Let `f:R rarr R ,` then `f(x) = 2 x + |cos x|` is:
(a).One-one into (b).One-one and onto
(c).May-one and into (d).Many-one and onto

Options

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

Answer

To determine the nature of the function \( f(x) = 2x + |\cos x| \), we will analyze its properties step by step.

### Step 1: Determine if the function is one-one (injective)

To check if \( f(x) \) is one-one, we need to analyze its derivative \( f'(x) \).

**Calculating the derivative:**
\[ f'(x) = \frac{d}{dx}(2x + |\cos x|) \]
The derivative of \( 2x \) is \( 2 \). For \( |\cos x| \), we need to consider two cases based on the value of \( \cos x \):

- **Case 1:** If \( \cos x \geq 0 \), then \( |\cos x| = \cos x \) and:
\[ f'(x) = 2 - \sin x \]

- **Case 2:** If \( \cos x < 0 \), then \( |\cos x| = -\cos x \) and:
\[ f'(x) = 2 + \sin x \]

**Analyzing the derivative:**
1. In **Case 1**:
- Since \( \sin x \) ranges from -1 to 1, we have:
\[ f'(x) = 2 - \sin x \geq 2 - 1 = 1 > 0 \]
Thus, \( f'(x) > 0 \) implies \( f(x) \) is increasing.

2. In **Case 2**:
- Similarly, we have:
\[ f'(x) = 2 + \sin x \geq 2 - 1 = 1 > 0 \]
Thus, \( f'(x) > 0 \) implies \( f(x) \) is also increasing.

Since \( f'(x) > 0 \) in both cases, \( f(x) \) is strictly increasing for all \( x \in \mathbb{R} \). Therefore, \( f(x) \) is one-one.

### Step 2: Determine if the function is onto (surjective)

To check if \( f(x) \) is onto, we need to analyze the range of \( f(x) \).

**Finding the range of \( f(x) \):**
1. The term \( 2x \) can take any real value as \( x \) varies over \( \mathbb{R} \).
2. The term \( |\cos x| \) ranges from \( 0 \) to \( 1 \).

Thus, the minimum value of \( f(x) \) occurs when \( |\cos x| = 0 \) (which happens at \( x = \frac{\pi}{2} + n\pi \), for \( n \in \mathbb{Z} \)):
\[ f(x) \geq 2x \quad \text{(as } |\cos x| \text{ is non-negative)} \]
As \( x \) approaches \( -\infty \), \( 2x \) approaches \( -\infty \). As \( x \) approaches \( +\infty \), \( 2x \) approaches \( +\infty \).

Therefore, the range of \( f(x) \) is:
\[ (-\infty, +\infty) \]
Since the codomain is also \( \mathbb{R} \), we conclude that \( f(x) \) is onto.

### Conclusion

Since \( f(x) \) is both one-one and onto, we conclude that the function \( f(x) = 2x + |\cos x| \) is:

**Answer: (b) One-one and onto.**

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