The number of solution (s) of equation `sin sin^(-1)([x])+cos^(-1)cosx=1` (where denotes the greatest integer function) is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To solve the equation \( \sin(\sin^{-1}(\lfloor x \rfloor)) + \cos^{-1}(\cos x) = 1 \), where \( \lfloor x \rfloor \) denotes the greatest integer function, we can follow these steps:

### Step 1: Simplify the Equation
We start with the equation:
\[ \sin(\sin^{-1}(\lfloor x \rfloor)) + \cos^{-1}(\cos x) = 1 \]
Using the property of inverse functions, we know that:
\[ \sin(\sin^{-1}(y)) = y \quad \text{for } y \in [-1, 1] \]
Thus, we can rewrite the equation as:
\[ \lfloor x \rfloor + \cos^{-1}(\cos x) = 1 \]

### Step 2: Simplify \( \cos^{-1}(\cos x) \)
The term \( \cos^{-1}(\cos x) \) gives us:
\[ \cos^{-1}(\cos x) = x \quad \text{if } x \in [0, \pi] \]
For \( x \) outside this range, it will take the principal value in the range \( [0, \pi] \).

### Step 3: Rewrite the Equation
Now we can express the equation as:
\[ \lfloor x \rfloor + x = 1 \]
Let \( x = n + f \), where \( n = \lfloor x \rfloor \) (an integer) and \( f \) is the fractional part of \( x \) (i.e., \( 0 \leq f < 1 \)). Substituting this into our equation gives:
\[ n + (n + f) = 1 \implies 2n + f = 1 \]

### Step 4: Analyze the Equation
From \( 2n + f = 1 \), we can isolate \( f \):
\[ f = 1 - 2n \]
Since \( f \) is the fractional part, it must satisfy \( 0 \leq f < 1 \). Therefore, we have:
\[ 0 \leq 1 - 2n < 1 \]

### Step 5: Solve the Inequalities
1. From \( 0 \leq 1 - 2n \):
\[ 2n \leq 1 \implies n \leq \frac{1}{2} \]
Since \( n \) is an integer, the only possible value is \( n = 0 \).

2. From \( 1 - 2n < 1 \):
\[ -2n < 0 \implies n > 0 \]
This condition contradicts the previous result since \( n \) cannot be both less than or equal to \( \frac{1}{2} \) and greater than \( 0 \).

### Conclusion
Since there are no integer values for \( n \) that satisfy both conditions, we conclude that there are no solutions to the equation.

Thus, the number of solutions is:
\[ \text{Number of solutions} = 0 \]

### Final Answer
The number of solutions is \( 0 \).

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