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WAEC 2008 · Paper 2 · Q9

  1. (a)

    A curve has gradient 3x2−2x+13x^2 - 2x + 1 at the point (x,y)(x, y). If it passes through the point (1,3)(1, 3), find its equation.

  2. (b)

    The first term of an arithmetic progression is 3 and the nnth term is 48. If the sum of the first nn terms is 255, find the: (i) value of nn; (ii) smallest value of rr for which the rrth term exceeds 149.

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

Worked solution (try it first)

(a)

  1. dydx=3x2−2x+1\dfrac{dy}{dx} = 3x^2 - 2x + 1, so integrate: y=x3−x2+x+cy = x^3 - x^2 + x + c.
  2. The curve passes through (1,3)(1, 3): 3=1−1+1+c3 = 1 - 1 + 1 + c, so c=2c = 2.
  3. The equation is y=x3−x2+x+2y = x^3 - x^2 + x + 2.

(b)(i)

  1. Use Sn=n2(a+l)S_n = \frac{n}{2}(a + l): 255=n2(3+48)255 = \frac{n}{2}(3 + 48).
  2. So 255=51n2255 = \frac{51n}{2}, giving n=51051=10n = \frac{510}{51} = 10.

(ii)

  1. Find dd from the 10th term: 3+9d=483 + 9d = 48, so d=5d = 5.
  2. The rrth term is 3+5(r−1)3 + 5(r - 1).
  3. It exceeds 149 when 5(r−1)>1465(r - 1) > 146, so r−1>29.2r - 1 > 29.2.
  4. The smallest whole number is r=31r = 31 (the 30th term is 148 and the 31st is 153).

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