The number of solutions of the equation `tan^(-1)(4{x})+cot^(-1)(x+[x]))=(pi)/(2)` is (where `[.]` denotes greatest integer function and `{.}` fractional part of function.

Options

A. 3
B. 2
C. 1
D. 4

Explanation

To solve the equation \( \tan^{-1}(4\{x\}) + \cot^{-1}(x + [x]) = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Understand the relationship between the functions
We know that:
\[ \tan^{-1}(a) + \cot^{-1}(b) = \frac{\pi}{2} \quad \text{if and only if} \quad a = b \]
Thus, we can set:
\[ 4\{x\} = x + [x] \]

### Step 2: Express \(x\) in terms of its integer and fractional parts
Let \(x = n + f\), where \(n = [x]\) (the greatest integer less than or equal to \(x\)) and \(f = \{x\}\) (the fractional part of \(x\)). Therefore, we can rewrite the equation as:
\[ 4f = (n + f) + n \]
This simplifies to:
\[ 4f = 2n + f \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ 4f - f = 2n \implies 3f = 2n \implies f = \frac{2n}{3} \]

### Step 4: Determine the range of \(f\)
Since \(f\) is the fractional part, it must satisfy:
\[ 0 \leq f < 1 \]
Substituting \(f = \frac{2n}{3}\) into this inequality gives:
\[ 0 \leq \frac{2n}{3} < 1 \]

### Step 5: Solve the inequalities
From \(0 \leq \frac{2n}{3}\), we get:
\[ n \geq 0 \]
From \(\frac{2n}{3} < 1\), we get:
\[ 2n < 3 \implies n < \frac{3}{2} \implies n \leq 1 \]
Thus, \(n\) can take the values \(0\) or \(1\).

### Step 6: Find corresponding values of \(x\)
1. **If \(n = 0\)**:
\[ f = \frac{2(0)}{3} = 0 \implies x = 0 + 0 = 0 \]

2. **If \(n = 1\)**:
\[ f = \frac{2(1)}{3} = \frac{2}{3} \implies x = 1 + \frac{2}{3} = \frac{5}{3} \]

### Step 7: Conclusion
The solutions to the equation are \(x = 0\) and \(x = \frac{5}{3}\). Therefore, the number of solutions is:
\[ \boxed{2} \]