A functional equation is an equation, which relates the values assumed by a function at two or more points, which are themselves related in a particular manner. For example, we define an odd function by the relation `f(-x) = -f(x)` for all x.This defination can be paraphrased to say that it is a function f(x), which satisfies the functional relation f(x) + f(y) = 0, whenever x+y = 0. Of course this does not identify the function uniquely, sometimes with some additionl information, a function satisfying a given functional equation can be identified uniquely.
Suppose a functional equation has a relation between f(x) and `f(1/x)`, then due to the reason that reciprocal of a reciprocal gives back the original number, we can substitute `1/x` for x. This will result into another equation and solving these two, we can find f(x) uniquely. Similarly, we can solve an equation which contains f(x) and f(-x). Such equations are of repetitive nature .
If for every `x in R`, the function f(x) satisfies the relation af(x) + bf(-x) = g(x), then

Options

A. f(x) can be uniquely determined if ag(x) - bg(-x) `ne` 0 and a = ± b
B. f(x) can have infinitely many values if ag(x) - bg(-x) = 0 and `a= pmb`
C. f(x) cannot be determined if `anepm b`
D. alI of these

Answer

To solve the functional equation given by \( af(x) + bf(-x) = g(x) \), we can follow these steps:

### Step 1: Write down the original equation
We start with the functional equation:
\[ af(x) + bf(-x) = g(x) \tag{1} \]

### Step 2: Substitute \(-x\) into the equation
Next, we substitute \(-x\) for \(x\) in equation (1):
\[ af(-x) + bf(x) = g(-x) \tag{2} \]

### Step 3: Multiply the equations
Now, we will multiply equation (1) by \(a\) and equation (2) by \(b\):
- From equation (1):
\[ a(af(x) + bf(-x)) = ag(x) \implies a^2f(x) + abf(-x) = ag(x) \tag{3} \]
- From equation (2):
\[ b(af(-x) + bf(x)) = bg(-x) \implies abf(-x) + b^2f(x) = bg(-x) \tag{4} \]

### Step 4: Subtract the equations
Now, we subtract equation (4) from equation (3):
\[ (a^2f(x) + abf(-x)) - (abf(-x) + b^2f(x)) = ag(x) - bg(-x) \]
This simplifies to:
\[ (a^2 - b^2)f(x) = ag(x) - bg(-x) \tag{5} \]

### Step 5: Analyze the results
From equation (5), we can express \(f(x)\) as:
\[ f(x) = \frac{ag(x) - bg(-x)}{a^2 - b^2} \tag{6} \]

### Step 6: Determine conditions for uniqueness
For \(f(x)\) to be uniquely determined, the denominator \(a^2 - b^2\) must not be zero. Therefore, we have:
\[ a^2 \neq b^2 \implies a \neq \pm b \]

### Conclusion
Thus, we conclude that:
- If \(a \neq \pm b\), then \(f(x)\) can be uniquely determined.
- If \(a = \pm b\), \(f(x)\) cannot be uniquely determined, leading to infinitely many solutions.