Question : If $m x^m-n x^n=0$, then what is the value of $\frac{1}{x^m+x^n}+\frac{1}{x^m-x^n}$ terms of $x^n$ is: Where $x,m,n$ are $>0$
Option 1: $\frac{2 m n}{x^n(n^2+m^2)}$
Option 2: $\frac{2 m n }{x^n(m^2-n^2)}$
Option 3: $\frac{2 m n}{x^n\left(m^2+n^2\right)}$
Option 4: $\frac{2 m n }{x^n(n^2-m^2)}$
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Correct Answer: $\frac{2 m n }{x^n(n^2-m^2)}$
Solution :
Given, $mx^m − nx^n = 0$
⇒ $mx^m = nx^n$
⇒ $\frac{x^m}{x^n} = \frac{n}{m}$
Now,
$\frac{1}{x^m+x^n}+\frac{1}{x^m-x^n}$
= $\frac{x^m-x^n + x^m+x^n}{(x^m+x^n)(x^m-x^n)}$
= $\frac{2x^m}{(x^m+x^n)(x^{2m}-x^n)}$
= $\frac{n-m+n+m}{(n+m)(n-m)}$
= $\frac{2n}{(n^2−m^2)}$
Now, the value of $\frac{1}{x^m+x^n}+\frac{1}{x^m−x^n}$ in terms of $x^n$
= $\frac{2n}{n^2-m^2}\times\frac{x^n}{x^n}$
= $\frac{2nm}{x^n(n^2-m^2)}$ [As $x^n=m$]
Hence, the correct answer is $\frac{2mn}{x^n(n^2-m^2)}$.
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