JEE 2023 — Mathematics PYQ

The values of $f(1),f(2),$ and $f(4)$ must come from set $B$. Since the smallest value in $B$ is $1$, we know $f(1)\ge 1$ and $f(2)\ge 1$. This means their sum $(f(1)+f(2))$ must be at least $2$.

From the equation $f(1)+f(2)=f(4)-1$, we can see that $f(4)$ must be large enough to satisfy this sum.

Step 2: Case-by-Case Breakdown

We check all possible values for $f(4)$ from set $B$:

• If $f(4)=3$: Then $f(1)+f(2)=2$. Only $(1,1)$ works. (1 way)

• If $f(4)=4$: Then $f(1)+f(2)=3$. Pairs: $(1,2),(2,1)$. (2 ways)

• If $f(4)=5$: Then $f(1)+f(2)=4$. Pairs: $(1,3),(3,1),(2,2)$. (3 ways)

• If $f(4)=6$: Then $f(1)+f(2)=5$. Pairs: $(1,4),(4,1),(2,3),(3,2)$. (4 ways)

(Note: $f(4)$ cannot be $1$ or $2$ because the sum would be $0$ or $1$, which isn't possible for $f(1)+f(2)$).

Total ways for $f(1),f(2),$ and $f(4)$ $=1+2+3+4=10$.

Step 3: Handle the Remaining Elements

There are two elements left in Domain $A$: $3$ and $5$.

• There are no conditions for $f(3)$ and $f(5)$.

• Each can be mapped to any of the $6$ elements in set $B$.

• Ways for $f(3)=6$

• Ways for $f(5)=6$

Step 4: Final Calculation

To get the total number of functions, we multiply the ways for all elements:

Final Answer: The number of such functions is $360$.