The number of integer `x` satisfying `sin^(-1)|x-2|+cos^(-1)(1-|3-x|)=pi/2` is 1 (b) 2 (c) 3 (d) 4

Options

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

Explanation

To solve the equation \( \sin^{-1} |x - 2| + \cos^{-1} (1 - |3 - x|) = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Understand the relationship between inverse trigonometric functions
We know that:
\[ \sin^{-1} a + \cos^{-1} b = \frac{\pi}{2} \quad \text{if and only if} \quad a = b \]
Thus, we can set:
\[ |x - 2| = 1 - |3 - x| \]

### Step 2: Solve the equation \( |x - 2| = 1 - |3 - x| \)
We will consider the two cases for the absolute values.

**Case 1:** \( x - 2 \geq 0 \) and \( 3 - x \geq 0 \)
This implies:
\[ x - 2 = 1 - (3 - x) \]
Simplifying gives:
\[ x - 2 = 1 - 3 + x \]
\[ x - 2 = x - 2 \]
This is always true for \( 2 \leq x \leq 3 \).

**Case 2:** \( x - 2 \geq 0 \) and \( 3 - x < 0 \)
This implies:
\[ x - 2 = 1 - (x - 3) \]
Simplifying gives:
\[ x - 2 = 1 - x + 3 \]
\[ x - 2 = 4 - x \]
\[ 2x = 6 \quad \Rightarrow \quad x = 3 \]

**Case 3:** \( x - 2 < 0 \) and \( 3 - x \geq 0 \)
This implies:
\[ -(x - 2) = 1 - (3 - x) \]
Simplifying gives:
\[ - x + 2 = 1 - 3 + x \]
\[ - x + 2 = -2 + x \]
\[ 2 + 2 = 2x \quad \Rightarrow \quad x = 2 \]

**Case 4:** \( x - 2 < 0 \) and \( 3 - x < 0 \)
This implies:
\[ -(x - 2) = 1 - (x - 3) \]
Simplifying gives:
\[ - x + 2 = 1 - x + 3 \]
\[ - x + 2 = 4 - x \]
This is again always true.

### Step 3: Determine the range of \( x \)
From the cases, we have the following:
- From Case 1, \( 2 \leq x \leq 3 \)
- From Case 2, \( x = 3 \)
- From Case 3, \( x = 2 \)
- From Case 4, \( x \) can be any value less than 2 or greater than 3, but we are only interested in the range \( 2 \leq x \leq 3 \).

### Step 4: Count the integer solutions
The integers \( x \) that satisfy \( 2 \leq x \leq 3 \) are:
- \( x = 2 \)
- \( x = 3 \)

Thus, there are **2 integer solutions**.

### Final Answer
The number of integer \( x \) satisfying the equation is **2**.

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