Injectivity Surjectivity and Symmetry of Logarithmic Trigonometric Function

To determine the correct statements about the function $f: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$ given by $f(x) = (\log(\sec x + \tan x))^3,$ we analyze its properties step-by-step:

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1. Symmetry (Odd / Even Function)

We compute $f(-x)$: $f(-x) = (\log(\sec(-x) + \tan(-x)))^3 = (\log(\sec x - \tan x))^3$

Using the trigonometric identity $\sec^2 x - \tan^2 x = 1$, we can write: $\sec x - \tan x = \frac{\sec^2 x - \tan^2 x}{\sec x + \tan x} = \frac{1}{\sec x + \tan x} = (\sec x + \tan x)^{-1}$

Substituting this back into the logarithmic expression: $\log(\sec x - \tan x) = \log\left((\sec x + \tan x)^{-1}\right) = -\log(\sec x + \tan x)$

Therefore: $f(-x) = (-\log(\sec x + \tan x))^3 = -(\log(\sec x + \tan x))^3 = -f(x)$

Since $f(-x) = -f(x)$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $f(x)$ is an odd function.
Thus, (A) is correct and (D) is incorrect.

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2. Injectivity (One-one Function)

Let $g(x) = \log(\sec x + \tan x)$. Differentiating $g(x)$ with respect to $x$: $g'(x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$

For all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\sec x > 0$, which means $g'(x) > 0$. Thus, $g(x)$ is a strictly increasing function on its domain.

Now, differentiate $f(x) = (g(x))^3$: $f'(x) = 3 (g(x))^2 \cdot g'(x) = 3 (\log(\sec x + \tan x))^2 \cdot \sec x$

Since $\sec x > 0$ and $(\log(\sec x + \tan x))^2 \ge 0$ (with equality holding only at $x = 0$), we have $f'(x) > 0$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \setminus \{0\}$.

A strictly increasing continuous function on an interval is strictly monotonic and therefore one-one (injective).
Thus, (B) is correct.

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3. Surjectivity (Onto Function)

We evaluate the limiting behavior of $f(x)$ at the boundaries of its domain:

1. As $x \to \frac{\pi}{2}^-$: $\sec x + \tan x \to \infty \implies \log(\sec x + \tan x) \to \infty \implies f(x) \to \infty$

2. As $x \to -\frac{\pi}{2}^+$: $\sec x + \tan x \to 0^+ \implies \log(\sec x + \tan x) \to -\infty \implies f(x) \to -\infty$

Since $f(x)$ is continuous on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, by the Intermediate Value Theorem, $f(x)$ takes all real values in $(-\infty, \infty)$.

Since $\text{Range}(f) = \mathbb{R} = \text{Codomain}$, the function is onto (surjective).
Thus, (C) is correct.

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Conclusion

The correct options are A, B, and C.