Let `f:(-pi/2,pi/2)-> RR` be given by `f(x) = (log(sec x + tan x))^3` Then which of the following is wrong?

Options

A. `f(x)` is an odd function
B. `f(x)` is a one-one function
C. `f(x)` is an onto function
D. `f(x)` is an even function

Explanation

To solve the problem, we need to analyze the function \( f(x) = (\log(\sec x + \tan x))^3 \) defined on the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) and determine which of the given options is incorrect.

### Step 1: Check if the function is odd

A function \( f(x) \) is odd if \( f(-x) = -f(x) \).

1. Calculate \( f(-x) \):
\[ f(-x) = (\log(\sec(-x) + \tan(-x)))^3 \]
Using the identities \( \sec(-x) = \sec(x) \) and \( \tan(-x) = -\tan(x) \):
\[ f(-x) = (\log(\sec x - \tan x))^3 \]

2. We know:
\[ \sec x - \tan x = \frac{1}{\sec x + \tan x} \]
Thus:
\[ f(-x) = \left(\log\left(\frac{1}{\sec x + \tan x}\right)\right)^3 = (-\log(\sec x + \tan x))^3 = -(\log(\sec x + \tan x))^3 = -f(x) \]

Since \( f(-x) = -f(x) \), the function \( f(x) \) is indeed odd.

### Step 2: Check if the function is one-to-one (injective)

A function is one-to-one if it is either strictly increasing or strictly decreasing.

1. The function \( f(x) \) is composed of \( \log(\sec x + \tan x) \), which is known to be increasing in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
2. Since the cube function \( x^3 \) is also increasing, the composition \( f(x) = (\log(\sec x + \tan x))^3 \) is increasing.

Thus, \( f(x) \) is one-to-one.

### Step 3: Check if the function is onto (surjective)

To check if the function is onto, we need to determine the range of \( f(x) \).

1. As \( x \) approaches \( -\frac{\pi}{2} \) from the right, \( \sec x + \tan x \) approaches \( 0 \), so \( \log(\sec x + \tan x) \) approaches \( -\infty \).
2. As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \sec x + \tan x \) approaches \( +\infty \), so \( \log(\sec x + \tan x) \) approaches \( +\infty \).
3. Therefore, the range of \( f(x) \) is \( (-\infty, +\infty) \).

Since the range of \( f(x) \) covers all real numbers, \( f(x) \) is onto.

### Conclusion

From the analysis:
- \( f(x) \) is odd.
- \( f(x) \) is one-to-one.
- \( f(x) \) is onto.

Thus, the only incorrect statement among the options provided is the one that claims \( f(x) \) is even.