If `f_(1)(x) = 2x + 3, f_(2)(x) = 3x^(2) + 5, f_(3)(x) = x + cos x` are defined from `R rarr R`, then `f_(1), f_(2)` and `f_(3)` are

Options

A. One-one onto
B. Many one into
C. One-one -into
D. Many one onto

Answer

To determine the nature of the functions \( f_1(x) = 2x + 3 \), \( f_2(x) = 3x^2 + 5 \), and \( f_3(x) = x + \cos x \), we will analyze each function one by one to check if they are one-one (injective) or many-one (not injective), and whether they are onto (surjective) or into (not surjective).

### Step 1: Analyze \( f_1(x) = 2x + 3 \)

1. **Determine if \( f_1 \) is one-one:**
- A function is one-one if it passes the horizontal line test. This means that for any two different values \( x_1 \) and \( x_2 \), \( f_1(x_1) \neq f_1(x_2) \).
- For \( f_1(x) = 2x + 3 \):
\[ f_1(x_1) = f_1(x_2) \implies 2x_1 + 3 = 2x_2 + 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2 \]
- Since \( x_1 = x_2 \), \( f_1 \) is one-one.

2. **Determine if \( f_1 \) is onto:**
- A function is onto if for every \( y \) in the codomain \( \mathbb{R} \), there exists an \( x \) in the domain such that \( f_1(x) = y \).
- For \( f_1(x) = y \):
\[ y = 2x + 3 \implies x = \frac{y - 3}{2} \]
- Since \( x \) can take any real value, \( f_1 \) is onto.

### Conclusion for \( f_1 \):
- \( f_1 \) is **one-one** and **onto**.

### Step 2: Analyze \( f_2(x) = 3x^2 + 5 \)

1. **Determine if \( f_2 \) is one-one:**
- For \( f_2(x_1) = f_2(x_2) \):
\[ 3x_1^2 + 5 = 3x_2^2 + 5 \implies 3x_1^2 = 3x_2^2 \implies x_1^2 = x_2^2 \]
- This implies \( x_1 = x_2 \) or \( x_1 = -x_2 \). Therefore, \( f_2 \) is not one-one (many-one).

2. **Determine if \( f_2 \) is onto:**
- The minimum value of \( f_2(x) \) occurs at \( x = 0 \):
\[ f_2(0) = 3(0)^2 + 5 = 5 \]
- Since \( f_2(x) \) can only take values \( y \geq 5 \), it is not onto.

### Conclusion for \( f_2 \):
- \( f_2 \) is **many-one** and **into**.

### Step 3: Analyze \( f_3(x) = x + \cos x \)

1. **Determine if \( f_3 \) is one-one:**
- Differentiate \( f_3 \):
\[ f_3'(x) = 1 - \sin x \]
- Set the derivative to zero to find critical points:
\[ 1 - \sin x = 0 \implies \sin x = 1 \implies x = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]
- The function \( f_3'(x) \) is positive except at points where \( \sin x = 1 \), indicating that \( f_3 \) is increasing everywhere else. Thus, \( f_3 \) is one-one.

2. **Determine if \( f_3 \) is onto:**
- As \( x \to -\infty \), \( f_3(x) \to -\infty \) and as \( x \to \infty \), \( f_3(x) \to \infty \). Therefore, \( f_3 \) can take all real values.

### Conclusion for \( f_3 \):
- \( f_3 \) is **one-one** and **onto**.

### Final Summary:
- \( f_1 \): One-one and onto
- \( f_2 \): Many-one and into
- \( f_3 \): One-one and onto