Question

Select the correct relation between $E$ and $F$. $E=\dfrac{x}{1+x}$ \; and \; $F=\dfrac{-x}{\,1-x\,}$, \; with $x>1$.

• A. $E>F$
• B. $E
• C. $E=F$
• D. $E<-F$

Hint:

When comparing two rational expressions with the same positive numerator pattern, reduce to one fraction: $A/B - A/C = A\,(C-B)/(BC)$. For $x>1$, signs are easy to track since all denominators are positive.

Answer 1

The Correct Option is B

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 $F-E>0$. \[ \boxed{F>E\ \Rightarrow\ E<F.} \]