Question

If \(\frac{x+1}{x-1}<2\), then \(x\) lies in the interval

• A. \((-\infty,-3)\cup(1,\infty)\)
• B. \((-\infty,-1)\cup(3,\infty)\)
• C. \((-\infty,1)\cup(3,\infty)\)
• D. \((-3,-1)\)
• E. \((1,3)\)

Hint:

For rational inequalities, always bring everything to one side and use sign analysis around critical points (where numerator or denominator is zero).

Answer 1

The Correct Option is C

Step 1: Write the inequality clearly.
We are given:
\[ \frac{x+1}{x-1}<2 \] We need to solve this inequality for \(x\).

Step 2: Bring all terms to one side.
\[ \frac{x+1}{x-1}-2<0 \] Take LCM:
\[ \frac{x+1-2(x-1)}{x-1}<0 \]

Step 3: Simplify the numerator.
\[ x+1-2x+2=-x+3 \] So the inequality becomes:
\[ \frac{-x+3}{x-1}<0 \]

Step 4: Rewrite for easier sign analysis.
\[ \frac{-(x-3)}{x-1}<0 \] Multiply both sides by \(-1\) (which reverses the inequality):
\[ \frac{x-3}{x-1}>0 \]

Step 5: Find critical points.
The expression changes sign at:
\[ x=1 \quad \text{and} \quad x=3 \] These divide the number line into intervals:
\[ (-\infty,1),\ (1,3),\ (3,\infty) \]

Step 6: Check the sign in each interval.
- For \(x<1\): both numerator and denominator are negative, so the fraction is positive.
- For \(1<x<3\): numerator negative, denominator positive → fraction negative.
- For \(x>3\): both numerator and denominator positive → fraction positive.
So the solution is:
\[ (-\infty,1)\cup(3,\infty) \]

Step 7: Match with the options.
This matches option \((3)\). Hence, the correct answer is:
\[ \boxed{(-\infty,1)\cup(3,\infty)} \]