WAEC 2014 · Paper 2 · Q10

  1. (a)

    If P=(42c3)P = \begin{pmatrix} 4 & 2 \\ c & 3 \end{pmatrix}, Q=(624d)Q = \begin{pmatrix} 6 & 2 \\ 4 & d \end{pmatrix} and PQ=QPPQ = QP, find the values of cc and dd.

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

  2. (b)

    Using the trapezium rule with five ordinates, calculate, correct to two decimal places, the approximate value of ∫26dxx\displaystyle\int_2^6 \frac{dx}{x}.

Worked solution (try it first)

(a)

  1. PQ=(328+2d6c+122c+3d)PQ = \begin{pmatrix} 32 & 8 + 2d \\ 6c + 12 & 2c + 3d \end{pmatrix} and QP=(24+2c1816+cd8+3d)QP = \begin{pmatrix} 24 + 2c & 18 \\ 16 + cd & 8 + 3d \end{pmatrix}.
  2. Match the top-left entries: 32=24+2c32 = 24 + 2c, so c=4c = 4.
  3. Match the top-right entries: 8+2d=188 + 2d = 18, so d=5d = 5.
  4. Check the others: 6c+12=36=16+cd6c + 12 = 36 = 16 + cd ✓ and 2c+3d=23=8+3d2c + 3d = 23 = 8 + 3d ✓.

(b)

  1. Five ordinates means four strips, so h=6−24=1h = \dfrac{6 - 2}{4} = 1.
  2. The ordinates of 1x\frac1x at x=2,3,4,5,6x = 2, 3, 4, 5, 6 are 0.5, 0.3333, 0.25, 0.2, 0.16670.5,\ 0.3333,\ 0.25,\ 0.2,\ 0.1667.
  3. 12[(0.5+0.1667)+2(0.3333+0.25+0.2)]=12(0.6667+1.5667)\frac12[(0.5 + 0.1667) + 2(0.3333 + 0.25 + 0.2)] = \frac12(0.6667 + 1.5667)
    =1.1167= 1.1167, about 1.121.12.

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