Consider a function `f:R rarr R` defined by `f(x)=x^(3)+4x+5`, then

Options

A. f is one - one but not onto
B. f is onto but not one - one
C. f is one - one and onto
D. f is neither one - one nor onto

Answer

To determine whether the function \( f(x) = x^3 + 4x + 5 \) is one-to-one (injective) and onto (surjective), we will follow these steps:

### Step 1: Check if the function is one-to-one

To check if \( f \) is one-to-one, we assume that \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in \mathbb{R} \).

\[ f(x_1) = f(x_2) \implies x_1^3 + 4x_1 + 5 = x_2^3 + 4x_2 + 5 \]

Subtracting \( 5 \) from both sides:

\[ x_1^3 + 4x_1 = x_2^3 + 4x_2 \]

Rearranging gives:

\[ x_1^3 - x_2^3 + 4(x_1 - x_2) = 0 \]

### Step 2: Factor the equation

Using the identity for the difference of cubes, we can factor \( x_1^3 - x_2^3 \):

\[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) + 4(x_1 - x_2) = 0 \]

Factoring out \( (x_1 - x_2) \):

\[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2 + 4) = 0 \]

### Step 3: Analyze the factors

For the product to be zero, either:

1. \( x_1 - x_2 = 0 \) which implies \( x_1 = x_2 \)
2. \( x_1^2 + x_1x_2 + x_2^2 + 4 = 0 \)

The second factor \( x_1^2 + x_1x_2 + x_2^2 + 4 \) is always positive since \( x_1^2 + x_1x_2 + x_2^2 \) is non-negative (as it is a sum of squares) and adding 4 makes it strictly positive. Thus, it cannot be zero.

Therefore, the only solution is \( x_1 = x_2 \), which shows that \( f \) is one-to-one.

### Step 4: Check if the function is onto

To check if \( f \) is onto, we need to determine if the range of \( f \) covers all real numbers.

1. As \( x \to +\infty \), \( f(x) \to +\infty \).
2. As \( x \to -\infty \), \( f(x) \to -\infty \).

Since \( f(x) \) is a continuous polynomial function (specifically a cubic function), it will take all values between \( -\infty \) and \( +\infty \).

### Conclusion

Since the range of \( f \) is all real numbers, which matches the codomain, we conclude that \( f \) is onto.

### Final Answer

The function \( f(x) = x^3 + 4x + 5 \) is both one-to-one and onto.