Suppose `a, b in R`. If the equation `(a)/(x) = (1)/( x - b) + (1)/( x + b) ` is not satisfied by any real value of x, then

Options

A. `0 lt a lt b lt 2`
B. `0 lt a lt 2`
C. `0lt b lt 2`
D. `0 lt b lt a lt 2`

Answer

To solve the given equation \(\frac{a}{x} = \frac{1}{x - b} + \frac{1}{x + b}\) and find the conditions under which it is not satisfied by any real value of \(x\), we can follow these steps:

### Step 1: Simplify the Right Side
We start by simplifying the right side of the equation:
\[ \frac{1}{x - b} + \frac{1}{x + b} = \frac{(x + b) + (x - b)}{(x - b)(x + b)} = \frac{2x}{x^2 - b^2} \]
Thus, the equation becomes:
\[ \frac{a}{x} = \frac{2x}{x^2 - b^2} \]

### Step 2: Cross Multiply
Next, we cross-multiply to eliminate the fractions:
\[ a(x^2 - b^2) = 2x^2 \]
Expanding this gives:
\[ ax^2 - ab^2 = 2x^2 \]

### Step 3: Rearrange the Equation
Rearranging the equation leads to:
\[ ax^2 - 2x^2 - ab^2 = 0 \]
This can be factored as:
\[ (a - 2)x^2 - ab^2 = 0 \]

### Step 4: Solve for \(x^2\)
From the rearranged equation, we can isolate \(x^2\):
\[ (a - 2)x^2 = ab^2 \]
Thus, we have:
\[ x^2 = \frac{ab^2}{a - 2} \]

### Step 5: Determine Conditions for No Real Solutions
For the equation to have no real solutions, the expression for \(x^2\) must be negative:
\[ \frac{ab^2}{a - 2} < 0 \]
This inequality can be analyzed by considering the signs of the numerator and denominator.

### Step 6: Analyze the Inequality
1. **Numerator**: \(ab^2\) is non-negative if \(b \neq 0\) (since \(b^2\) is always non-negative).
2. **Denominator**: \(a - 2\) must be negative for the fraction to be negative, which implies:
\[ a - 2 < 0 \implies a < 2 \]

### Step 7: Conclusion
For the inequality \(\frac{ab^2}{a - 2} < 0\) to hold, we need:
- \(b^2 > 0\) (which means \(b \neq 0\))
- \(a < 2\)

Thus, the condition for the equation to not be satisfied by any real value of \(x\) is:
\[ a < 2 \quad \text{and} \quad b \neq 0 \]