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Suppose $A:=\{x\in [0,1]: x=\frac{p}{2^q},p\in \mathbb Z, q\in \mathbb N\}$ and $B:=\mathbb Z[\sqrt 2]\bigcap[0,1]$. Let $f:[0,1]\to \mathbb R$ be a function defined by $$\begin{align} f(x) = \begin{cases} x+\frac{3}{2}, & \text{if $x\in A$}\\ 2x+1, & \text{if $x\in B$}\\ 2 ,&\text{elsewhere} \end{cases} \end{align} $$ Then which of the following is or are correct?

$a$). $f(x) $ is continuous at all points $x\notin A\bigcup B$.

$b$). $f(x)$ is discontinuous at each point of its domain.

$c).$ $f(x)$ is differentiable at exactly one point of its domain.

$d).$ $f(x)$ is continuous at exactly one point but differentiable nowhere.

The question's appearance made me to link the procedure of solving with Proof of continuity of Thomae Function at irrationals. and Proving Thomae's function is nowhere differentiable..

$1.$ To check the continuity:

I do such questions by defining $A_{\alpha}=\{(x_n):(x_n)\to x, x_n\in S-\{\alpha \}\}$ whenever $f:S\subset \mathbb R\to \mathbb R $ and $x \in S'$, $B_{\alpha}=\{(f(x_n)):(x_n)\in A_{\alpha}\}$ and $C_{\alpha}=\text{The set of all limit points of the sequence in}$ $ B_\alpha$.

Now here we have $f:[0,1]\to\mathbb R$, choose $(a_n)\in A_{\alpha}$ then consequently my $f(x_n)\to f(x)$ here I will have:

$$\begin{align} f(x_n) \to \begin{cases} x_n+\frac{3}{2}, & \text{if $x_n\in A$}\\ 2x_n+1, & \text{if $x_n\in B$}\\ 2 ,&\text{elsewhere} \end{cases} \end{align}$$ i.e.

$$\begin{align} f(x_n) \to \begin{cases} x+\frac{3}{2}\\ 2x+1\\ 2\\ \end{cases} \end{align}$$

Since continuity is a property on the members of domain so firstly I checked it at $x=0$ for the set $A$ $\lim_{x\to 0}(x+\frac{3}{2})\to \frac{3}{2}$.

And similarly for the set $B$ we have that $\lim_{x\to0}(2x+1)\to 1$.

Clearly without even ckecking what $f(0)$ is, we can claim that $f(x)$ is discontinuous at $x=0$. Similar is the argument for $x=1$.

Now to see the point of at which $f(x)$ is continuous we will have $x+\frac{3}{2}=2x+1$ that gives $x=\frac{1}{2}$.

Consequently, at $x=\frac{1}{2}$ we clearly have the limit $2$ and $f(\frac{1}{2})$ i.e. for set $A$, we have $p=1,q=1$ again its $2$, for $B$, it will look like $B=\{a+b\sqrt2:a,b\in\mathbb Z\}\cap[0,1]$ here we can choose $a=\frac{1}{2},b=0$. So clearly my function is continuous exactly at one point.

$2$.To check the differentiability:

Points where my function is discontinuous, my function will not be differentiable. So I will directly check at $x=\frac{1}{2}$.

For set $A$ :

$\lim_{x\to\frac{1}{2}}\frac{f(x)-f(\frac{1}{2})}{x-\frac{1}{2}}=1$.

For set $B$:

$\lim_{x\to\frac{1}{2}}\frac{f(x)-f(\frac{1}{2})}{x-\frac{1}{2}}=2$.

As both of these are not equal, thus my function is not differentiable at $x=\frac{1}{2}$. So $d).$ is the only correct option.

Is there any correction required?

Thanks.

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Is the point of the exercise strictly to pick which of a, b, c, or d are correct? If so, then you are mostly good. However, if you are expected to demonstrate that d is true, then you have not done so. What you have done is show a, b, c are false. You have not shown that d is true.

Why you have not shown d is true:

(1) you assume without proof that $A, B$, and $[0,1] \setminus (A \cup B)$ are all dense. If those have already been proven elsewhere in your course (in the book, in class, or as another exercise), that is okay. But if not, you need to add proofs for them. That $A$ and $[0,1] \setminus (A \cup B)$ are dense are pretty easy (for the latter, note that $A\cup B$ is countable). You can actually prove all the results except continuity at $\frac 12$ from those two alone. To prove continuity at $\frac 12$, you only need that the piecewise function for $B$ is the restriction of a continuous function on $[0,1]$ to $B$, and that function agrees with the other two at $\frac 12$. The density of $B$ is not required (but you still have to phrase it so you are not assuming $B$ is dense).

(2) You have not shown that $f$ is discontinuous anywhere except at $0$.

(3) Your argument that $f$ is continuous at $\frac 12$ is fragmentary at best. You make no mention of the behavior of the third function, though it is also needed. And you offer no argument as to why it is sufficient that the three functions agree at $\frac 12$.

(4) $\frac 12 \notin \Bbb Z$, so this statement is false:

for $B$, it will look like $B=\{a+b\sqrt2:a,b\in\mathbb Z\}\cap[0,1]$ here we can choose $a=\frac{1}{2},b=0$.

In fact, $\frac 12\notin B$. Every element of $B$ is irrational except $0,1$, while every element of $A$ is rational. Since $0,1 \in A, B$, $f$ is not even well-defined at those points, as the $A$ and $B$ functions provide different values for them.

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