`f: (0,oo) to (-pi/2,pi/2)" be defined as, "f(x)=tan^(-1) (log_(e)x)`.
The above function can be classified as :

Options

A. injective but not surjective
B. surjective but not injective
C. neither injective nor surjective
D. both injective as well as surjective

Answer

To classify the function \( f: (0, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) defined as \( f(x) = \tan^{-1}(\log_e x) \), we will analyze its injectivity and surjectivity.

### Step 1: Determine the Range of \( f(x) \)

1. **Domain of \( x \)**: The function is defined for \( x \) in the interval \( (0, \infty) \).
2. **Logarithmic Transformation**: As \( x \) varies from \( 0 \) to \( \infty \), \( \log_e x \) varies from \( -\infty \) to \( \infty \).
3. **Behavior of \( \tan^{-1} \)**: The function \( \tan^{-1}(y) \) maps \( y \) from \( -\infty \) to \( \infty \) to the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

Thus, as \( x \) approaches \( 0 \), \( \log_e x \) approaches \( -\infty \) and \( f(x) \) approaches \( -\frac{\pi}{2} \). As \( x \) approaches \( \infty \), \( \log_e x \) approaches \( \infty \) and \( f(x) \) approaches \( \frac{\pi}{2} \). Therefore, the range of \( f(x) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

### Step 2: Check Surjectivity

1. **Co-domain**: The co-domain of the function is also \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
2. **Range and Co-domain Comparison**: Since the range of \( f(x) \) is equal to its co-domain, we conclude that \( f(x) \) is surjective.

### Step 3: Check Injectivity

1. **Differentiate \( f(x) \)**: We find the derivative \( f'(x) \):
\[ f'(x) = \frac{1}{1 + (\log_e x)^2} \cdot \frac{1}{x} \]
2. **Analyze the Derivative**:
- The term \( \frac{1}{x} \) is positive for \( x > 0 \).
- The term \( 1 + (\log_e x)^2 \) is always positive since it is a sum of 1 and a square.
- Therefore, \( f'(x) > 0 \) for all \( x > 0 \).

Since \( f'(x) > 0 \), the function \( f(x) \) is strictly increasing. A strictly increasing function is injective (one-to-one).

### Conclusion

Since \( f(x) \) is both injective and surjective, we classify the function as a **bijective function**.