Q. For every integer n, let`a_(n)` and `b_(n)` be real numbers. Let function `f:R->R` be given by a `f(x)={a_n+sin pix, for x in [2n,2n+1]`, `b_n+cos pix, for x in (2n+1,2n)` for all integers n.

Options

A. `a_(n-1) - b_(n-1) = 0`
B. `a_(n) - b_(n) = 1`
C. `a_(n) - b_(n+1) = 1`
D. `a_(n-1) -b_(n) = -1`

Answer

To solve the problem, we need to analyze the function \( f \) defined piecewise for every integer \( n \):

\[ f(x) = \begin{cases} a_n + \sin(\pi x) & \text{for } x \in [2n, 2n+1] \\ b_n + \cos(\pi x) & \text{for } x \in (2n+1, 2n+2) \end{cases} \]

We will check the continuity of \( f \) at the points \( x = 2n \) and \( x = 2n + 1 \).

### Step 1: Continuity at \( x = 2n \)

1. **Left-Hand Limit (LHL)** as \( x \) approaches \( 2n \):
\[ \text{LHL} = \lim_{x \to 2n^-} f(x) = \lim_{h \to 0} f(2n - h) = b_n + \cos(\pi(2n - h)) \]
Since \( \cos(\pi(2n - h)) = \cos(\pi h) \) and \( \cos(0) = 1 \):
\[ \text{LHL} = b_n + 1 \]

2. **Right-Hand Limit (RHL)** as \( x \) approaches \( 2n \):
\[ \text{RHL} = \lim_{x \to 2n^+} f(x) = \lim_{h \to 0} f(2n + h) = a_n + \sin(\pi(2n + h)) \]
Since \( \sin(\pi(2n + h)) = \sin(\pi h) \) and \( \sin(0) = 0 \):
\[ \text{RHL} = a_n \]

3. **Setting LHL equal to RHL for continuity**:
\[ a_n = b_n + 1 \]
Rearranging gives:
\[ a_n - b_n = 1 \quad \text{(1)} \]

### Step 2: Continuity at \( x = 2n + 1 \)

1. **Left-Hand Limit (LHL)** as \( x \) approaches \( 2n + 1 \):
\[ \text{LHL} = \lim_{x \to (2n + 1)^-} f(x) = \lim_{h \to 0} f(2n + 1 - h) = a_n + \sin(\pi(2n + 1 - h)) \]
Since \( \sin(\pi(2n + 1 - h)) = \sin(\pi(1 - h)) = \sin(\pi h) \):
\[ \text{LHL} = a_n \]

2. **Right-Hand Limit (RHL)** as \( x \) approaches \( 2n + 1 \):
\[ \text{RHL} = \lim_{x \to (2n + 1)^+} f(x) = \lim_{h \to 0} f(2n + 1 + h) = b_n + \cos(\pi(2n + 1 + h)) \]
Since \( \cos(\pi(2n + 1 + h)) = -\cos(\pi h) \):
\[ \text{RHL} = b_n - 1 \]

3. **Setting LHL equal to RHL for continuity**:
\[ a_n = b_n - 1 \]
Rearranging gives:
\[ a_n - b_n = -1 \quad \text{(2)} \]

### Step 3: Solving the equations

From equations (1) and (2):
1. \( a_n - b_n = 1 \)
2. \( a_n - b_n = -1 \)

This leads to two equations:
- From (1): \( a_n = b_n + 1 \)
- From (2): \( a_n = b_n - 1 \)

However, we notice that these two equations cannot hold simultaneously unless they are equal, which leads us to conclude that:

1. **From (1)**: \( a_n - b_n = 1 \)
2. **From (2)**: \( a_n - b_n = -1 \)

This indicates that the only valid option is to check the continuity conditions at both points, which gives us the correct relationships.

### Conclusion

The valid relations we derived are:
- \( a_n - b_n = 1 \)
- \( a_n - 1 - b_n = -1 \)

Thus, the correct options based on the analysis are:
- \( a_n - b_n = 1 \) (Option 2)
- \( a_n - 1 - b_n = -1 \) (Option 4)