Test Index

MHTCET 5 May 2025 Shift 2 Paper

© examsnet.com

Question : 109 of 150

Marks: +1, -0

The function

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

\_\_\_\_

function

even
odd
neither even nor odd
square

Solution:

Concept:

A function is even if

f(-x) = f(x)

for all

x

.

Explanation:

Given:

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

Replace

x

by

-x

:

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

Since

\sqrt{1 + x^2} - x = \frac{1}{\sqrt{1 + x^2} + x}

, we get:

f(-x) = \sec \left[\log \left(\frac{1}{\sqrt{1 + x^2} + x}\right)\right]

Using

\log \left(\frac{1}{a}\right) = -\log(a)

:

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

Since

\sec(-\theta) = \sec(\theta)

:

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

Thus,

f(x)

is an even function.

Answer:

f(x)

is an even function. Correct option: A. even

© examsnet.com

Go to Question:

Prev Question

Next Question