JEE Mathematics 2025 Question with Solution

Given: $A = \{1,2,3,4\}$ and $B = \{1,4,9,16\}$. We need the number of many-one functions $f: A \to B$ such that $1 \in f(A)$.

Find: The number of non-injective functions satisfying the given image condition.

Total number of functions from $A$ to $B$ is

$$4^4 = 256$$

Functions for which $1 \notin f(A)$ must map every element of $A$ into $\{4,9,16\}$. Their number is

$$3^4 = 81$$

So the number of functions with $1 \in f(A)$ is

$$256 - 81 = 175$$

Now subtract the one-to-one functions among these. Since $|A| = |B| = 4$, every injective function is a bijection, and every bijection contains $1$ in its image. Hence the number of injective functions is

$$4! = 24$$

Therefore the number of many-one functions is

$$175 - 24 = 151$$

Therefore, the correct option is D. The number of such functions is $151$.

Given: $f: A \to B$ where $A = \{1,2,3,4\}$ and $B = \{1,4,9,16\}$.

Find: Count functions that are many-one and satisfy $1 \in f(A)$.

First count all possible functions:

$$|B|^{|A|} = 4^4 = 256$$

Next, count functions that do not contain $1$ in the image. Then each element of $A$ has only $3$ choices:

$$3^4 = 81$$

Hence functions with $1$ appearing in the image are

$$256 - 81 = 175$$

A many-one function means the function is not one-to-one. So remove all injective functions from this count.

The injective functions from a $4$-element set to another $4$-element set are exactly all bijections:

$${}^4P_4 = 4! = 24$$

Thus the required number is

$$175 - 24 = 151$$

the solution itself concludes that the required count is $151$, even though the answer key lists a different option. Therefore, using the solution, the correct option is D.