Question

Which of the following is not an odd function

• A. $\ln \left(\frac{x^4+x^2+1}{(x^2+x+1{)}^2}\right)$
• B. $\sin \left(\sin \right(x\left)\right)$
• C. $\sin \left(\tan x\right)$
• D. $f\left(x\right)$, where $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)\forall x\in R-\left[0\right]$ and $f\left(2\right)=33$

Answer 1

Answer: (A), (D)

The correct options are
A $\ln \left(\frac{x^4+x^2+1}{(x^2+x+1{)}^2}\right)$
D $f\left(x\right)$, where $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)\forall x\in R-\left[0\right]$ and $f\left(2\right)=33$

(i) $ln\left(\frac{x^4+x^2+1}{{(x^2+x^{}+1)}^2}\right)=f\left(x\right)$

$f(-x)=\ln \left(\frac{x^4+x^2+1}{{(x^2-x^{}+1)}^2}\right)\ne -f\left(x\right)$

$\therefore$ $\ln \left(\frac{x^4+x^2+1}{{(x^2+x^{}+1)}^2}\right)$ is not an odd function

(ii) $\sin \left(\sin x\right)=f\left(x\right)$

$f(-x)=\sin \left(\sin -x\right)=\sin (-\sin x)=-\sin \left(\sin x\right)=-f\left(x\right)$

$\therefore f(-x)=-f\left(x\right)$

$\therefore$ $\sin \left(\sin x\right)$ is odd

(iii) $\sin \left(\tan x\right)=f\left(x\right)=\sin (-\tan x)=-\sin \left(\tan x\right)=-f\left(x\right)$

$\therefore$ $\sin \left(\tan x\right)$ is odd

iv) $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right).f\left(\frac{1}{x}\right)....\left(1\right)$

$f\left(2\right)=33$

$33+f\left(\frac{1}{2}\right)=33.f\left(\frac{1}{2}\right)\Rightarrow 32f\left(\frac{1}{2}\right)=33\Rightarrow f\left(\frac{1}{2}\right)=\frac{33}{32}$

if $f\left(x\right)$ is odd

$-f\left(x\right)=f(-x)$

$f(-x)+f(-\frac{1}{x})=f(-x).f(-\frac{1}{x})\Rightarrow -f(x-+f\left(\frac{1}{x}\right)=f(x).f\left(\frac{1}{x}\right).....(2)$

from (1) and (2)

$f\left(x\right)+f\left(\frac{1}{x}\right)=0$

$f\left(x\right)=-f\left(\frac{1}{x}\right)$

but $f\left(2\right)\ne -f\left(\frac{1}{2}\right)$

$\therefore$ $f$ is not odd