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WAEC 2016 · Paper 2 · Q11

  1. (a)

    If kP2=72{}^kP_2 = 72, find the value of kk.

  2. (b)

    Solve the equation 2cos⁡2θ−5cos⁡θ=32\cos^2\theta - 5\cos\theta = 3, for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1.  kP2=k(k−1)=72\,{}^kP_2 = k(k - 1) = 72, so k2−k−72=0k^2 - k - 72 = 0.
  2. Factorise: (k−9)(k+8)=0(k - 9)(k + 8) = 0.
  3. kk must be a positive whole number, so k=9k = 9.

(b)

  1. Let c=cos⁡θc = \cos\theta: 2c2−5c−3=02c^2 - 5c - 3 = 0, so (2c+1)(c−3)=0(2c + 1)(c - 3) = 0.
  2. cos⁡θ=3\cos\theta = 3 is impossible, so cos⁡θ=−12\cos\theta = -\frac12.
  3. Cosine is negative in the second and third quadrants: θ=180∘−60∘=120∘\theta = 180^\circ - 60^\circ = 120^\circ or θ=180∘+60∘=240∘\theta = 180^\circ + 60^\circ = 240^\circ.

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