Mathematics - Functions Question with Solution | TestHub

Let $x=I+F$, where $I=\lfloor x\rfloor \in \mathbb{Z}$ and $F=\{x\}\in [0,1)$. So $f(x)=I+\sqrt{F}$.

A) To check injectivity, assume $f(x_1)=f(x_2)$. Let $x_1=I_1+F_1$ and $x_2=I_2+F_2$. Then $I_1+\sqrt{F_1}=I_2+\sqrt{F_2}$. Since $\sqrt{F_1},\sqrt{F_2}\in [0,1)$, we must have $I_1=I_2$. (If $I_1\ne I_2$, say $I_1>I_2$, then $I_1-I_2\ge 1$. But $\sqrt{F_2}-\sqrt{F_1}\in (-1,1)$, so $I_1-I_2=\sqrt{F_2}-\sqrt{F_1}$ is impossible). Since $I_1=I_2$, it follows that $\sqrt{F_1}=\sqrt{F_2}⟹F_1=F_2$. Therefore, $x_1=x_2$. So $f(x)$ is an injective function. This statement is correct. B) For any integer $k$, consider $x\in [k,k+1)$. Then $\lfloor x\rfloor =k$ and $\{x\}=x-k$. So $f(x)=k+\sqrt{x-k}$. As $x$ varies from $k$ to $k+1$, $x-k$ varies from $0$ to $1$. Thus, $\sqrt{x-k}$ varies from $0$ to $1$. So, for $x\in [k,k+1)$, the range of $f(x)$ is $[k,k+1)$. The union of all such intervals for $k\in \mathbb{Z}$ is ${\bigcup }_{k\in \mathbb{Z}}[k,k+1)=\mathbb{R}$. So the range of $f(x)$ is $\mathbb{R}$. This statement is correct.

C) We need to check continuity at integer points, as $\lfloor x\rfloor$ and $\{x\}$ are potentially discontinuous there. Let $k\in \mathbb{Z}$.$f(k)=\lfloor k\rfloor +\sqrt{\{k\}}=k+\sqrt{0}=k$.$li{m}_{x\to k^{+}}f(x)={\lim }_{x\to k^{+}}(\lfloor x\rfloor +\sqrt{\{x\}})={\lim }_{x\to k^{+}}(k+\sqrt{x-k})=k+\sqrt{0}=k$.$li{m}_{x\to k^{-}}f(x)={\lim }_{x\to k^{-}}(\lfloor x\rfloor +\sqrt{\{x\}})={\lim }_{x\to k^{-}}((k-1)+\sqrt{x-(k-1)})$. As $x\to k^{-}$, $x-(k-1)\to k-(k-1)=1$. So, $li{m}_{x\to k^{-}}f(x)=(k-1)+\sqrt{1}=k-1+1=k$.

Since $f(k)={\lim }_{x\to k^{+}}f(x)={\lim }_{x\to k^{-}}f(x)=k$, $f(x)$ is continuous at all integer points. Between integer points, $\lfloor x\rfloor$ is constant and $\{x\}$ is $x-k$, so $f(x)$ is continuous. Thus, $f(x)$ is a continuous function for all $x\in \mathbb{R}$. This statement is correct.

D) A periodic function $f(x)$ must satisfy $f(x+T)=f(x)$ for some $T>0$. Since $f(x)$ is injective (from A), it cannot be periodic unless its domain is a single point, which is not the case here. Also, $\lfloor x\rfloor$ is not periodic, which prevents $f(x)$ from being periodic. This statement is incorrect.