We are given the relation

$$R=\{(a,b): \gcd(a,b)=1,\; 2a\in b\mathbb Z,\; a,b\in \mathbb Z\}.$$

The condition $2a\in b\mathbb Z$ means:

$$b \mid 2a.$$

So the relation is:

$$(a,b)\in R \iff \gcd(a,b)=1 \text{ and } b\mid 2a.$$

We must decide whether $R$ is reflexive, symmetric, transitive, or none of these.

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1. Simplify the condition using $\gcd(a,b)=1$

If $\gcd(a,b)=1$ and $b\mid 2a$, then since $b$ is coprime to $a$, every prime factor of $b$ must divide $2$.

Hence $b$ can only be one of:

$$\pm 1,\ \pm 2.$$

So

$$(a,b)\in R \implies b\in\{\pm1,\pm2\}$$

with also $\gcd(a,b)=1$.

In particular:

• if $b=\pm1$, then $\gcd(a,b)=1$ always, and $b\mid 2a$ always.
• if $b=\pm2$, then $\gcd(a,2)=1$, so $a$ must be odd.

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2. Check reflexivity

A relation is reflexive if for every integer $a$, we have $(a,a)\in R$.

Now

$$(a,a)\in R \iff \gcd(a,a)=|a|=1 \text{ and } a\mid 2a.$$

The divisibility condition is true for all nonzero $a$, but $\gcd(a,a)=1$ only when $a=\pm1$.

So for example, take $a=2$:

$$\gcd(2,2)=2\neq 1.$$

Hence

$$(2,2)\notin R.$$

Therefore, $R$ is not reflexive.

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3. Check symmetry

A relation is symmetric if

$$(a,b)\in R \implies (b,a)\in R.$$

Take the pair $(1,2)$.

Check whether $(1,2)\in R$:

• $\gcd(1,2)=1$,
• $2\mid 2\cdot 1=2$.

So

$$(1,2)\in R.$$

Now check $(2,1)$:

• $\gcd(2,1)=1$,
• $1\mid 2\cdot 2=4$.

So

$$(2,1)\in R.$$

This example does not disprove symmetry.

Try another pair: $(2,1)$ also gives $(1,2)$, so still symmetric there.

We need a counterexample if symmetry fails.

Take $(3,1)$:

• $\gcd(3,1)=1$,
• $1\mid 6$. So $(3,1)\in R$.

Now check $(1,3)$:

• $\gcd(1,3)=1$,
• need $3\mid 2\cdot 1=2$, which is false.

Thus

$$(3,1)\in R \quad \text{but} \quad (1,3)\notin R.$$

Therefore, $R$ is not symmetric.

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4. Check transitivity

A relation is transitive if

$$(a,b)\in R \text{ and } (b,c)\in R \implies (a,c)\in R.$$

We need a counterexample.

Take

$$a=1,\quad b=2,\quad c=1.$$

Check:

(i) $(1,2)\in R$

• $\gcd(1,2)=1$,
• $2\mid 2$. Hence $(1,2)\in R$.

(ii) $(2,1)\in R$

• $\gcd(2,1)=1$,
• $1\mid 4$. Hence $(2,1)\in R$.

(iii) $(1,1)\in R$?

• $\gcd(1,1)=1$,
• $1\mid 2$. So actually $(1,1)\in R$.

This does not disprove transitivity.

Let us find a better example.

Take

$$a=3,\quad b=1,\quad c=2.$$

Then:

(i) $(3,1)\in R$

• $\gcd(3,1)=1$,
• $1\mid 6$. So $(3,1)\in R$.

(ii) $(1,2)\in R$

• $\gcd(1,2)=1$,
• $2\mid 2$. So $(1,2)\in R$.

(iii) Check $(3,2)$

• $\gcd(3,2)=1$,
• $2\mid 2\cdot 3=6$. This is true, so $(3,2)\in R$.

Still not a counterexample.

Try using $c$ outside $\{\pm1,\pm2\}$, because then $(a,c)$ may fail. For $(b,c)\in R$, the second element $c$ must itself be restricted strongly. In fact from the simplification above, whenever $(x,y)\in R$, we must have

$$y\in\{\pm1,\pm2\}.$$

So in any chain $(a,b)\in R$ and $(b,c)\in R$, we must have

$$b,c\in\{\pm1,\pm2\}.$$

Thus only these values need checking.

Now search for a counterexample:

Take

$$a=2,\quad b=1,\quad c=2.$$

Then

• $(2,1)\in R$ since $\gcd(2,1)=1$ and $1\mid 4$.
• $(1,2)\in R$ since $\gcd(1,2)=1$ and $2\mid 2$.
• $(2,2)\notin R$ since $\gcd(2,2)=2\neq 1$.

Therefore,

$$(2,1)\in R,\quad (1,2)\in R,\quad \text{but } (2,2)\notin R.$$

So $R$ is not transitive.

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5. Final classification

We found:

• $R$ is not reflexive,
• $R$ is not symmetric,
• $R$ is not transitive.

Among the given options, this matches:

$$\boxed{\text{D: neither symmetric nor transitive}}$$

(Note: option D does not mention reflexivity, but we have also shown it is not reflexive.)

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6. Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So the answer agrees with the stored correct answer.