If `f(x) =2 [x]+ cos x`, then `f: R ->R` is: (where [ ] denotes greatest integer function)

Options

A. one-one and onto
B. one-one and into
C. many-one and into
D. many-one and onto

Explanation

To determine the nature of the function \( f(x) = 2[x] + \cos x \), where \([x]\) denotes the greatest integer function, we will analyze its domain, range, and whether it is one-to-one (injective) and onto (surjective).

### Step 1: Determine the Domain
The function \( f(x) = 2[x] + \cos x \) is defined for all real numbers \( x \). The greatest integer function \([x]\) is defined for all \( x \), and since \(\cos x\) is also defined for all \( x\), we conclude:
- **Domain**: \( \mathbb{R} \)

### Step 2: Determine the Co-domain
The co-domain is the set of values that \( f(x) \) can take. Since we are considering \( f: \mathbb{R} \to \mathbb{R} \), we initially assume:
- **Co-domain**: \( \mathbb{R} \)

### Step 3: Analyze the Function
The function can be expressed as:
\[ f(x) = 2[x] + \cos x \]
Here, \([x]\) takes integer values, and \(\cos x\) oscillates between -1 and 1.

### Step 4: Determine the Range
1. For any integer \( n \) (where \( n = [x] \)), the term \( 2[x] \) will take values \( 2n \).
2. The term \( \cos x \) will add values between -1 and 1 to \( 2n \).

Thus, for \( n \) being any integer:
\[ f(x) = 2n + \cos x \quad \text{where } n \in \mathbb{Z} \]
This means:
\[ f(x) \text{ can take values in the intervals } [2n - 1, 2n + 1] \]
for each integer \( n \).

### Step 5: Determine if the Function is One-to-One
To check if \( f \) is one-to-one, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

1. Suppose \( f(x_1) = f(x_2) \).
2. This implies:
\[ 2[x_1] + \cos x_1 = 2[x_2] + \cos x_2 \]
3. If \( [x_1] = [x_2] \), then \( 2[x_1] = 2[x_2] \), and we have:
\[ \cos x_1 = \cos x_2 \]
4. The cosine function is periodic and not one-to-one; thus, \( x_1 \) could be different from \( x_2 \) while having the same cosine value.

Therefore, \( f(x) \) is **not one-to-one**.

### Step 6: Determine if the Function is Onto
To check if \( f \) is onto, we need to see if every real number \( y \) can be expressed as \( f(x) \).

1. For any integer \( n \), the values \( f(x) \) can take are in the intervals \( [2n - 1, 2n + 1] \).
2. As \( n \) varies over all integers, the intervals cover all real numbers.

Thus, the function is **onto**.

### Conclusion
The function \( f(x) = 2[x] + \cos x \) is:
- **Domain**: \( \mathbb{R} \)
- **Co-domain**: \( \mathbb{R} \)
- **Range**: \( \mathbb{R} \)
- **One-to-One**: No
- **Onto**: Yes

### Final Answer
The function \( f: \mathbb{R} \to \mathbb{R} \) is onto but not one-to-one.