WAEC 2009 · Paper 2 · Q12

  1. (a)

    Use the trapezium rule with five ordinates to evaluate ∫01x1+x2 dx\displaystyle\int_0^1 \frac{x}{1 + x^2}\,dx, correct to four significant figures.

  2. (b)

    If A=(2132)A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, find the image of the point (1,2)(1, 2) under the linear transformation A2+A+2IA^2 + A + 2I, where II is the 2×22 \times 2 unit matrix.

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

Worked solution (try it first)

(a)

  1. Five ordinates means four strips, so h=1−04=0.25h = \frac{1 - 0}{4} = 0.25.
  2. Ordinates of y=x1+x2y = \dfrac{x}{1 + x^2} at x=0,0.25,0.5,0.75,1x = 0, 0.25, 0.5, 0.75, 1: 00, 0.235290.23529, 0.40.4, 0.480.48, 0.50.5.
  3. Trapezium rule: h2[y0+y4+2(y1+y2+y3)]=0.125[0+0.5+2(1.11529)]\frac{h}{2}[y_0 + y_4 + 2(y_1 + y_2 + y_3)] = 0.125[0 + 0.5 + 2(1.11529)].
  4. =0.125×2.73059=0.34132= 0.125 \times 2.73059 = 0.34132, which is 0.34130.3413 to 4 significant figures.

(b)

  1. A2=(2132)(2132)A^2 = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}
    =(74127)= \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix}.
  2. Add AA and 2I2I: A2+A+2I=(7+2+24+112+37+2+2)A^2 + A + 2I = \begin{pmatrix} 7 + 2 + 2 & 4 + 1 \\ 12 + 3 & 7 + 2 + 2 \end{pmatrix}
    =(1151511)= \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}.
  3. Multiply by (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix}: (11+1015+22)=(2137)\begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}.
  4. The image is (21,37)(21, 37).

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