Question

If the real valued function \[F(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}}\] is even , then n is equal to :
A) $2$
B) $\dfrac{2}{3}$
C) $\dfrac{1}{4}$
D) $1$

Answer 1

Hint: In this question it is given that $F(x)$ is even function hence form the even function definition we know that $F(x) = F( - x)$means that \[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}\] solve the equation and get the value of n .

Complete step by step answer:
As in the question \[F(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}}\] it is given ,
Hence for the function is even it means that $F(x) = F( - x)$ . So for this we have to put $ - x$ in the place of $x$ ,
\[F( - x) = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}\]
So from the even function definition we know that $F(x) = F( - x)$
hence ,
\[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}\]
Now we know that ${a^{ - x}} = \dfrac{1}{{{a^x}}}$ so put this into RHS , we get
\[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{\dfrac{1}{{{a^x}}} - 1}}{{{{( - x)}^n}\left( {\dfrac{1}{{{a^x}}} + 1} \right)}}\]
\[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{\dfrac{{1 - {a^x}}}{{{a^x}}}}}{{{{( - x)}^n}\left( {\dfrac{{1 + {a^x}}}{{{a^x}}}} \right)}}\]
So in the above equation in RHS ${a^x}$ is common in both numerator and denominator so it will cancel out the remaining equation become
\[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{1 - {a^x}}}{{{{( - x)}^n}({a^x} + 1)}}\]
\[\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{ - 1({a^x} - 1)}}{{{{( - x)}^n}({a^x} + 1)}}\]
Hence in the equation \[\dfrac{{{a^x} - 1}}{{{a^x} + 1}}\] is common in both LHS and RHS hence it will cancel out the remaining equation become
\[\dfrac{1}{{{x^n}}} = \dfrac{{ - 1}}{{{{( - x)}^n}}}\]
Take common ${( - 1)^n}$ in RHS from denominator side , the equation become
\[\dfrac{1}{{{x^n}}} = \dfrac{{ - 1}}{{{{( - 1)}^n}{{(x)}^n}}}\]
\[\dfrac{1}{1} = \dfrac{{ - 1}}{{{{( - 1)}^n}}}\]
Now on cross multiplication we get ${( - 1)^n} = - 1$ hence $n = 1$

$\therefore $ n=1, Hence, option D is correct answer.

Note:
As in the question, even function is given if in the place of even, the odd function is given then we apply $F(x) = - F(x)$ and the rest of things are the same solve this and the value for n.
As in the last step ${( - 1)^n} = - 1$ the possible value of $n = 1,3,5,7,9....$ but we take only $n = 1$ because in the option only $n = 1$ is given .