Question

If the graph of $f(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}}$ is symmetrical about the $y$ axis, then $n = $
A.$2$
B.$\dfrac{2}{3}$
C.$\dfrac{1}{4}$
D.$\dfrac{{ - 1}}{3}$

Answer 1

Hint: We are given with function of the graph and its geometrical interpretation and are asked to find the value of $n$. Thus, we will use the concept of even and odd function. Finally, we will evaluate the value of $n$.

Complete step-by-step answer:
Given,
$f(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}}$
Also, we are told that the graph is symmetrical about the $y$ axis.
Thus, the function is clearly an even function.
Thus, $f( - x) = f(x)$
Now, $f( - x) = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}$
Further, we get $f( - x) = \dfrac{{{a^{ - x}} - 1}}{{{{( - 1)}^n}{{(x)}^n}({a^{ - x}} + 1)}}$
Now, $f(x) = f( - x)$
Thus, we get
$\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{{a^{ - x}} - 1}}{{{{( - 1)}^n}{{(x)}^n}({a^{ - x}} - 1)}}$
Further, we get
${( - 1)^n} = ( - 1)$
Thus,
For this to be true, when $n$ is odd
Thus,
$n = \dfrac{2}{3}$ Or $n = \dfrac{{ - 1}}{3}$
But, when, $n = \dfrac{2}{3}$
${( - 1)^{2/3}} = 1$
Thus, $(1) \ne ( - 1)$
Hence, $n \ne \dfrac{2}{3}$
Hence, clearly, $n = \dfrac{{ - 1}}{3}$

Additional Information:
When the graph of a function is symmetric about an axis, then we consider the function an even function. But, if the graph of the function is symmetric about the origin, we consider the function an odd one.
For an even function,
$f( - x) = f(x)$
For an odd function,
$f( - x) = - f(x)$
Some examples of even function are: $f(x) = {x^2}$, $f(x) = {x^4}$
Some examples of odd function are: $f(x) = {x^3}$
When a graph is symmetric about an axis, then the axis acts as the mirror for the graph. And if the graph is symmetric about the origin, then the mirror is placed taking the origin as the reference line.

Note: We have used the fundamental property of an even function so as to equate and evaluate the value of the parameter. But if the function was odd, then the property under consideration would also change.