Let {x} and [x] denote the fractional and integral part of a real number x respectively. The value (s) of x satisfying 4{x} = x + [x] is/are

Options

A. 0
B. `(5)/(3)`
C. `(1)/(3)`
D. 1

Answer

To solve the equation \( 4\{x\} = x + [x] \), where \(\{x\}\) is the fractional part of \(x\) and \([x]\) is the greatest integer (or integral part) of \(x\), we can follow these steps:

### Step 1: Understand the Definitions
The fractional part \(\{x\}\) is defined as:
\[ \{x\} = x - [x] \]
where \([x]\) is the greatest integer less than or equal to \(x\).

### Step 2: Substitute the Definition into the Equation
Substituting the definition of the fractional part into the equation gives:
\[ 4(x - [x]) = x + [x] \]

### Step 3: Simplify the Equation
Expanding the left side:
\[ 4x - 4[x] = x + [x] \]
Now, rearranging the equation:
\[ 4x - x = [x] + 4[x] \]
This simplifies to:
\[ 3x = 5[x] \]

### Step 4: Express \(x\) in Terms of \([x]\)
From the equation \(3x = 5[x]\), we can express \(x\) as:
\[ x = \frac{5}{3}[x] \]

### Step 5: Let \([x] = n\) (where \(n\) is an integer)
Let \([x] = n\), then:
\[ x = \frac{5}{3}n \]

### Step 6: Determine the Range of \(x\)
Since \([x] = n\), we have:
\[ n \leq x < n + 1 \]
Substituting for \(x\):
\[ n \leq \frac{5}{3}n < n + 1 \]

### Step 7: Solve the Inequalities
1. From \(n \leq \frac{5}{3}n\):
\[ 0 \leq \frac{5}{3}n - n \implies 0 \leq \frac{2}{3}n \implies n \geq 0 \]

2. From \(\frac{5}{3}n < n + 1\):
\[ \frac{5}{3}n - n < 1 \implies \frac{2}{3}n < 1 \implies n < \frac{3}{2} \]
Since \(n\) is an integer, the possible values for \(n\) are \(0\) and \(1\).

### Step 8: Find Corresponding Values of \(x\)
1. If \(n = 0\):
\[ x = \frac{5}{3} \cdot 0 = 0 \]

2. If \(n = 1\):
\[ x = \frac{5}{3} \cdot 1 = \frac{5}{3} \]

### Conclusion
The values of \(x\) that satisfy the equation \(4\{x\} = x + [x]\) are:
\[ x = 0 \quad \text{and} \quad x = \frac{5}{3} \]