Mathematics

1. 1.

   Find the area between y = 3x^2 + 2 and the x-axis from x = 0 to x = 2.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The curve is positive throughout the interval, so its definite integral is the required area.
   2. Increase each power by one and divide by the new power: integral(3x^2 + 2) = x^3 + 2x + C.
   3. Subtract the lower-limit value from the upper-limit value: (8 + 4) - 0 = 12.

   Marking points

   • An antiderivative is x^3 + 2x.
   • Evaluate [x^3 + 2x] from 0 to 2.
   • The area is 12 square units.

   Examiner tip: Area uses square units. The constant cancels in a definite integral.

2. 2.

   A particle has velocity v(t) = t^2 - 4t + 3 m s^-1 for 0 <= t <= 3 seconds. Find its displacement and total distance travelled in this interval.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Displacement counts direction; distance counts the magnitude of movement. First find where velocity is zero.
   2. Velocity is positive on (0, 1) and negative on (1, 3). F(0) = 0, F(1) = 4/3 and F(3) = 0.
   3. The signed contributions cancel, but their magnitudes add: |4/3| + |-4/3| = 8/3 m.

   Marking points

   • v(t) = (t - 1)(t - 3), so the sign changes at t = 1 inside the interval.
   • An antiderivative is F(t) = t^3/3 - 2t^2 + 3t.
   • Displacement = F(3) - F(0) = 0 m.
   • Distance = 4/3 + 4/3 = 8/3 m.

   Examiner tip: Zero displacement does not mean zero distance. Split the integral at every change in velocity sign.

3. 3.

   A model satisfies dP/dt = 0.2P(1 - P/100), with P(0) = 20. By separation of variables, obtain P(t) and find the time when P = 50. Time is in years.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Rewrite the separated fraction as 1/P + 1/(100 - P). The second term integrates to -ln(100 - P).
   2. At t = 0 the ratio P/(100 - P) is 20/80 = 1/4. Exponentiate and rearrange to isolate P.
   3. At P = 50 the ratio is 1, so e^(0.2t) = 4. The model approaches 100 rather than growing without bound.

   Marking points

   • Separate as 100/[P(100 - P)] dP = 0.2 dt.
   • Integration gives ln(P/(100 - P)) = 0.2t + C for 0 < P < 100.
   • Use P(0) = 20 to obtain P/(100 - P) = e^(0.2t)/4.
   • P(t) = 100/(1 + 4e^(-0.2t)).
   • P = 50 gives t = ln(4)/0.2, approximately 6.93 years.

   Examiner tip: Keep the minus sign when integrating 1/(100 - P). This is advanced concept practice; check your specification for logistic models.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.