QJEEM25Ma2054: JEE Main 2025 Mathematics - Class 11 JEEM25Ma2054 Solution: Let and . Then the number of many-one functions such that is…

$\textbf{Step 1: Total Functions with } 1 \in f(A)$

Any function $f: A \to B$ where $A = \{1, 2, 3, 4\}$ and $B = \{1, 4, 9, 16\}$ is defined by choosing one of the four elements of $B$ for each element of $A$ . Thus, the total number of functions is

$4^4 = 256.$

To count those functions where $1$ appears at least once in the set $f(A)$ , we can use the complementary counting method: subtract the functions that never use $1$ . If $1$ is excluded, each element of $A$ has only 3 choices (namely, $\{4, 9, 16\}$ ), so the number of such functions is

$3^4 = 81.$

Thus, the number of functions such that $1 \in f(A)$ is

$256 - 81 = 175.$

$\textbf{Step 2: Counting Many-One Functions}$

In this context, "many-one functions" are understood to be non-injective functions. Since an injective (one-to-one) function from $A$ to $B$ must be a permutation (because both sets have 4 elements), the number of one-to-one functions is

$4! = 24.$

It is important to note that every injective function $f: A \to B$ has $f(A) = B$ (a full permutation) which automatically means $1 \in f(A)$ .

Thus, the number of many-one (non-injective) functions $f: A \to B$ with $1 \in f(A)$ is found by subtracting the one-to-one functions from the total functions that include $1$ :

$175 - 24 = 151.$

$\boxed{151}$

This detailed explanation shows that the number of many-one functions $f: A \rightarrow B$ such that $1 \in f(A)$ is indeed $151$ .