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CAT 2024 Slot 2 — QA Question 15

Solving InequalitiesEasy

The value of x satisfying the inequality 1x+5 ≤ 12x-3 are

Answer & solution

  • A

    -5 < x < 3/2 or 3/2 < x ≤ 8

  • B

    x < -5 or x > 3/2

  • x < -5 or 3/2 < x ≤ 8

  • D

    -5 < x < 3/2 or x > 3/2

Solution

Medium

Never cross-multiply an inequality with variable denominators. Move everything to one side, combine into a single rational expression, and use a sign chart on the critical points.

1

Combine into one fraction. Rearrange 1x+5≤12x−3\dfrac{1}{x+5}\le\dfrac{1}{2x-3}.

1x+5−12x−3≤0⇒ (2x−3)−(x+5)(x+5)(2x−3)≤0(common denominator)⇒ x−8(x+5)(2x−3)≤0\begin{aligned} &\frac{1}{x+5}-\frac{1}{2x-3}\le 0\\ &\Rightarrow\ \frac{(2x-3)-(x+5)}{(x+5)(2x-3)}\le 0 \quad\text{(common denominator)}\\ &\Rightarrow\ \frac{x-8}{(x+5)(2x-3)}\le 0 \end{aligned}
2

Critical points and sign chart. Zeros/undefined at x=−5, 32, 8x=-5,\ \tfrac32,\ 8. Test each interval (x=−5x=-5 and x=32x=\tfrac32 excluded as denominators vanish; x=8x=8 included since the expression is 00).

\begin{aligned} &x<-5:\ \tfrac{(-)}{(-)(-)}<0\ \checkmark\\ &-50\ \times\\ &\tfrac328:\ \tfrac{(+)}{(+)(+)}>0\ \times \end{aligned}
3

Collect the solution. Keep the intervals where the expression is ≤0\le 0, including x=8x=8.

\begin{aligned} &x<-5\quad\text{or}\quad \tfrac32
x<-5\ \text{ or }\ \tfrac32
CAT 2024 Slot 2 QA Q15: The value of x satisfying the inequality 1 x + 5 &le; 1 2 x - 3 are — Solution | TheCATExam