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 .
Suppose that for every `x !=0` af(x)+`bf(1/x)=1/x-5`, where `aneb` then the value of the integral `int_(1)^(2)f(x)dx` is

Options

A. `(2a"ln"2-10a+7b)/(2(a^(2)-b^(2)))`
B. `(pi)/(2sqrt2)`
C. `(7a"ln"+10b-2a"ln"2)/(2(a^(2)-b^(2)))`
D. none of these

Answer

To solve the functional equation given by \( af(x) + bf\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \) for \( x \neq 0 \), we will follow these steps:

### Step 1: Substitute \( x \) with \( \frac{1}{x} \)
We start by substituting \( x \) with \( \frac{1}{x} \) in the original equation:

\[ af\left(\frac{1}{x}\right) + bf(x) = x - 5 \]

### Step 2: Write the two equations
Now we have two equations:
1. \( af(x) + bf\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \) (Equation 1)
2. \( af\left(\frac{1}{x}\right) + bf(x) = x - 5 \) (Equation 2)

### Step 3: Multiply and rearrange
Next, we will multiply Equation 1 by \( a \) and Equation 2 by \( b \):

From Equation 1:
\[ a^2f(x) + abf\left(\frac{1}{x}\right) = \frac{a}{x} - 5a \]

From Equation 2:
\[ baf\left(\frac{1}{x}\right) + b^2f(x) = bx - 5b \]

### Step 4: Subtract the equations
Now we will subtract the second modified equation from the first modified equation:

\[ (a^2 - b^2)f(x) + (ab - ba)f\left(\frac{1}{x}\right) = \left(\frac{a}{x} - 5a\right) - (bx - 5b) \]

Since \( ab - ba = 0 \), we simplify to:

\[ (a^2 - b^2)f(x) = \frac{a}{x} - bx - 5a + 5b \]

### Step 5: Solve for \( f(x) \)
Now we can isolate \( f(x) \):

\[ f(x) = \frac{1}{a^2 - b^2}\left(\frac{a}{x} - bx - 5a + 5b\right) \]

### Step 6: Evaluate the integral
We need to evaluate the integral \( \int_1^2 f(x) \, dx \):

\[ \int_1^2 f(x) \, dx = \int_1^2 \frac{1}{a^2 - b^2}\left(\frac{a}{x} - bx - 5a + 5b\right) \, dx \]

### Step 7: Break down the integral
We can break this down into separate integrals:

\[ = \frac{1}{a^2 - b^2} \left( a \int_1^2 \frac{1}{x} \, dx - b \int_1^2 x \, dx - 5a \int_1^2 1 \, dx + 5b \int_1^2 1 \, dx \right) \]

### Step 8: Calculate each integral
Calculating each integral:
1. \( \int_1^2 \frac{1}{x} \, dx = \ln(2) \)
2. \( \int_1^2 x \, dx = \left[\frac{x^2}{2}\right]_1^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \)
3. \( \int_1^2 1 \, dx = 2 - 1 = 1 \)

### Step 9: Substitute back into the integral
Substituting these values back, we have:

\[ = \frac{1}{a^2 - b^2} \left( a \ln(2) - b \cdot \frac{3}{2} - 5a + 5b \right) \]

### Final Answer
Thus, the value of the integral \( \int_1^2 f(x) \, dx \) is:

\[ \frac{1}{a^2 - b^2} \left( a \ln(2) - \frac{3b}{2} - 5a + 5b \right) \]