Let`f(x)=x^(135)+x^(125)-x^(115)+x^(5)+1`. If `f(x)` divided by `x^(3)-x`, then the remainder is some function of `x` say `g(x)`. Then `g(x)` is an

Options

A. a)one-one function
B. b)many one function
C. c)into function
D. d)onto function

Explanation

To solve the problem, we need to find the remainder \( g(x) \) when \( f(x) = x^{135} + x^{125} - x^{115} + x^5 + 1 \) is divided by \( x^3 - x \).

### Step-by-Step Solution:

1. **Identify the Degree of the Remainder**:
When dividing a polynomial \( f(x) \) by another polynomial \( d(x) \) of degree \( n \), the remainder \( g(x) \) will have a degree less than \( n \). Here, \( d(x) = x^3 - x \) has a degree of 3. Therefore, the remainder \( g(x) \) will be of degree less than 3, which means it can be expressed as:
\[ g(x) = ax^2 + bx + c \]
where \( a, b, c \) are constants.

**Hint**: The degree of the remainder is always less than the degree of the divisor.

2. **Use Polynomial Long Division**:
We can express the relationship:
\[ f(x) = (x^3 - x)q(x) + g(x) \]
for some polynomial \( q(x) \).

3. **Evaluate \( f(x) \) at Specific Points**:
To find \( g(x) \), we can evaluate \( f(x) \) at the roots of \( x^3 - x = 0 \). The roots are \( x = 0, 1, -1 \).

- **At \( x = 0 \)**:
\[ f(0) = 0^{135} + 0^{125} - 0^{115} + 0^5 + 1 = 1 \]
Thus, \( g(0) = c = 1 \).

- **At \( x = 1 \)**:
\[ f(1) = 1^{135} + 1^{125} - 1^{115} + 1^5 + 1 = 1 + 1 - 1 + 1 + 1 = 3 \]
Therefore, \( g(1) = a(1^2) + b(1) + c = a + b + 1 = 3 \).
Simplifying gives:
\[ a + b = 2 \quad \text{(Equation 1)} \]

- **At \( x = -1 \)**:
\[ f(-1) = (-1)^{135} + (-1)^{125} - (-1)^{115} + (-1)^5 + 1 = -1 - 1 + 1 - 1 + 1 = -1 \]
Thus, \( g(-1) = a(-1)^2 + b(-1) + c = a - b + 1 = -1 \).
Simplifying gives:
\[ a - b = -2 \quad \text{(Equation 2)} \]

4. **Solve the System of Equations**:
We now have two equations:
- \( a + b = 2 \) (Equation 1)
- \( a - b = -2 \) (Equation 2)

Adding these two equations:
\[ 2a = 0 \implies a = 0 \]
Substituting \( a = 0 \) into Equation 1:
\[ 0 + b = 2 \implies b = 2 \]
Thus, we have \( a = 0, b = 2, c = 1 \).

5. **Write the Remainder**:
Therefore, the remainder \( g(x) \) is:
\[ g(x) = 0 \cdot x^2 + 2x + 1 = 2x + 1 \]

6. **Determine the Type of Function**:
The function \( g(x) = 2x + 1 \) is a linear function. A linear function is both one-one (injective) and onto (surjective) when considered over the real numbers.

### Conclusion:
The function \( g(x) \) is a **one-one function**.