Question

Which of the following is not an odd function?
This question has multiple correct options
(a) $\ln \left( \dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}} \right)$
(b) $\sin \left( \sin \left( x \right) \right)$
(c) $\sin \left( \tan x \right)$
(d) $f\left( x \right)$, where $f\left( x \right)+f\left( \dfrac{1}{x} \right)=f\left( x \right)\cdot f\left( \dfrac{1}{x} \right)\forall x\in R-\left[ 0 \right]$ and $f\left( 2 \right)=33$

Answer 1

Hint: In this question, we need to check every option by using the principle formula of the odd function, $f\left( -x \right)=-f\left( x \right)$. If we check every option according to this condition, we will be able to figure out which option is an odd function or which options are not an odd function.

Complete step-by-step solution
We will be checking all the options to check whether they are an odd function or they are not an odd function.
To be an odd function, the particular function should satisfy the condition, $f\left( -x \right)=-f\left( x \right)$.
For option (a):
$f\left( x \right)=\ln \left( \dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}} \right)$
$f\left( -x \right)=\ln \left[ \dfrac{{{\left( -x \right)}^{4}}+{{\left( -x \right)}^{2}}+1}{{{\left( {{\left( -x \right)}^{2}}+\left( -x \right)+1 \right)}^{2}}} \right]$
Now, we know that, if we have ${{\left( -x \right)}^{n}}$, the term $x$ is positive when $n$ is an even number and it is negative when $n$ is an odd number.
We get,
$\begin{align}
  & f\left( -x \right)=\ln \left[ \dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}} \right] \\
 & f\left( -x \right)\ne -f\left( x \right)
\end{align}$
Therefore, the option (a) is not an odd function.
For option (b):
$f\left( x \right)=\sin \left( \sin \left( x \right) \right)$
We know, that $\sin \left( -x \right)=-\sin \left( x \right)$ means sine is an odd function, therefore,
$\begin{align}
  & f\left( -x \right)=\sin \left( \sin \left( -x \right) \right) \\
 & =\sin \left( -\sin \left( x \right) \right) \\
 & =-\sin \left( \sin \left( x \right) \right)
\end{align}$
$f\left( -x \right)=-f\left( x \right)$
For option (c):
$f\left( x \right)=\sin \left( \tan \left( x \right) \right)$
$f\left( -x \right)=\sin \left( \tan \left( -x \right) \right)$………….…. (1)
Now, we know, $\tan x=\dfrac{\sin x}{\cos x}$
If we consider $f\left( -x \right)=\tan \left( -x \right)$
$\begin{align}
  & \tan \left( -x \right)=\dfrac{\sin \left( -x \right)}{\cos \left( -x \right)} \\
 & =\dfrac{-\sin x}{\cos x} \\
 & =-\tan x
\end{align}$
Therefore, $f\left( -x \right)=-f\left( x \right)$
From (1), we get
$\begin{align}
  & f\left( -x \right)=\sin \left( -\tan x \right) \\
 & =-\sin \left( \tan x \right)
\end{align}$
Therefore, $f\left( -x \right)=-f\left( x \right)$
Therefore, option (c) is also an odd function.
For option (d):
$f\left( x \right)+f\left( \dfrac{1}{x} \right)=f\left( x \right)\cdot f\left( \dfrac{1}{x} \right)$ ……………….… (2)
we also have $f\left( 2 \right)=33$
Let us substitute $x=2$ we get
$\begin{align}
  & f\left( 2 \right)+f\left( \dfrac{1}{2} \right)=f\left( 2 \right)\cdot f\left( \dfrac{1}{2} \right) \\
 & \left( 33 \right)+f\left( \dfrac{1}{2} \right)=33\cdot f\left( \dfrac{1}{2} \right) \\
 & 33=33f\left( \dfrac{1}{2} \right)-f\left( \dfrac{1}{2} \right) \\
 & 33=f\left( \dfrac{1}{2} \right)\left[ 33-1 \right] \\
 & 33=32f\left( \dfrac{1}{2} \right) \\
 & f\left( \dfrac{1}{2} \right)=\dfrac{33}{32}
\end{align}$
Now, let us consider, $f\left( x \right)$ is an odd function
$\begin{align}
  & f\left( -x \right)+f\left( \dfrac{1}{-x} \right)=f\left( -x \right)\cdot f\left( \dfrac{1}{-x} \right) \\
 & -f\left( x \right)-f\left( \dfrac{1}{x} \right)=-f\left( x \right)\times -f\left( \dfrac{1}{x} \right) \\
 & -f\left( x \right)-f\left( \dfrac{1}{x} \right)=f\left( x \right)\cdot f\left( \dfrac{1}{x} \right)
\end{align}$
From equation (2), we get
$\begin{align}
  & -f\left( x \right)-f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right) \\
 & f\left( x \right)+f\left( \dfrac{1}{x} \right)+f\left( x \right)+f\left( \dfrac{1}{x} \right)=0 \\
 & 2\left[ f\left( x \right)+f\left( \dfrac{1}{x} \right) \right]=0 \\
 & f\left( x \right)+f\left( \dfrac{1}{x} \right)=0 \\
 & f\left( x \right)=-f\left( \dfrac{1}{x} \right)
\end{align}$
But we know, $f\left( 2 \right)\ne -f\left( \dfrac{1}{2} \right)$
Therefore, this option is not an odd function.
Hence, options (a) and (d) are the functions which are not odd functions.

Note: Odd function has a condition to satisfy, which is $f\left( -x \right)=-f\left( x \right)$. Similarly, to check whether a function is an even function, you simply need to follow $f\left( -x \right)=f\left( x \right)$, $\cos x$ which we used in the question is the most commonly known even function.