CAT 2024 Slot 2 · Quant · Question 20

Never cross-multiply an inequality whose denominators can change sign — move everything to one side instead.

Rewriting, \frac{1}{2x-3}-\frac{1}{x+5}\ge 0.

Combining over the common denominator, \frac{\left( x+5 \right)-\left( 2x-3 \right)}{\left( 2x-3 \right)\left( x+5 \right)}\ge 0.

The numerator simplifies to x+5-2x+3=8-x, so the condition is \frac{8-x}{\left( 2x-3 \right)\left( x+5 \right)}\ge 0.

The critical values are where the expression is zero or undefined: x=8, x=\frac{3}{2} and x=-5.

These cut the number line into four stretches, and the sign is constant on each.

For x<-5, say x=-6: the numerator is positive, and both bracketed factors are negative so their product is positive — the whole expression is positive, and these values are accepted.

For -5<x<\frac{3}{2}, say x=0: the numerator is positive, 2x-3 is negative and x+5 is positive, so the expression is negative — rejected.

For \frac{3}{2}<x<8, say x=2: numerator positive and denominator positive, so the expression is positive — accepted.

For x>8, say x=9: the numerator is negative while the denominator is positive — rejected.

At x=8 the expression is exactly zero, and the inequality allows equality, so x=8 is included.

At x=-5 and x=\frac{3}{2} the expression is undefined, so both are excluded.

The solution is x<-5 or \frac{3}{2}<x\le 8, the second option.

Shortcut. Options 3 and 4 both admit values just above -5, where the left side is a huge positive number and the right side is negative — plainly false. Testing x=0 alone kills half the options in seconds.