Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

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

Options:

Even

Odd

Constant

None of these

Correct Answer:

Even

Explanation:

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

so $f(-x)=\sec(\log(-x+\sqrt{1+x^2}×\frac{x+\sqrt{1+x^2}}{x+\sqrt{1+x^2}})$

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

$f(-x)=\sec(\log(x+\sqrt{1+x^2}))$ even function