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

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

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

Options:

An even function

An odd function

A periodic function

Neither an even nor an odd function

Correct Answer:

An odd function

Explanation:

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

$=\log\left(-x+\sqrt{x^2+1}×\frac{(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}\right)$

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

so $f(-x)=-f(x)$ an odd function