Key Concepts and Formulas

• One-one (Injective) Function: A function $h: A \rightarrow B$ is one-one if distinct elements in the domain map to distinct elements in the codomain. Mathematically, $x_1 \neq x_2 \implies h(x_1) \neq h(x_2)$, or equivalently, $h(x_1) = h(x_2) \implies x_1 = x_2$.
• Onto (Surjective) Function: A function $h: A \rightarrow B$ is onto if every element in the codomain $B$ has at least one pre-image in the domain $A$. This means the range of the function is equal to its codomain.
• Prime Factorization: Every integer greater than 1 can be uniquely represented as a product of prime numbers.

Step-by-Step Solution

We are given functions $f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$. The domain is $\mathbb{N}-\{1\} = \{2, 3, 4, \dots\}$. The codomain is $\mathbb{N} = \{1, 2, 3, \dots\}$.

The function $f(a)$ is defined as the maximum power of any prime $p$ such that $p^\alpha$ divides $a$. The function $g(a)$ is defined as $g(a) = a+1$.

We need to analyze the properties (one-one and onto) of the function $(f+g)(a) = f(a) + g(a) = f(a) + a + 1$.

Step 1: Understanding the function $f(a)$

Let's analyze the behavior of $f(a)$ with examples:

• If $a$ is a prime number, say $a=p$, then $a = p^1$. The maximum power of any prime dividing $a$ is $1$. So, $f(p)=1$.
  • $f(2) = 1$
  • $f(3) = 1$
  • $f(5) = 1$
• If $a$ is a composite number:
  • $f(4)$: $4 = 2^2$. The maximum power is $2$. So, $f(4)=2$.
  • $f(6)$: $6 = 2^1 \cdot 3^1$. The powers of prime factors are $1$ and $1$. The maximum is $1$. So, $f(6)=1$.
  • $f(8)$: $8 = 2^3$. The maximum power is $3$. So, $f(8)=3$.
  • $f(12)$: $12 = 2^2 \cdot 3^1$. The powers are $2$ and $1$. The maximum is $2$. So, $f(12)=2$.

From these examples, we observe that $f(a) \ge 1$ for all $a \in \mathbb{N}-\{1\}$.

Step 2: Analyzing the one-one property of $(f+g)(a)$

To check if $(f+g)$ is one-one, we look for two distinct inputs $a_1, a_2$ such that $(f+g)(a_1) = (f+g)(a_2)$.

Let's compute $(f+g)(a)$ for some small values of $a$:

• For $a=2$:

  • $f(2) = 1$
  • $g(2) = 2+1 = 3$
  • $(f+g)(2) = f(2) + g(2) = 1 + 3 = 4$.

• For $a=3$:

  • $f(3) = 1$
  • $g(3) = 3+1 = 4$
  • $(f+g)(3) = f(3) + g(3) = 1 + 4 = 5$.

• For $a=4$:

  • $f(4) = 2$ (since $4=2^2$)
  • $g(4) = 4+1 = 5$
  • $(f+g)(4) = f(4) + g(4) = 2 + 5 = 7$.

• For $a=5$:

  • $f(5) = 1$ (since $5$ is prime)
  • $g(5) = 5+1 = 6$
  • $(f+g)(5) = f(5) + g(5) = 1 + 6 = 7$.

We found that $(f+g)(4) = 7$ and $(f+g)(5) = 7$. Since $4 \neq 5$ but their images under $(f+g)$ are equal, the function $(f+g)$ is not one-one.

Step 3: Analyzing the onto property of $(f+g)(a)$

To check if $(f+g)$ is onto, we need to see if its range covers the entire codomain $\mathbb{N} = \{1, 2, 3, \dots\}$. This means for every $y \in \mathbb{N}$, there must exist an $a \in \mathbb{N}-\{1\}$ such that $(f+g)(a) = y$.

Let's determine the minimum possible value of $(f+g)(a)$.

• We know $f(a) \ge 1$ for all $a \in \mathbb{N}-\{1\}$.
• The domain for $a$ is $\{2, 3, 4, \dots\}$. Therefore, $a \ge 2$.
• This implies $g(a) = a+1 \ge 2+1 = 3$.

So, for any $a \in \mathbb{N}-\{1\}$: $(f+g)(a) = f(a) + g(a) \ge 1 + 3 = 4$ The minimum value that $(f+g)(a)$ can take is $4$.

This means the range of $(f+g)$ is a subset of $\{4, 5, 6, \dots\}$. The values $1, 2,$ and $3$ are in the codomain $\mathbb{N}$ but are not in the range of $(f+g)$. For instance, there is no $a \in \mathbb{N}-\{1\}$ such that $(f+g)(a) = 1$, $(f+g)(a) = 2$, or $(f+g)(a) = 3$. Therefore, the function $(f+g)$ is not onto.

Common Mistakes & Tips

• Misinterpreting $f(a)$: Carefully read the definition of $f(a)$. It's the maximum power of any prime factor. For example, for $a=12=2^2 \cdot 3^1$, $f(12)=2$, not $2+1=3$ or $2 \cdot 1 = 2$.
• Checking for Onto: When proving a function is not onto, it's sufficient to find just one element in the codomain that is not in the range. Analyzing the minimum value of the function is a common strategy.
• Finding Counterexamples for One-one: If you suspect a function is not one-one, try testing small composite numbers or numbers with different prime factorizations that might lead to the same output.

Summary

We analyzed the function $(f+g)(a) = f(a) + a + 1$. By testing specific values, we found that $(f+g)(4) = 7$ and $(f+g)(5) = 7$, demonstrating that the function is not one-one. By examining the minimum possible values of $f(a)$ and $g(a)$, we determined that $(f+g)(a) \ge 4$ for all $a$ in the domain, meaning the values $1, 2, 3$ in the codomain are not reached, proving the function is not onto.

The final answer is \boxed{A}.