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Review practice · Section 1 of 2

Questions 1–6 in this collection

Separate continuous payoff time from monthly rounding · Select the nonnegative branch of an integrated equation · +4 more

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  1. 1

    In a loan scheme, the amount owed is MM dollars at time tt months; both variables are continuous. Interest increases the amount at a rate proportional to the amount owed, while repayment occurs at a constant rate of $500 per month. Interest and repayment balance when M=10000M=10000. An initial loan of $8000 is taken on 1 March 2013.

    (i) Show dM/dt=(M−10000)/20dM/dt=(M-10000)/20. marks

    (ii) Find MM in terms of tt. marks

    (iii) Find the date when the loan is cleared. marks

    [2013/PJC/II/3]

    CJC · Review problems · Question 1

  2. 2

    Show that the general solution of y(dy/dx)=2x−1y(dy/dx)=2x-1, y≥0y\ge0, is y2=2x2−2x+Cy^2=2x^2-2x+C, where CC is arbitrary. Sketch the curve passing through (0,1)(0,1).

    CJC · Review problems · Question 2

  3. 3

    (i) Use z=(2x−1)dy/dxz=(2x-1)dy/dx to express (2x−1)d2y/dx2=2(1−dy/dx)(2x-1)d^2y/dx^2=2(1-dy/dx) as an equation in z,xz,x. marks

    (ii) Solve with y=2y=2 and dy/dx=2dy/dx=2 at x=0x=0. marks

    (iii) Describe the gradient as (a) x→±∞x\to\pm\infty mark and (b) x→1/2x\to1/2. mark

    (iv) Sketch the solution from (ii). marks

    [2013/IJC/II/4(modified)]

    CJC · Review problems · Question 3

  4. 4

    Using y=vxy=vx, solve dy/dx=y(x+y)/[x(y−x)]dy/dx=y(x+y)/[x(y-x)].

    CJC · Review problems · Question 4

  5. 5

    (a) Find y=f(x)y=f(x) through (1,0)(1,0) and (−1,2)(-1,2) with dy/dx=3x2−x−2+Cdy/dx=3x^2-x^{-2}+C. marks

    (b) Perry jogs a distance xx kilometres in tt hours. The difference between his initial speed and his speed at time tt is proportional to distance covered. He starts at 55 km/h and travels at 44 km/h after 22 km. Find the exact time for a 55 km jog. marks

    [2013/YJC/I/11 (modified)]

    CJC · Review problems · Question 5

  6. 6

    A town has population 30003000. The number infected after tt days satisfies dI/dt=kI(3000−I)dI/dt=kI(3000-I), k>0k>0.

    (i) Given I(0)=30I(0)=30, show I=3000/(1+99e−3000kt)I=3000/(1+99e^{-3000kt}). marks

    (ii) Given I(7)=240I(7)=240, find when 90%90\% of the population is infected. marks

    (iii) State one contextual assumption needed for the model. mark

    [2013/JJC/II/2]

    CJC · Review problems · Question 6

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