WAEC 2017 · Paper 2 · Q11

  1. (a)

    If f(x)=∫(4x−x2) dxf(x) = \displaystyle\int(4x - x^2)\,dx and f(3)=21f(3) = 21, find f(x)f(x).

  2. (b)

    The second, fourth and eighth terms of an Arithmetic Progression (A.P.) form the first three consecutive terms of a Geometric Progression (G.P.). The sum of the third and fifth terms of the A.P. is 20. Find the: (i) first four terms of the A.P.; (ii) sum of the first ten terms of the A.P.

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

Worked solution (try it first)

(a)

  1. Integrate: f(x)=2x2−x33+cf(x) = 2x^2 - \dfrac{x^3}{3} + c.
  2. Use f(3)=21f(3) = 21: 18−9+c=2118 - 9 + c = 21, so c=12c = 12.
  3. Then f(x)=2x2−x33+12f(x) = 2x^2 - \dfrac{x^3}{3} + 12.

(b)(i)

  1. The 2nd, 4th and 8th terms of the A.P. are a+da + d, a+3da + 3d and a+7da + 7d.
  2. They form a G.P., so (a+3d)2=(a+d)(a+7d)(a + 3d)^2 = (a + d)(a + 7d).
  3. Expand: a2+6ad+9d2=a2+8ad+7d2a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2, so 2d2=2ad2d^2 = 2ad and d(d−a)=0d(d - a) = 0.
  4. An A.P. with d=0d = 0 would be constant, so d=ad = a.
  5. The 3rd and 5th terms add to 20: (a+2d)+(a+4d)=20(a + 2d) + (a + 4d) = 20, so 2a+6d=202a + 6d = 20.
  6. With d=ad = a: 8a=208a = 20, so a=d=52a = d = \frac52.
  7. The first four terms are 52,5,152,10\frac52, 5, \frac{15}{2}, 10.

(ii)

  1. S10=102[2(52)+9(52)]S_{10} = \dfrac{10}{2}\left[2\left(\frac52\right) + 9\left(\frac52\right)\right]
    =5(5+452)= 5\left(5 + \dfrac{45}{2}\right)
    =137.5= 137.5.

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