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Even and odd Nature of Functions

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EVEN / ODD FUNCTIONS

A function is even if \(f(x) = f( - x).\) This means that the function is the same for \(\text{+ve } x\text{-axis}\) and \(\text{-ve } x\text{-axis,}\) or graphically, symmetric about the \(y\text{-axis}\)

For example, \(f(x) = {x^2}\) is even

A function is even about \('a'\) if it is symmetric about the line \(x=a.\) For example, \(f(x) = {(x - 1)^2}\) is even about \(x=1.\)

For such a function \(f(a - x) = f(a + x)\)

A function is odd if \(f(x)=  f(-x),\) that is, the function on one side of \(x\text{-axis}\) is sign inverted with respect to the other side or graphically, symmetric about the origin.

For example, \(f(x) = {x^3}\) is odd.

As in the even case, \(f(x)\) can be odd about \('a'.\) For example, \(f(x) = {(x + 2)^3}\) is odd about \(x=-2.\)

A function can also be neither even nor odd.

For example, \(f(x)={x^2}+{x^3},\text{ as }f(x)\ne f(-x)\text{ and }f(x)\ne- f(-x)\)

Example - 34

Check whether the following functions are even or odd.

\(\begin {align} {\bf{(a)}} \ \; f(x) = \sin x - \cos x\qquad \qquad {\bf{(b)}} \ \; f(x) = \frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1 \qquad \qquad {\bf{(c)}} \ \; f(x) = \sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{{{(1 + x)}^2}}}\end {align}\)

Solution:

(a)  \(f(-x)=\sin(- x)-\cos(-x)=-{\text{sin}}x-{\text{cos}}x\)

We see that  \(f( - x) \ne f(x)\)   and   \(f( - x) \ne  - f(x)\)

Hence, this function is neither even nor odd.

(b)  \(f( - x) =\begin{align} \frac{{ - x}}{{{e^{ - x}} - 1}} - \frac{x}{2} + 1= \frac{{x{e^x}}}{{{e^x} - 1}} - \frac{x}{2} + 1\end{align}\)

\(= \begin{align}\frac{{x{e^x} + x}}{{2({e^x} - 1)}} + 1=\frac{{x({e^x} - 1) + 2x}}{{2({e^x} - 1)}} + 1=\frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1 = f(x)\end{align}\)

Therefore, this function is even.

 (c)  \(f( - x) = f(x)\), which is obvious just by observation.

This function is even.

Example – 35

If the function \(f\) satisfies \(f(x + y) + f(x - y) = 2f(x).\,f(y)\)  \(\forall    x, y  \;  in\;  \mathbb{R} \), and  f(0) \( \ne\) 0, prove that \(f(x)\) is even

Solution:

Substituting \(x=y=0,\)

we get   \(2f(0) = 2f{(0)^2}\)      \( \Rightarrow \)    \(f(0)\,\,(f(0) - 1) = 0\)

Since   \(f(0)\) \( \ne\) 0,    we have   \(f(0)\)  = 1

Now substitute \(x=0,\)

to get        \(f(y) + f( - y) = 2f(0)f(y) = 2f(y)\)

\( \Rightarrow \)         \(f(y) = f( - y)\)

\( \Rightarrow \)         \(f\) is even.

Example – 36

If  \(f(x + y) = f(x).f(y)\forall \)  real  \(x,\,y\)  and  \(f(0) \;\ne\,0,\)  then prove that the function \(f(x) = \begin{align}\frac{{f(x)}}{{1 + {{\left( {f(x)} \right)}^2}}}\end{align}\) is an even function.

Solution:

As in the previous question, we first substitute  \(x=y=0\)  to get  \(f(0)=1\)  \(\Rightarrow \) \(f(x) =\begin{align} \frac{1}{{f( - x)}}\end{align}\)

(Note that the approach in such questions is always to try out some sample substitutions that yield significant results).

\[Now \qquad\qquad\qquad f( - x) = \frac{{f( - x)}}{{1 + {{\left( {f( - x)} \right)}^2}}} = \frac{{\frac{1}{{f(x)}}}}{{1 + {{\left( {\frac{1}{{f(x)}}} \right)}^2}}} = \frac{{f(x)}}{{1 + {{\left( {f(x)} \right)}^2}}} = f(x).\]

Hence, \(f(x)\) is even.

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