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Find out whether the given function is even, odd or neither even or odd, where
\[f\left( x \right)=\left\{ \begin{matrix}
   x\left| x \right| & x\le -1 \\
   \left[ 1+x \right]+\left[ 1-x \right] & -1< x<1 \\
   -x\left| x \right| & x\ge 1 \\
\end{matrix} \right\}\]
Where $\left| {} \right|$ and $\left[ {} \right]$ represent the modulus and greatest integral functions.
A. $f\left( x \right)$ is even
B. $f\left( x \right)$ is odd
C. $f\left( x \right)$ is neither even nor odd
D. $f\left( x \right)$ is both even and odd

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Last updated date: 13th Jun 2024
Total views: 402k
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Answer
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Hint: For this question we will define what are even functions and odd functions, then we will take the given function in the question and then apply the definition of modulus function and the greatest integer functions and then find the value of $f\left( x \right)$ and $f\left( -x \right)$ , and see what definitions it satisfy an even function or an odd function.

Complete step-by-step answer:
First we will understand what are even functions and odd functions.
So, a function is an even function when: $f\left( x \right)=f\left( -x \right)$ , for all $x$ . In other words there is a symmetry about the \[y-\text{axis}\] like a reflection. For example, let $f\left( x \right)={{x}^{2}}+1$ : $\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{2}}+1,f\left( -x \right)={{\left( -x \right)}^{2}}+1={{x}^{2}}+1 \\
 & \Rightarrow f\left( x \right)=f\left( -x \right) \\
\end{align}$
Therefore, it is an even function. Let’s look at the graph of $f\left( x \right)={{x}^{2}}+1$ :
seo images

We see that this function has a symmetry about \[x-\text{axis}\] therefore it’s an even function.
So, a function is an odd function when: $-f\left( x \right)=f\left( -x \right)$ , for all $x$ . In other words there is an origin symmetry. For example, let $f\left( x \right)={{x}^{3}}-x$ :
 $\begin{align}
  & \Rightarrow f\left( x \right)={{x}^{3}}-x,-f\left( x \right)=-\left( {{x}^{3}}-x \right)=-{{x}^{3}}+x,f\left( -x \right)={{\left( -x \right)}^{3}}-\left( -x \right)=-{{x}^{3}}+x \\
 & \Rightarrow -f\left( x \right)=f\left( -x \right) \\
\end{align}$
Therefore, it is an odd function. Let’s look at the graph of $f\left( x \right)={{x}^{3}}-x$ :
seo images

We see that this function has origin symmetry; therefore it’s an odd function.
When a function does not satisfy both the above conditions then the function is neither odd nor even.
Now let’s take a look at the question, we have:
\[f\left( x \right)=\left\{ \begin{matrix}
   x\left| x \right| & x\le -1 \\
   \left[ 1+x \right]+\left[ 1-x \right] & -1< x<1 \\
   -x\left| x \right| & x\ge 1 \\
\end{matrix} \right\}\]
Now, let's see what a modulus function is: so, a modulus function is a function which gives the absolute value of a number or variable. It produces the magnitude of the number of variables. It is also termed as an absolute value function. The outcome of this function is always positive, no matter what input has been given to the function. It is represented as $y=\left| x \right|$ . Modulus Function is basically defined as the real valued function, say \[f:\mathbb{R}\to \mathbb{R}\] where $y=\left| x \right|$ for each $x\in \mathbb{R}\text{ or }f\left( x \right)=\left| x \right|$ . This function can be defined using modulus operation as follows:
$f\left( x \right)=\left\{ \begin{matrix}
   -x & x<0 \\
   x & x\ge 0 \\
\end{matrix} \right\}$
Now let’s define the greatest integer function. So, the real function $f:P\to P$ defined by $f\left( a \right)=\left[ a \right],\text{ }a\in P$ assumes the value of the greatest integer less than or equal to \[a\] , is called the greatest integer function. Therefore,
$\begin{align}
  & f\left( a \right)=\left[ a \right]=-1\text{ for }-1\le a<0 \\
 & f\left( a \right)=\left[ a \right]=0\text{ for 0}\le a<1 \\
 & f\left( a \right)=\left[ a \right]=1\text{ for }1\le a<2 \\
\end{align}$

Therefore, using these definitions we will get:

\[\begin{align}
  & \Rightarrow f\left( x \right)=\left\{ \begin{matrix}
   x\left| x \right| & x\le -1 \\
   \left[ 1+x \right]+\left[ 1-x \right] & -1< x<1 \\
   -x\left| x \right| & x\ge 1 \\
\end{matrix} \right\}\Rightarrow f\left( x \right)=\left\{ \begin{matrix}
   x\left( -x \right) & x\le -1 \\
   2+\left[ x \right]+\left[ -x \right] & -1< x<1 \\
   -x\left( x \right) & x\ge 1 \\
\end{matrix} \right\} \\
 & \Rightarrow f\left( x \right)=\left\{ \begin{matrix}
   {{x}^{2}} & x\le -1 \\
   2+\left[ x \right]+\left[ -x \right] & -1< x<1 \\
   -{{x}^{2}} & x\ge 1 \\
\end{matrix} \right\} \\
\end{align}\]
Now let’s see the value of $f\left( -x \right)$ :
\[\begin{align}
  & \Rightarrow f\left( -x \right)=\left\{ \begin{matrix}
   {{\left( -x \right)}^{2}} & x\le -1 \\
   2+\left[ -x \right]+\left[ -\left( -x \right) \right] & -1< x<1 \\
   -{{\left( -x \right)}^{2}} & x\ge 1 \\
\end{matrix} \right\} \\
 & \Rightarrow f\left( -x \right)=\left\{ \begin{matrix}
   {{x}^{2}} & x\le -1 \\
   2+\left[ -x \right]+\left[ x \right] & -1< x<1 \\
   -{{x}^{2}} & x\ge 1 \\
\end{matrix} \right\} \\
\end{align}\]
Now, we see that \[f\left( x \right)=f\left( -x \right)\] , therefore it is an even function.

So, the correct answer is “Option A”.

Note: Students may make the mistake while applying the properties of modulus functions and the greatest integer functions. In the questions involving modulus functions, we must understand the fundamental properties of the absolute value, that are as follows:
A. Non-negativity $\Rightarrow \left| a \right|\ge 0$
B. Positive – definiteness $\Rightarrow \left| a \right|=0\Leftrightarrow a=0$
C. Multiplicativity $\Rightarrow \left| ab \right|=\left| a \right|\left| b \right|$
D. Subadditivity $\Rightarrow \left| a+b \right|\le \left| a \right|+\left| b \right|$