number of solutions of the equation f(x-1)+f(x+1)=sinα, 0

f(x)=0,if |x|>1 is

To determine the number of solutions for the equation \( f(x-1) + f(x+1) = \sin \alpha \) given the function \( f(x) = 0 \) for \( |x| > 1 \), we need to analyze the behavior of the function \( f(x) \) within the specified domain and how it interacts with the sine function.

Understanding the Function's Behavior

The function \( f(x) \) is defined as follows:

• For \( |x| \leq 1 \), \( f(x) \) can take on any values depending on its specific definition within this interval.
• For \( |x| > 1 \), \( f(x) = 0 \).

This means that the function is non-zero only in the interval \([-1, 1]\) and becomes zero outside of this range. Therefore, we need to focus our analysis on this interval.

Evaluating the Equation

We rewrite the equation \( f(x-1) + f(x+1) = \sin \alpha \). The values of \( x \) that we can consider must ensure that both \( x-1 \) and \( x+1 \) fall within the interval \([-1, 1]\) where \( f(x) \) is non-zero. This leads us to the following conditions:

• For \( x-1 \) to be in \([-1, 1]\), we need \( -1 \leq x-1 \leq 1 \), which simplifies to \( 0 \leq x \leq 2 \).
• For \( x+1 \) to be in \([-1, 1]\), we need \( -1 \leq x+1 \leq 1 \), which simplifies to \( -2 \leq x \leq 0 \).

Combining these two conditions, we find that \( x \) must lie within the interval \([0, 2]\) and \([-2, 0]\). The overlap of these intervals is \( x = 0 \) only. Thus, we need to evaluate \( f(-1) \) and \( f(1) \) at this point.

Analyzing the Function Values

At \( x = 0 \):

• We have \( f(-1) \) and \( f(1) \) which are both defined values of \( f(x) \) within the interval.
• The equation becomes \( f(-1) + f(1) = \sin \alpha \).

Now, the number of solutions to this equation depends on the specific values of \( f(-1) \) and \( f(1) \). If we denote \( f(-1) = a \) and \( f(1) = b \), then we are looking for the number of pairs \( (a, b) \) such that \( a + b = \sin \alpha \).

Considering the Range of Sine

The sine function oscillates between -1 and 1. Therefore, the equation \( a + b = \sin \alpha \) will have solutions depending on the values of \( a \) and \( b \). If \( a \) and \( b \) are constrained to certain ranges, we can determine how many pairs satisfy the equation:

• If both \( a \) and \( b \) can take values such that their sum can equal any value in \([-1, 1]\), then there could be infinitely many solutions.
• If \( a \) and \( b \) are fixed or limited to specific values, the number of solutions will depend on those values.

Final Thoughts

In summary, the number of solutions to the equation \( f(x-1) + f(x+1) = \sin \alpha \) is contingent upon the specific definitions of \( f(x) \) within the interval \([-1, 1]\). If \( f(-1) \) and \( f(1) \) can take on a range of values, then the equation can potentially have multiple solutions. However, if they are fixed, the solutions will be limited accordingly. Therefore, without additional information about the specific form of \( f(x) \) in the interval, we cannot definitively state the number of solutions.