For a real number x, let [x] denote the greatest integer less than or equal to x. Let f: R`->` R be defined as `f(x)= 2x + [x]+ sin x cos x` then f is

Options

A. One-one but not onto
B. onto but not one-one
C. Both one-one and onto
D. Neither one-one nor onto

Answer

To determine the properties of the function \( f(x) = 2x + [x] + \sin x \cos x \), where \([x]\) is the greatest integer less than or equal to \(x\), we will check if the function is one-to-one (injective) and onto (surjective).

### Step 1: Analyze the function components

The function consists of three parts:
1. \(2x\) - a linear function that is increasing.
2. \([x]\) - the greatest integer function, which is a step function and is constant over intervals of the form \([n, n+1)\) for integers \(n\).
3. \(\sin x \cos x\) - a periodic function with a range of \([-0.5, 0.5]\).

### Step 2: Check if \(f(x)\) is one-to-one

To check if \(f(x)\) is one-to-one, we need to see if it is strictly increasing or strictly decreasing.

- For \(x\) in the interval \([n, n+1)\) (where \(n\) is an integer):
- \([x] = n\)
- Thus, \(f(x) = 2x + n + \sin x \cos x\)

The derivative of \(f(x)\) with respect to \(x\) is:
\[ f'(x) = 2 + \frac{d}{dx}(\sin x \cos x) \]
Using the product rule:
\[ \frac{d}{dx}(\sin x \cos x) = \cos^2 x - \sin^2 x \]
Thus,
\[ f'(x) = 2 + \cos^2 x - \sin^2 x \]
Since \(\cos^2 x - \sin^2 x\) oscillates between -1 and 1, \(f'(x)\) is always greater than or equal to 1. Therefore, \(f(x)\) is strictly increasing in each interval \([n, n+1)\).

### Step 3: Check if \(f(x)\) is onto

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

- Since \([x]\) takes integer values and \(\sin x \cos x\) oscillates between \(-0.5\) and \(0.5\), the function \(f(x)\) will take on values that are not continuous due to the step nature of \([x]\).
- As \(x\) approaches \(n+1\) from the left, \(f(x)\) approaches \(2(n+1) + n + \sin(n+1)\cos(n+1)\), while as \(x\) approaches \(n\) from the right, \(f(x)\) jumps to \(2n + (n+1) + \sin n \cos n\). This means there are gaps in the values of \(f(x)\).

### Conclusion

1. **One-to-One**: The function \(f(x)\) is one-to-one because it is strictly increasing in each interval \([n, n+1)\).
2. **Onto**: The function \(f(x)\) is not onto because the discontinuities introduced by the greatest integer function create gaps in the range of \(f(x)\).

Thus, the final answer is that \(f(x)\) is **one-to-one but not onto**.