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TMUA 2017 · Paper 1 · Question 5 of 20

TMUA 2017 Paper 1 Question 5

Algebra and functions — Combining two inequalities into one quadratic. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Algebra and functionsCombining two inequalities into one quadratic8 options
S is the complete set of values of x which satisfy both the inequalities x2 − 8x + 12 < 0  and  2x + 1 > 9 The set S can also be represented as a single inequality.

Which one of the following single inequalities represents the set S?

  1. A(x2 − 8x + 12)(2x + 1) < 0
  2. B(x2 − 8x + 12)(2x + 1) > 0
  3. Cx2 − 10x + 24 < 0
  4. Dx2 − 10x + 24 > 0
  5. Ex2 − 6x + 8 < 0
  6. Fx2 − 6x + 8 > 0
  7. Gx < 2
  8. Hx > 6
Show the answer and worked solution
answer · C
  1. A(x2 − 8x + 12)(2x + 1) < 0
  2. B(x2 − 8x + 12)(2x + 1) > 0
  3. Cx2 − 10x + 24 < 0
  4. Dx2 − 10x + 24 > 0
  5. Ex2 − 6x + 8 < 0
  6. Fx2 − 6x + 8 > 0
  7. Gx < 2
  8. Hx > 6
The first inequality factorises as (x−2)(x−6) < 0, giving 2 < x < 6; the second gives x > 4. Both hold exactly when 4 < x < 6. An interval like that is the solution set of an upward quadratic with roots 4 and 6 taken negative: (x−4)(x−6) < 0, that is x2 − 10x + 24 < 0. Multiplying the two original left-hand sides together does not work, because a product is also negative when the factors have opposite signs.