The answer
(a) \(\dfrac{2p^3}{3r^2}\)
(b)(i) \(-10\)
(ii) \(b = \dfrac{5a - 4c}{a + 3}\)
(c)(i) \(-\dfrac{13}{4} + \left(x - \dfrac{7}{2}\right)^2\)
(ii) \(\left(\dfrac{7}{2}, -\dfrac{13}{4}\right)\)
(d) \(x = 5\) or \(x = 2.5\)
O-Level E-Math 2019 Paper 2 Question 1 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) Simplify \(\dfrac{4p^2 r}{3} \div \dfrac{2r^3}{p}\). [1]
(b) \(a = \dfrac{3b + 4c}{5 - b}\). (i) Evaluate \(a\) when \(b = 6\) and \(c = -2\). [1] (ii) Express \(b\) in terms of \(a\) and \(c\). [2]
(c) (i) Express \(9 - 7x + x^2\) in the form \(p + (q + x)^2\). [2] (ii) Write down the coordinates of the minimum point of the graph of \(9 - 7x + x^2\). [1]
(d) Solve \(\dfrac{1}{x - 3} + \dfrac{6}{x - 1} = 2\). [4]
(a) Dividing by a fraction is multiplying by its reciprocal: \[\frac{4p^2 r}{3} \div \frac{2r^3}{p} = \frac{4p^2 r}{3} \times \frac{p}{2r^3} = \frac{4p^3 r}{6r^3} = \frac{2p^3}{3r^2}.\]
(b)(i) \(a = \dfrac{3(6) + 4(-2)}{5 - 6} = \dfrac{18 - 8}{-1} = \dfrac{10}{-1} = -10\).
(ii) Multiply both sides by \((5 - b)\): \(a(5 - b) = 3b + 4c \Rightarrow 5a - ab = 3b + 4c\). Collect the \(b\) terms on one side: \(5a - 4c = 3b + ab = b(3 + a)\). Hence \(b = \dfrac{5a - 4c}{a + 3}\).
(c)(i) Write \(9 - 7x + x^2 = x^2 - 7x + 9\). Completing the square: \(\left(x - \dfrac{7}{2}\right)^2 - \dfrac{49}{4} + 9 = \left(x - \dfrac{7}{2}\right)^2 - \dfrac{13}{4}\). This is \(p + (q + x)^2\) with \(p = -\dfrac{13}{4}\) and \(q = -\dfrac{7}{2}\).
(ii) The squared term is least (zero) when \(x = \dfrac{7}{2}\), giving the minimum value \(-\dfrac{13}{4}\). Minimum point \(\left(\dfrac{7}{2}, -\dfrac{13}{4}\right) = (3.5, -3.25)\).
(d) Multiply through by \((x - 3)(x - 1)\): \[(x - 1) + 6(x - 3) = 2(x - 3)(x - 1).\] \(x - 1 + 6x - 18 = 2(x^2 - 4x + 3) \Rightarrow 7x - 19 = 2x^2 - 8x + 6 \Rightarrow 2x^2 - 15x + 25 = 0\). Factorising: \((2x - 5)(x - 5) = 0\), so \(x = \dfrac{5}{2}\) or \(x = 5\). (Both are valid; neither makes a denominator zero.)
Answer: (a) \(\dfrac{2p^3}{3r^2}\)
(b)(i) \(-10\)
(ii) \(b = \dfrac{5a - 4c}{a + 3}\)
(c)(i) \(-\dfrac{13}{4} + \left(x - \dfrac{7}{2}\right)^2\)
(ii) \(\left(\dfrac{7}{2}, -\dfrac{13}{4}\right)\)
(d) \(x = 5\) or \(x = 2.5\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a division of algebraic fractions question from Algebraic manipulation, worth 11 marks: 1 + 1 + 2 + 2 + 1 + 4.
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