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Question: If the real valued function \[F(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}}\] is even , then n is e...

If the real valued function F(x)=ax−1xn(ax+1)F(x) = \dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} is even , then n is equal to :
A) 22
B) 23\dfrac{2}{3}
C) 14\dfrac{1}{4}
D) 11

Explanation

Solution

In this question it is given that F(x)F(x) is even function hence form the even function definition we know that F(x)=F(−x)F(x) = F( - x)means that ax−1xn(ax+1)=a−x−1(−x)n(a−x+1)\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)=ax−1xn(ax+1)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)F(x) = F( - x) . So for this we have to put −x - x in the place of xx ,
F(−x)=a−x−1(−x)n(a−x+1)F( - x) = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}
So from the even function definition we know that F(x)=F(−x)F(x) = F( - x)
hence ,
ax−1xn(ax+1)=a−x−1(−x)n(a−x+1)\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{{a^{ - x}} - 1}}{{{{( - x)}^n}({a^{ - x}} + 1)}}
Now we know that a−x=1ax{a^{ - x}} = \dfrac{1}{{{a^x}}} so put this into RHS , we get
ax−1xn(ax+1)=1ax−1(−x)n(1ax+1)\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{\dfrac{1}{{{a^x}}} - 1}}{{{{( - x)}^n}\left( {\dfrac{1}{{{a^x}}} + 1} \right)}}
ax−1xn(ax+1)=1−axax(−x)n(1+axax)\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 ax{a^x} is common in both numerator and denominator so it will cancel out the remaining equation become
ax−1xn(ax+1)=1−ax(−x)n(ax+1)\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{1 - {a^x}}}{{{{( - x)}^n}({a^x} + 1)}}
ax−1xn(ax+1)=−1(ax−1)(−x)n(ax+1)\dfrac{{{a^x} - 1}}{{{x^n}({a^x} + 1)}} = \dfrac{{ - 1({a^x} - 1)}}{{{{( - x)}^n}({a^x} + 1)}}
Hence in the equation ax−1ax+1\dfrac{{{a^x} - 1}}{{{a^x} + 1}} is common in both LHS and RHS hence it will cancel out the remaining equation become
1xn=−1(−x)n\dfrac{1}{{{x^n}}} = \dfrac{{ - 1}}{{{{( - x)}^n}}}
Take common (−1)n{( - 1)^n} in RHS from denominator side , the equation become
1xn=−1(−1)n(x)n\dfrac{1}{{{x^n}}} = \dfrac{{ - 1}}{{{{( - 1)}^n}{{(x)}^n}}}
11=−1(−1)n\dfrac{1}{1} = \dfrac{{ - 1}}{{{{( - 1)}^n}}}
Now on cross multiplication we get (−1)n=−1{( - 1)^n} = - 1 hence n=1n = 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)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{( - 1)^n} = - 1 the possible value of n=1,3,5,7,9....n = 1,3,5,7,9.... but we take only n=1n = 1 because in the option only n=1n = 1 is given .