JEE Main 2026 — Functions Question with Solution

Given $A=\{1,2,3,4,5,6\}$. Since $f:A\to A$ is a one-one function, $f(x)$ takes distinct values from $A$ for each $x\in A$.

We are given the condition $f(2)+f(3)=5$. The possible pairs for $(f(2),f(3))$ from the set $A$ are:
$(1,4),(2,3),(3,2),(4,1)$

Notice that for all these pairs, the condition $f(3)\le 4$ is automatically satisfied.

We also have the condition $f(1)\ge 3$. Since $f$ is one-one, $f(1)$ cannot take the values already assigned to $f(2)$ and $f(3)$. Let us find the number of choices for $f(1)$ in each case:

Case 1: $(f(2),f(3))=(1,4)$
$f(1)\ge 3$ and $f(1)\notin \{1,4\}\Rightarrow f(1)\in \{3,5,6\}$. (3 choices)

Case 2: $(f(2),f(3))=(2,3)$
$f(1)\ge 3$ and $f(1)\notin \{2,3\}\Rightarrow f(1)\in \{4,5,6\}$. (3 choices)

Case 3: $(f(2),f(3))=(3,2)$
$f(1)\ge 3$ and $f(1)\notin \{3,2\}\Rightarrow f(1)\in \{4,5,6\}$. (3 choices)

Case 4: $(f(2),f(3))=(4,1)$
$f(1)\ge 3$ and $f(1)\notin \{4,1\}\Rightarrow f(1)\in \{3,5,6\}$. (3 choices)

In each of the 4 cases, there are exactly 3 choices for $f(1)$.

After assigning values to $f(1),f(2),$ and $f(3)$, there are exactly 3 elements left in the codomain to be assigned to $f(4),f(5),$ and $f(6)$. Since $f$ is one-one, these can be assigned in $3!=6$ ways.

Total number of such functions = (Number of pairs) $\times$ (Choices for $f(1)$) $\times$ (Arrangements for remaining elements)
Total = $4\times 3\times 6=72$

Answer: $72$