Q.Let $A = \{1, 2, 3, 4\}$ and $B = \{p, q, r, s, t\}$. Diagram

Check each diagram against the two function conditions — every element of $A$ has an arrow, and each has only ONE arrow — before reading off the domain, co-domain and range.

Diagram (i): $1\to p$, $2\to q$, $3\to r$, $4\to q$.

• Check condition 1 (every element of $A$ has an arrow): $1,2,3,4$ each have exactly one arrow leaving them. ✓
• Check condition 2 (no element has more than one arrow): each of $1,2,3,4$ has exactly one outgoing arrow, no duplicates. ✓
• Conclusion: this IS a function. Domain $= A = \{1,2,3,4\}$. Co-domain $= B = \{p,q,r,s,t\}$. Range $=$ the elements of $B$ actually hit by an arrow $= \{p,q,r\}$ — note $q$ is hit twice (by both $2$ and $4$) but is listed only once in the range, and $s,t$ are never hit, so they belong to the co-domain but not the range.

Diagram (ii): $1\to p$, $2\to q$, $2\to r$, $3\to s$ (element $4$ unmapped).

• Check condition 1: element $4$ has NO outgoing arrow at all. ✗ — this alone already breaks the function requirement. …

Confusing co-domain with range — the co-domain is the ENTIRE target set $B$ as stated in the problem, while the range is only the part of $B$ that actually gets an arrow; a common error is reporting $\{p,q,r,s,t\}$ as the range in diagram (i) instead of the correct, smaller $\{p,q,r\}$. Also, in diagram (ii), spotting only ONE of the two sep …