2025 ACJC P2 Q5

2025 ACJC P2 Q5

Junior College 2
12 marks
  1. By using the substitution \({x = a\tan\theta}\), show that
    \({\displaystyle\displaystyle \displaystyle\int \frac{1}{x\sqrt{x^2 + a^2}} \mathrm{d}x = \frac{1}{a}\ln\frac{x}{\sqrt{x^2 + a^2} + a} + c,}\)
    where \({x > 0}\) and \({a}\) is a positive constant.[4]
  2. At time \({t}\) in a chemical reaction, \({x}\) kg of substance \(X\) and \({y}\) kg of substance \(Y\) are present. Initially, there is \(4\) kg of \(X\) and \(5\) kg of \(Y\).
    The variables \({x}\) and \({y}\) satisfy the equations \({\frac{\mathrm{d}x}{\mathrm{d}t} = xy}\) and \({\frac{\mathrm{d}y}{\mathrm{d}t} = x^2}\).
    1. Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\), and by solving this differential equation, show that \(y = \sqrt{x^2+9}.\)[4]
    2. Obtain a differential equation in terms of \(x\) and \(t\) only and hence find an expression of \(t\) in terms of \(x\).[4]

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Answer:(a) \(\frac{1}{a}\ln\left(\frac{x}{\sqrt{x^2+a^2}+a}\right)+c\) (b)(i) \(y=\sqrt{x^2+9}\) (b)(ii) \(t=\frac{1}{3}\ln\left(\frac{x}{\sqrt{x^2+9}+3}\right)+\frac{1}{3}\ln2\)

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