Question

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

• A. even
• B. odd
• C. neither even nor odd
• D. square

Hint:

$\log(x+\sqrt{1+x^2})$ is a classic Odd function. Applying an Even function (like $\sec$) to an Odd function results in an Even function.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine if the given composite function is even (\( f(-x) = f(x) \)) or odd (\( f(-x) = -f(x) \)).
Step 3: Detailed Explanation:
Let \( g(x) = \log(x + \sqrt{1 + x^2}) \).
Consider \( g(-x) \):
\[ g(-x) = \log(-x + \sqrt{1 + (-x)^2}) = \log(\sqrt{1 + x^2} - x) \]
Multiply and divide by the conjugate:
\[ g(-x) = \log \left( \frac{(\sqrt{1+x^2} - x)(\sqrt{1+x^2} + x)}{\sqrt{1+x^2} + x} \right) \]
\[ g(-x) = \log \left( \frac{1+x^2 - x^2}{\sqrt{1+x^2} + x} \right) = \log \left( \frac{1}{\sqrt{1+x^2} + x} \right) \]
\[ g(-x) = -\log(x + \sqrt{1+x^2}) = -g(x) \]
Thus, \( g(x) \) is an odd function.
Now, \( f(x) = \sec(g(x)) \).
\[ f(-x) = \sec(g(-x)) = \sec(-g(x)) \]
Since secant is an even function (\( \sec(-\theta) = \sec \theta \)):
\[ f(-x) = \sec(g(x)) = f(x) \]
Therefore, \( f(x) \) is an even function.
Step 4: Final Answer:
The function is an even function.