Stepwise
Start practisingPractise

WAEC 2009 · Paper 2 · Q11

  1. (a)

    Evaluate ∫12x5−x2 dx\displaystyle\int_1^2 \frac{x}{\sqrt{5 - x^2}}\,dx.

  2. (b)

    (i) Evaluate ∣2−3101−212−3∣\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}. (ii) Using your answer in (b)(i), solve the simultaneous equations 2x−3y+z=102x - 3y + z = 10, y−2z=−7y - 2z = -7, x+2y−3z=−9x + 2y - 3z = -9.

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

Worked solution (try it first)

(a)

  1. Substitute t2=5−x2t^2 = 5 - x^2.
  2. Then 2t dt=−2x dx2t\,dt = -2x\,dx, so x dx=−t dtx\,dx = -t\,dt.
  3. Change the limits: x=1x = 1 gives t=2t = 2, and x=2x = 2 gives t=1t = 1.
  4. The integral becomes ∫21−tt dt=∫21(−1) dt\displaystyle\int_2^1 \frac{-t}{t}\,dt = \int_2^1 (-1)\,dt.
  5. Evaluate: [−t]21=−1−(−2)=1[-t]_2^1 = -1 - (-2) = 1.

(b)(i)

  1. Expand along the top row: 2(−3+4)+3(0+2)+1(0−1)2(-3 + 4) + 3(0 + 2) + 1(0 - 1), which is 2+6−1=72 + 6 - 1 = 7.

(ii)

  1. By Cramer's rule, replace each column in turn by (10,−7,−9)(10, -7, -9).
  2. Δx=∣10−31−71−2−92−3∣\Delta_x = \begin{vmatrix} 10 & -3 & 1 \\ -7 & 1 & -2 \\ -9 & 2 & -3 \end{vmatrix}
    =14= 14, Δy=−7\Delta_y = -7 and Δz=21\Delta_z = 21.
  3. So x=147=2x = \frac{14}{7} = 2, y=−77=−1y = \frac{-7}{7} = -1 and z=217=3z = \frac{21}{7} = 3.
  4. Check in the third equation: 2−2−9=−92 - 2 - 9 = -9 ✓.

Report a problem with this question