Question

Which of the following is an odd function?

• A. $|x|+1$
• B. $\sin x+\cos x$
• C. $x^2\sec x+x{\tan }^{2}x$
• D. $x^2\cot x+4x^4\text{cosec x+}{\text{x}}^5$

Answer 1

Answer: (D)

Let $f(x)=|x|+1$
$\therefore$ $f(-x)=|-x|+1$
$=f(x),$ even function [b] Let $f(x)=\sin x+\cos x$
$\therefore$ $f(-x)=\sin (-x)+\cos (-x)$
$=-\sin x+\cos x$ neither even nor odd function
Let $f(x)=x^2\sec x+x{\tan }^{2}x$
$\therefore$ $f(-x)=x^2\sec x-x{\tan }^{2}x$ neither even nor odd function
Let $f(x)=x^2\cot x+4x^2\text{cosec x+}{\text{x}}^5$
$\therefore$ $f(-x)=-x^2\cot x-4x^4\text{cosec x-}{\text{x}}^5$
$=-f(x),$ odd function
Hence, option is correct.