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WAEC 2016 · Paper 2 · Q11

  1. (a)

    Without using mathematical tables or a calculator, solve 2+log⁡10x−log⁡1020=log⁡10(x2+4)2 + \log_{10} x - \log_{10} 20 = \log_{10}(x^2 + 4).

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

  2. (b)

    The equation of a circle is given by x2+y2+2x−6y+n=0x^2 + y^2 + 2x - 6y + n = 0, where nn is a constant. If the circle has a radius of 2 units, find the value of nn.

  3. (c)

    Find the equation of the tangent to the curve y=7x−4x2y = 7x - 4x^2 at the point where x=1x = 1.

    Show the answer

    x+y−4=0x + y - 4 = 0

Worked solution (try it first)

(a)

  1. Write 2 as a log: 2=log⁡101002 = \log_{10} 100.
  2. The left side is then log⁡10100x20=log⁡105x\log_{10}\dfrac{100x}{20} = \log_{10} 5x.
  3. Drop the logs: 5x=x2+45x = x^2 + 4, so x2−5x+4=0x^2 - 5x + 4 = 0.
  4. Factorise: (x−1)(x−4)=0(x - 1)(x - 4) = 0, so x=1x = 1 or x=4x = 4.
  5. Both make every log positive.

(b)

  1. Complete the squares: (x+1)2+(y−3)2=1+9−n(x + 1)^2 + (y - 3)^2 = 1 + 9 - n.
  2. The radius squared is 10−n=22=410 - n = 2^2 = 4, so n=6n = 6.

(c)

  1. At x=1x = 1: y=7−4=3y = 7 - 4 = 3.
  2. The gradient is dydx=7−8x\dfrac{dy}{dx} = 7 - 8x, which is −1-1 at x=1x = 1.
  3. The tangent is y−3=−1(x−1)y - 3 = -1(x - 1), so y=−x+4y = -x + 4, that is x+y−4=0x + y - 4 = 0.

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