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TMUA 2019 · Paper 1 · Question 2 of 20

TMUA 2019 Paper 1 Question 2

Algebra and functions — Discriminant · quadratic positive for all x. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Algebra and functionsDiscriminant · quadratic positive for all x8 options
Find the complete set of values of the real constant k for which the expression x2 + kx + 2x + 1 − 2k is positive for all real values of x.
  1. A−12 < k < 0
  2. Bk < −12 or k > 0
  3. C−√6 − 3 < k < √6 − 3
  4. Dk < −√6 − 3 or k > √6 − 3
  5. E−2 < k < 12
  6. Fk < −2 or k > 12
  7. G0 < k < 4
  8. Hk < 0 or k > 4
Show the answer and worked solution
answer · A
  1. A−12 < k < 0
  2. Bk < −12 or k > 0
  3. C−√6 − 3 < k < √6 − 3
  4. Dk < −√6 − 3 or k > √6 − 3
  5. E−2 < k < 12
  6. Fk < −2 or k > 12
  7. G0 < k < 4
  8. Hk < 0 or k > 4
Collect the x terms first: the expression is x2 + (k+2)x + (1 − 2k). The coefficient of x2 is positive, so the quadratic is positive everywhere exactly when it has no real root, that is when the discriminant is negative. That gives (k+2)2 − 4(1 − 2k) < 0, so k2 + 4k + 4 − 4 + 8k < 0, i.e. k2 + 12k < 0. Factorising, k(k + 12) < 0, so −12 < k < 0.