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TMUA 2020 · Paper 1 · Question 3 of 20

TMUA 2020 Paper 1 Question 3

Algebra and functions — Simultaneous inequalities · cubic and quadratic. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Algebra and functionsSimultaneous inequalities · cubic and quadratic7 options
Find the complete set of values of x for which (x+4)(x+3)(1−x) > 0   and   (x+2)(x−2) < 0
  1. A1 < x < 2
  2. B−2 < x < 1
  3. C−2 < x < 2
  4. Dx < −2 or x > 1
  5. Ex < −4 or x > 2
  6. Fx < −4 or −3 < x < 1
  7. G−4 < x < −2 or x > 1
Show the answer and worked solution
answer · B
  1. A1 < x < 2
  2. B−2 < x < 1
  3. C−2 < x < 2
  4. Dx < −2 or x > 1
  5. Ex < −4 or x > 2
  6. Fx < −4 or −3 < x < 1
  7. G−4 < x < −2 or x > 1
The second inequality is the easy one: (x+2)(x−2) < 0 gives −2 < x < 2. For the first, flip the last bracket to get a positive leading coefficient: the inequality is equivalent to (x+4)(x+3)(x−1) < 0. That cubic has roots −4,  −3,  1 and is negative on x < −4 and on −3 < x < 1. Intersecting with −2 < x < 2 leaves only −2 < x < 1. Option F is the first inequality on its own, which is the tempting mistake.