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WAEC 2020 · Paper 2 · Q10

  1. (a)

    If 15y×5(2y−2)×9(y−1)=115^y \times 5^{(2y - 2)} \times 9^{(y - 1)} = 1, find the value of yy.

  2. (b)

    Solve sin⁡2x+cos⁡2x=116\sin^2x + \cos2x = \frac{1}{16}, 0∘≤x≤360∘0^\circ \le x \le 360^\circ (2 d.p.).

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

Worked solution (try it first)

(a)

  1. Split into primes: 15y=3y5y15^y = 3^y5^y and 9y−1=32y−29^{y - 1} = 3^{2y - 2}.
  2. Collect the powers of 3: 3y×32y−2=33y−23^y \times 3^{2y - 2} = 3^{3y - 2}.
  3. Collect the powers of 5: 5y×52y−2=53y−25^y \times 5^{2y - 2} = 5^{3y - 2}.
  4. So the left side is 33y−253y−2=153y−23^{3y - 2}5^{3y - 2} = 15^{3y - 2}, and 1=1501 = 15^0.
  5. Compare the indices: 3y−2=03y - 2 = 0, so y=23y = \frac23.

(b)

  1. Use cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x: the left side is sin⁡2x+cos⁡2x−sin⁡2x=cos⁡2x\sin^2 x + \cos^2 x - \sin^2 x = \cos^2 x.
  2. So cos⁡2x=116\cos^2 x = \frac{1}{16} and cos⁡x=14\cos x = \frac14 or cos⁡x=−14\cos x = -\frac14.
  3. cos⁡x=14\cos x = \frac14: x=75.52∘x = 75.52^\circ or 360∘−75.52∘=284.48∘360^\circ - 75.52^\circ = 284.48^\circ.
  4. cos⁡x=−14\cos x = -\frac14: x=180∘−75.52∘x = 180^\circ - 75.52^\circ
    =104.48∘= 104.48^\circ or 180∘+75.52∘=255.52∘180^\circ + 75.52^\circ = 255.52^\circ.

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