Animated Solution for Mathematics - Functions: Let $f,g:\mathbb{N}-\{1\}\to \mathbb{N}$ be functions defined by $f(a)=\alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha }$ divides $a$, and $g(a)=a+1$, for all $a\in \mathbb{N}-\{1\}$. Then, the function $f+g$ is

Welcome, fellow explorer of the mathematical universe! Today, we are going to dissect a problem that might seem like a simple function mapping, but it is actually a beautiful dance between the structure of integers and the behavior of functions.

We are looking at two functions, $f$ and $g$, defined on the set of natural numbers excluding one, $\mathbb{N}-\{1\}$. Our mission is to understand the nature of their sum, $h(a)=f(a)+g(a)$.

Let us first demystify $f(a)$. The problem defines $f(a)=\alpha$, where $\alpha$ is the maximum power of a prime $p$ such that $p^{\alpha }$ divides $a$.

Think of this as the 'prime-power signature' of a number. If you take a number like $12$, its prime factorization is $2^2\cdot 3^1$. The exponents are $2$ and $1$. The maximum of these is $2$, so $f(12)=2$.

If you take a prime number like $5$, it is just $5^1$, so $f(5)=1$. This function essentially extracts the 'heaviest' prime component of a number's DNA.

Now, we introduce $g(a)=a+1$. Our combined function is defined as:

To understand if this function is one-one or onto, we must test it. Let us start with the smallest possible input, $a=2$.

Here, $f(2)=1$ (since $2=2^1$), so $h(2)=1+2+1=4$. Next, for $a=3$, $f(3)=1$ (since $3=3^1$), so $h(3)=1+3+1=5$.

Is it One-One?

But here is where the plot thickens. Let us look at $a=4$. The prime factorization of $4$ is $2^2$. The maximum prime power is $2$, so $f(4)=2$.

Then, $h(4)=2+4+1=7$. Now, look at $a=5$. Since $5$ is prime, $f(5)=1$. Then, $h(5)=1+5+1=7$.

Stop right there! Do you see it? We have $h(4)=7$ and $h(5)=7$. Two distinct inputs, $4$ and $5$, have collided into the same output, $7$.

In the world of functions, this is a 'many-to-one' relationship. Because we found two different inputs that map to the same output, the function fails the test for being one-one. It is not injective.

Is it Onto?

Now, let us tackle the second part: is the function onto? For a function to be onto, every element in the codomain must be 'hit' by at least one element from the domain.

Our codomain is the set of all natural numbers $\mathbb{N}=\{1,2,3,4,\dots \}$. Let us find the minimum value of $h(a)$. We know $a\ge 2$ and $f(a)\ge 1$ for all $a\in \mathbb{N}-\{1\}$.

Therefore, the minimum value of $h(a)=f(a)+a+1$ is:

This means the range of our function is a subset of $\{4,5,6,\dots \}$. The values $1,2,$ and $3$ are sitting in the codomain, but no input $a$ can ever produce them.

Since the range does not equal the codomain, the function is not surjective.

We have walked through the logic, tested the values, and uncovered the truth. The function $f+g$ is neither one-one nor onto.

This leads us directly to the conclusion that the function is neither injective nor surjective. Remember, in JEE Advanced, it is not just about the calculation; it is about the investigation. Keep exploring, keep questioning, and keep falling in love with the logic behind the math!