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WAEC 2009 · Paper 2 · Q9

  1. (a)

    Simplify x+2x−2−x+3x−1\dfrac{x + 2}{x - 2} - \dfrac{x + 3}{x - 1}.

  2. (b)(i)

    The graph of the equation y=Ax2+Bx+Cy = Ax^2 + Bx + C passes through the points (0,0)(0, 0), (1,4)(1, 4) and (2,10)(2, 10). Find the value of CC.

  3. (b)(ii)

    Find the values of AA and BB.

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

  4. (b)(iii)

    Find the coordinates of the other point where the graph cuts the xx-axis.

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

Worked solution (try it first)

(a)

  1. Use the common denominator (x−2)(x−1)(x - 2)(x - 1): the top is (x+2)(x−1)−(x+3)(x−2)(x + 2)(x - 1) - (x + 3)(x - 2).
  2. Expand: (x2+x−2)−(x2+x−6)(x^2 + x - 2) - (x^2 + x - 6).
  3. Subtract: the top is 44, so the answer is 4(x−2)(x−1)\dfrac{4}{(x - 2)(x - 1)}.

(b)(i)

  1. Put in (0,0)(0, 0): 0=A(0)+B(0)+C0 = A(0) + B(0) + C, so C=0C = 0.

(ii)

  1. Put in (1,4)(1, 4): A+B=4A + B = 4.
  2. Put in (2,10)(2, 10): 4A+2B=104A + 2B = 10, so 2A+B=52A + B = 5.
  3. Subtract the first equation from the second: A=1A = 1.
  4. Then B=4−1=3B = 4 - 1 = 3.

(iii)

  1. The curve is y=x2+3xy = x^2 + 3x.
  2. On the xx-axis y=0y = 0: x(x+3)=0x(x + 3) = 0, so x=0x = 0 or x=−3x = -3.
  3. (0,0)(0, 0) is given, so the other point is (−3,0)(-3, 0).

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