If `f:R rarr R` and f(x)=g(x)+h(x) where g(x) is a polynominal and h(x) is a continuous and differentiable bounded function on both sides, then f(x) is one-one, we need to differentiate f(x). If f'(x) changes sign in domain of f, then f, if many-one else one-one.
If `f:R rarr R` and f(x)=2ax +sin2x, then the set of values of a for which f(x) is one-one and onto is

Options

A. `a in (-1/2,1/2)`
B. `a in (-1,1)`
C. `a in R -(-1/2,1/2)`
D. `a in R-(-1,1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2ax + \sin(2x) \) and determine the values of \( a \) for which \( f(x) \) is one-one and onto.

### Step 1: Differentiate \( f(x) \)
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2ax + \sin(2x)) \]
Using the rules of differentiation, we get:
\[ f'(x) = 2a + \frac{d}{dx}(\sin(2x)) = 2a + 2\cos(2x) \]
Thus, we can simplify this to:
\[ f'(x) = 2(a + \cos(2x)) \]

### Step 2: Analyze the derivative \( f'(x) \)
Next, we need to analyze when \( f'(x) \) changes sign. The term \( \cos(2x) \) oscillates between -1 and 1. Therefore, we can express the range of \( f'(x) \):
\[ f'(x) = 2(a + \cos(2x)) \]
The critical points occur when \( f'(x) = 0 \):
\[ 2(a + \cos(2x)) = 0 \implies a + \cos(2x) = 0 \implies \cos(2x) = -a \]

### Step 3: Determine the values of \( a \)
Since \( \cos(2x) \) can take values in the interval \([-1, 1]\), we need:
\[ -1 \leq -a \leq 1 \]
This gives us two inequalities:
1. \( -1 \leq -a \) which simplifies to \( a \leq 1 \)
2. \( -a \leq 1 \) which simplifies to \( a \geq -1 \)

Thus, we have:
\[ -1 \leq a \leq 1 \]

### Step 4: Conclusion about one-one and onto
- If \( a \) is in the interval \( (-1, 1) \), then \( f'(x) \) does not change sign (since \( f'(x) \) will always be positive), meaning \( f(x) \) is a one-one function.
- If \( a = -1 \) or \( a = 1 \), then \( f'(x) \) can equal zero for some \( x \), which means \( f(x) \) could be many-one.

### Final Answer
The set of values of \( a \) for which \( f(x) \) is one-one and onto is:
\[ a \in (-1, 1) \]