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: Symmetry Check
Replace $x$ with $-x$:
$f(-x) = \sec \left[ \log \left( -x + \sqrt{1 + (-x)^2} \right) \right] = \sec \left[ \log \left( \sqrt{1+x^2} - x \right) \right]$.

Step 2: Properties of Log
Note that $(\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x) = 1+x^2-x^2 = 1$.
So, $\sqrt{1+x^2}-x = \frac{1}{\sqrt{1+x^2}+x} = (x+\sqrt{1+x^2})^{-1}$.

Step 3: Calculation
$f(-x) = \sec [ -\log(x+\sqrt{1+x^2}) ]$.
Since $\sec(-\theta) = \sec \theta$, we have $f(-x) = \sec [ \log(x+\sqrt{1+x^2}) ] = f(x)$.

Step 4: Conclusion
As $f(-x) = f(x)$, the function is Even.
Final Answer: (A)