $f(x)=1-y$. If $\frac{1}{2}<y<1$, $f(x)=y$. The equation $f(x)=\frac{1}{2}$ implies $1-y=\frac{1}{2}$ (for $0\le y\le \frac{1}{2}$) or $y=\frac{1}{2}$ (for $\frac{1}{2}<y<1$, which is not possible).

So, $\{x\}=\frac{1}{2}$. This means $x=n+\frac{1}{2}$ for any integer $n$, giving infinitely many solutions (B is correct).

For the range, as $y$ varies in $[0,\frac{1}{2}]$, $1-y$ varies in $[\frac{1}{2},1]$. As $y$ varies in $(\frac{1}{2},1)$, $y$ varies in $(\frac{1}{2},1)$. Combining these, the range of $f(x)$ is $[\frac{1}{2},1)$ (D is correct). For $f(x)=\frac{3}{4}$: If $0\le \{x\}\le \frac{1}{2}$, then $1-\{x\}=\frac{3}{4}\Rightarrow \{x\}=\frac{1}{4}$. If $\frac{1}{2}<\{x\}<1$, then $\{x\}=\frac{3}{4}$.

So, in $[0,1)$, there are two solutions: $x=\frac{1}{4}$ and $x=\frac{3}{4}$. Due to periodicity, in $[0,2)$, there will be four solutions: $\frac{1}{4},\frac{3}{4},1+\frac{1}{4}=\frac{5}{4},1+\frac{3}{4}=\frac{7}{4}$. Thus, C is incorrect.