Let f(x) be defined by `f(x) = x- [x], 0!=x in R`, where [x] is the greatest integer less than or equal to x then the number of solutions of `f(x) +f(1/x) =1`

Options

A. 0
B. infinite
C. 1
D. 2

Answer

To solve the equation \( f(x) + f\left(\frac{1}{x}\right) = 1 \) where \( f(x) = x - [x] \) (the fractional part of \( x \)), we will follow these steps:

### Step 1: Understand the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = x - [x] \]
This means \( f(x) \) gives the fractional part of \( x \), which is denoted as \( \{x\} \). Hence, we can rewrite the equation as:
\[ \{x\} + \left\{\frac{1}{x}\right\} = 1 \]

### Step 2: Analyze the intervals
Since \( x \) cannot be zero, we will consider two intervals for \( x \):
1. \( 0 < x < 1 \)
2. \( x \geq 1 \)

### Step 3: Case 1: \( 0 < x < 1 \)
In this interval, the fractional part \( f(x) = \{x\} = x \). Now, we need to find \( f\left(\frac{1}{x}\right) \):
- Since \( x < 1 \), \( \frac{1}{x} > 1 \).
- Therefore, \( f\left(\frac{1}{x}\right) = \left\{\frac{1}{x}\right\} = \frac{1}{x} - \left[\frac{1}{x}\right] \).

Let \( n = \left[\frac{1}{x}\right] \), then:
\[ \left\{\frac{1}{x}\right\} = \frac{1}{x} - n \]

Now substituting into the equation:
\[ x + \left(\frac{1}{x} - n\right) = 1 \]
This simplifies to:
\[ x + \frac{1}{x} - n = 1 \implies x + \frac{1}{x} = n + 1 \]

### Step 4: Analyze \( n \)
Since \( n \) is an integer, we can write:
\[ n = x + \frac{1}{x} - 1 \]
To find valid solutions, we need \( n \) to be a non-negative integer.

### Step 5: Case 2: \( x \geq 1 \)
In this case, \( f(x) = \{x\} = x - n \) where \( n = [x] \). Now, \( \frac{1}{x} < 1 \):
- Thus, \( f\left(\frac{1}{x}\right) = \left\{\frac{1}{x}\right\} = \frac{1}{x} \).

The equation now becomes:
\[ (x - n) + \frac{1}{x} = 1 \]
This simplifies to:
\[ x - n + \frac{1}{x} = 1 \implies x + \frac{1}{x} = n + 1 \]

### Step 6: Finding solutions
From both cases, we need to check for integer values of \( n \) that satisfy \( x + \frac{1}{x} = n + 1 \).

1. For \( 0 < x < 1 \):
- \( x + \frac{1}{x} \) is minimized at \( x = 1 \) (where it equals 2) and increases as \( x \) approaches 0.
- Thus, \( n + 1 \) must be at least 2, meaning \( n \geq 1 \).

2. For \( x \geq 1 \):
- \( n \) can take any integer value as \( x \) increases.

### Conclusion
After checking both cases, we find that there are no values of \( x \) that satisfy \( f(x) + f\left(\frac{1}{x}\right) = 1 \) for either interval. Thus, the number of solutions is:

\[ \text{Number of solutions} = 0 \]