Let A = {7, 8, 9, 10, 11, 12, 13; 14, 15, 16} and let f ∶ A → N be defined by f(x) = the highest prime factor of x. How many elements are there in the range of f?
6
The problem asks us to determine the number of elements in the range of a function f, defined over a specific set A. The function f(x) gives the highest prime factor of the element x.
First, let's list the elements of the set A:
\(A = \{7, 8, 9, 10, 11, 12, 13, 14, 15, 16\}\)
The function is defined as \(f: A \rightarrow N\), where \(f(x) = \text{the highest prime factor of } x\).
The range of the function f is the set of all possible values of f(x) for each x in the domain A. To find the range, we need to calculate f(x) for every element in A and then collect the unique values.
Let's calculate the highest prime factor for each element in A:
| Element \(x\) | Prime Factorization of \(x\) | Highest Prime Factor \(f(x)\) |
|---|---|---|
| 7 | 7 | 7 |
| 8 | \(2 \times 2 \times 2\) | 2 |
| 9 | \(3 \times 3\) | 3 |
| 10 | \(2 \times 5\) | 5 |
| 11 | 11 | 11 |
| 12 | \(2 \times 2 \times 3\) | 3 |
| 13 | 13 | 13 |
| 14 | \(2 \times 7\) | 7 |
| 15 | \(3 \times 5\) | 5 |
| 16 | \(2 \times 2 \times 2 \times 2\) | 2 |
The values of f(x) for \(x \in A\) are {7, 2, 3, 5, 11, 3, 13, 7, 5, 2}.
The range of f is the set of these values, considering only the unique elements. Let's list the unique values:
The set of unique values in the range of f is {2, 3, 5, 7, 11, 13}.
To find the number of elements in the range, we simply count the number of unique values we found.
Number of elements in the range = 6.
Therefore, there are 6 elements in the range of the function f.
| Concept | Description | Example |
|---|---|---|
| Domain | The set of all possible input values for a function. | For \(f(x)\) in this problem, the domain is set A. |
| Range | The set of all possible output values of a function. | For \(f(x)\) in this problem, the range is the set of unique highest prime factors of elements in A. |
| Prime Factor | A prime number that divides a given number exactly. | Prime factors of 12 are 2 and 3. |
| Highest Prime Factor | The largest prime number that divides a given number. | Highest prime factor of 12 is 3. |
This problem combines concepts from set theory (understanding sets and functions) and number theory (prime factorization).
Consider the following statements:
Statement 1: The function f : R → R such that f(x) = x 3for all x ∈ R is one-one
Statement 2: f(a) = f(b) ⇒ a = b for all a, b ∈ R if the function f is one-one.
Which one of the following is correct in respect of the above statements?
If \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}},\) then what is \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) equal to?
Let R be a relation from N to N defined by R = {(x, y): x, y ∈ N and x 2 = y 3}. Which of the following are not correct?
1. (x, x) ∈ R for all x ∈ N
2. (x, y) ∈ R ⇒ (y, x) ∈ R
3. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R
Select the correct answer using the code given below :
If \(\displaystyle\sum_{x=2}^n\) f(x) = 2044, then what is the value of n ?
What is \(\displaystyle\sum_{x=1}^5\) f(2x − 1) equal to ?
What is \(\displaystyle\sum_{x=1}^{6} 2^{x}\,f(x)\) equal to?
If f(α) = \(\sqrt{\sec^2\alpha−1}\) , then what is \(\frac{f(\alpha)+f(\beta)}{1−f(\alpha) f(\beta)}\) equal to ?
Let f(x) be a function such that f'(x) = g(x) and f''(x) = −f(x). Let h(x) = {f(x)} 2+ {g(x)} 2. Then consider the following statements :
1. h'(3) = 0
2. h(1) = h(2)
Which of the statements given above is/are correct ?
Consider the following statements in respect of the function f(x) = \(\left\{\begin{array}{rc}|x|+1, & 0<|x| \leqslant 3 \\ 1, & x = 0\end{array}\right.\)
1. The function attains maximum value only at x = 3
2. The function attains local minimum only at x = 0
Which of the statements given above is/are correct ?
If the function of \(f(x)=\dfrac{x}{x-1}\) express f(3x) in terms of f(x)
If \(f(x)-\dfrac{1}{1+2^{1/x}}\) then at x = 0 the function is:
If f(x) is a periodic function and a is a positive real number such that f(x + 2α) + f(x) = 0 for all x ∈ ℝ, then the period of f(x) is:
If \(f(x)=\frac{1}{1+x}\), g(x) = f{f(x)} and h(x) = f[f{f(x)}], then the value of f(x).g(x).h(x) is:
If \(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right]\), \(x \in \left[ { - \frac{1}{2},1} \right]\), then f(x) will be