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WAEC 2019 · Paper 2 · Q10

  1. (a)

    Find the range of values of pp for which 4x2−px+1=04x^2 - px + 1 = 0 has real roots.

    Show the answer

    p≤−4p \le -4 or p≥4p \ge 4

  2. (b)

    (i) Expand (1+3x)6(1 + 3x)^6 in ascending powers of xx. (ii) Using the expansion in (b)(i), find, correct to four significant figures, the value of (1.03)6(1.03)^6.

Worked solution (try it first)

(a)

  1. Read off the coefficients: a=4a = 4, b=−pb = -p, c=1c = 1.
  2. Real roots need b2−4ac≥0b^2 - 4ac \ge 0: p2−16≥0p^2 - 16 \ge 0.
  3. Factorise: (p−4)(p+4)≥0(p - 4)(p + 4) \ge 0, with roots p=−4p = -4 and p=4p = 4.
  4. "≥0\ge 0" is outside the roots: p≤−4p \le -4 or p≥4p \ge 4.

(b)(i)

  1. Each term is (6r)(3x)r\binom6r(3x)^r: the coefficients are 11, 6×36 \times 3, 15×915 \times 9, 20×2720 \times 27, 15×8115 \times 81, 6×2436 \times 243, 729729.
  2. So (1+3x)6=1+18x+135x2+540x3+1215x4+1458x5+729x6(1 + 3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6.

(ii)

  1. Find xx: 1+3x=1.031 + 3x = 1.03 gives x=0.01x = 0.01.
  2. Substitute: 1+0.18+0.0135+0.00054+0.00001215+…=1.19405…1 + 0.18 + 0.0135 + 0.00054 + 0.00001215 + \ldots = 1.19405\ldots
  3. So (1.03)6=1.194(1.03)^6 = 1.194 to 4 significant figures.

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