A calculus problem by A Former Brilliant Member

Let $g(x) = \log \left(x + \sqrt {1+x^2} \right)$ , then we have:

$\begin{aligned} g(x) + g(-x) & = \log \left(x+\sqrt{1+x^2} \right) + \log \left( -x+\sqrt{1+(-x)^2} \right) \\ & = \log \left(x+\sqrt{1+x^2} \right) + \log \left( -x+\sqrt{1+x^2} \right) \\ & = \log \left(x+\sqrt{1+x^2} \right) \left( -x+\sqrt{1+(-x)^2} \right) \\ & = \log \left( -x^2+1+x^2 \right) \\ & = \log 1 \\ & = 0 \\ \implies g(-x) & = - g(x) \end{aligned}$

Therefore, $g(x)$ is an odd function. And now, we have:

$\begin{aligned} f(x) & = \sec \left[ \log \left(x + \sqrt{1+x^2} \right) \right] \\ & = \sec \left[ g(x) \right] \\ \implies f(-x) & = \sec [g(-x)] \\ & = \color{#3D99F6} \sec [-g(x)] & \small \color{#3D99F6} \text{Since } \sec (x) \text{ is an even function.} \\ & = \color{#3D99F6} \sec [g(x)] \\ \implies f(-x) & = f(x) \end{aligned}$

Therefore, $f(x)$ is $\boxed{\text{Even}}$ .