Which one of the following function contains only one integer in its range ?
[ Note sgn (k) denotes the signum function of k.]

Options

A. a. `f(x) = 1/2 cos^(-1) ((1-x^(2))/(1+x^(2)))`
B. b. `g(x) = sgn (x + 1/x)`
C. c. `h(x) = sin^(2) x + 2 sin x + 2`
D. d. ` k(x) = cos^(-1) ( x^(2) - 2x + 2)`

Answer

To solve the problem of identifying which function contains only one integer in its range, we will analyze each function step by step.

### Step 1: Analyze the first function \( f(x) = \frac{1}{2} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \)

1. Rewrite the function:
\[ f(x) = \frac{1}{2} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
This can be transformed into:
\[ f(x) = \begin{cases} 2 \tan^{-1}(x) & \text{if } x \geq 0 \\ -2 \tan^{-1}(x) & \text{if } x < 0 \end{cases} \]

2. Determine the range:
- For \( x \geq 0 \), \( f(x) \) ranges from \( 0 \) to \( \frac{\pi}{2} \).
- For \( x < 0 \), \( f(x) \) ranges from \( -\frac{\pi}{2} \) to \( 0 \).
- Thus, the overall range of \( f(x) \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which includes the integers \( 0 \) and \( 1 \).

### Step 2: Analyze the second function \( g(x) = \text{sgn}(x) + \frac{1}{x} \)

1. Rewrite the function:
\[ g(x) = \text{sgn}(x) + \frac{1}{x} \]

2. Determine the range:
- For \( x > 0 \), \( g(x) \) approaches \( +\infty \) as \( x \) approaches \( 0 \) and approaches \( 1 \) as \( x \) approaches \( +\infty \).
- For \( x < 0 \), \( g(x) \) approaches \( -\infty \) as \( x \) approaches \( 0 \) and approaches \( -1 \) as \( x \) approaches \( -\infty \).
- Thus, the range of \( g(x) \) includes the integers \( -1 \) and \( 1 \).

### Step 3: Analyze the third function \( h(x) = \sin^2(x) + 2\sin(x) + 2 \)

1. Rewrite the function:
\[ h(x) = \sin^2(x) + 2\sin(x) + 2 = (\sin(x) + 1)^2 + 1 \]

2. Determine the range:
- Since \( \sin(x) \) ranges from \( -1 \) to \( 1 \), \( \sin(x) + 1 \) ranges from \( 0 \) to \( 2 \).
- Therefore, \( h(x) \) ranges from \( 1 \) (when \( \sin(x) = -1 \)) to \( 5 \) (when \( \sin(x) = 1 \)).
- Thus, the range of \( h(x) \) includes the integers \( 1, 2, 3, 4, 5 \).

### Step 4: Analyze the fourth function \( k(x) = \cos^{-1}(x^2 - 2x + 2) \)

1. Rewrite the function:
\[ k(x) = \cos^{-1}(x^2 - 2x + 2) \]

2. Determine the range:
- The expression \( x^2 - 2x + 2 \) can be rewritten as \( (x - 1)^2 + 1 \), which is always \( \geq 1 \).
- Therefore, \( k(x) \) is defined only when \( x^2 - 2x + 2 = 1 \), which occurs at \( x = 1 \).
- At \( x = 1 \), \( k(1) = \cos^{-1}(1) = 0 \).
- Thus, the range of \( k(x) \) contains only one integer, which is \( 0 \).

### Conclusion

The function \( k(x) = \cos^{-1}(x^2 - 2x + 2) \) contains only one integer in its range, which is \( 0 \). Therefore, the answer is:

**Option D: \( k(x) \)**