Select the correct relation between $E$ and $F$.
$E=\dfrac{x}{1+x}$ \; and \; $F=\dfrac{-x}{\,1-x\,}$, \; with $x>1$.
Show Hint
- $E>F$
- $E
- $E=F$
- $E<-F$
The Correct Option is B
Solution and Explanation
Step 1: Simplify $F$.
\[
F=\frac{-x}{1-x}=\frac{x}{x-1}(\text{multiply numerator and denominator by }-1).
\]
Step 2: Compare $F$ and $E$ by subtraction.
For $x>1$, denominators $x-1$ and $x+1$ are positive. Compute
\[
F-E=\frac{x}{x-1}-\frac{x}{x+1}
=\frac{x\big[(x+1)-(x-1)\big]}{(x-1)(x+1)}
=\frac{2x}{x^2-1}.
\]
Since $x>1\Rightarrow x^2-1>0$, we have $F-E>0$.
\[
\boxed{F>E\ \Rightarrow\ E<F.}
\]
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