Which of the following functions is one-one?

Options

A. `fR to R " is given by"f(x)=2x^(1)+1"For all " x in R`
B. `g:Z to Z " given by"g(x)=x^(4)"For all " x in Z`
C. `h:R to R " given h"(x)=x^(3)+4"For all " x in R`
D. `phi:C to C " given by "phi(z)=z^(3)+4"For all " z in C`

Explanation

To determine which of the given functions is one-one, we will analyze each function step by step. A function is one-one (injective) if it never assigns the same value to two different domain elements.

### Step-by-Step Solution:

1. **Option A: \( f(x) = 2x^2 + 1 \)**
- Let's evaluate \( f(1) \) and \( f(-1) \):
\[ f(1) = 2(1)^2 + 1 = 2 + 1 = 3 \]
\[ f(-1) = 2(-1)^2 + 1 = 2(1) + 1 = 2 + 1 = 3 \]
- Here, \( f(1) = f(-1) = 3 \) but \( 1 \neq -1 \).
- Therefore, this function is **not one-one**.

2. **Option B: \( g(x) = x^4 \)**
- Let's evaluate \( g(1) \) and \( g(-1) \):
\[ g(1) = (1)^4 = 1 \]
\[ g(-1) = (-1)^4 = 1 \]
- Here, \( g(1) = g(-1) = 1 \) but \( 1 \neq -1 \).
- Therefore, this function is **not one-one**.

3. **Option C: \( h(x) = x^3 + 4 \)**
- Assume \( h(x) = h(y) \):
\[ x^3 + 4 = y^3 + 4 \]
- Subtracting 4 from both sides:
\[ x^3 = y^3 \]
- Taking the cube root:
\[ x = y \]
- Since \( x = y \) is the only solution, this function is **one-one**.

4. **Option D: \( z(x) = z^3 + 4 \)**
- Let's evaluate \( z(\omega) \) and \( z(\omega^2) \) where \( \omega \) is a cube root of unity:
\[ z(\omega) = \omega^3 + 4 = 1 + 4 = 5 \]
\[ z(\omega^2) = (\omega^2)^3 + 4 = 1 + 4 = 5 \]
- Here, \( z(\omega) = z(\omega^2) = 5 \) but \( \omega \neq \omega^2 \).
- Therefore, this function is **not one-one**.

### Conclusion:
The only function that is one-one among the options is **Option C: \( h(x) = x^3 + 4 \)**.