Question

The function $f\left(x\right)=\sec \left(\log \left(x+\sqrt{1+x^2}\right)\right)$ is

• A. Odd
• B. Even
• C. Neither odd nor even
• D. Constant

Answer 1

The correct option is B

Even

Explanation for correct answer

We know that in even function $f\left(x\right)=f\left(-x\right)$

$f\left(x\right)=\sec \left(\log \left(x+\sqrt{1+x^2}\right)\right)$

$$\begin{gathered} \Rightarrow f\left(-x\right)=\sec \left(\log \left(\left(-x\right)+\sqrt{1+{\left(-x\right)}^2}\right)\right) \\ =\sec \left(\log \left(\sqrt{1+x^2}-x\right)\right) \\ =\sec \left(\log \left(\frac{\left(\sqrt{1+x^2}-x\right)\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}\right)\right) \\ =\sec \left(\log \left(\frac{1+x^2-x^2}{\left(\sqrt{1+x^2}+x\right)}\right)\right) \\ =\sec \left(\log \left(\frac{1}{\left(\sqrt{1+x^2}+x\right)}\right)\right) \\ =\sec \left(\log \left(1\right)-\log \left(\left(\sqrt{1+x^2}+x\right)\right)\right)\left[\because \log \left(\frac{a}{b}\right)=\log \left(a\right)-\log \left(b\right)\right] \\ =\sec \left(-\log \left(\left(\sqrt{1+x^2}+x\right)\right)\right)\left[\because \log \left(1\right)=0\right] \\ =\sec \left(\log \left(\left(\sqrt{1+x^2}+x\right)\right)\right)\left[\because \sec \left(-\theta \right)=\sec \left(\theta \right)\right] \end{gathered}$$

Therefore $f\left(x\right)=f\left(-x\right)$, so, the function is even.

Hence, $\mathrm{Option}\left(\mathrm{B}\right)$ is correct.