The answer
(a) \(-3, -2, -1, 0\)
(b) \(\dfrac{15pq^5}{2r^3}\)
(c) \(\dfrac{16b^8}{a^6}\)
(d) \(\dfrac{5x-2}{(x+2)(4-x)}\)
O-Level E-Math 2024 Paper 2 Question 1 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) Find all the integers that satisfy the inequality \(-4 \leqslant 3 + 2x < 5\). [3]
(b) Simplify \(\dfrac{20p^3q}{7r^2} \div \dfrac{8p^2r}{21q^4}\). [3]
(c) Simplify \(\left(\dfrac{a^9}{64b^{12}}\right)^{-\frac{2}{3}}\). [2]
(d) Write as a single fraction in its simplest form \(\dfrac{x}{x+2} + \dfrac{3}{4-x} - 1\). [4]
(a) Split the double inequality. \(-4 \leqslant 3+2x \Rightarrow 2x \geqslant -7 \Rightarrow x \geqslant -3.5\); and \(3+2x < 5 \Rightarrow 2x < 2 \Rightarrow x < 1\). So \(-3.5 \leqslant x < 1\), giving integers \(-3, -2, -1, 0\).
(b) Dividing flips the second fraction: \(\dfrac{20p^3q}{7r^2} \times \dfrac{21q^4}{8p^2r} = \dfrac{420\,p^3q^5}{56\,p^2r^3} = \dfrac{15pq^5}{2r^3}\).
(c) A negative index inverts the bracket; the power \(\tfrac23\) is a cube root then square: \(\left(\dfrac{64b^{12}}{a^9}\right)^{\frac{2}{3}} = \dfrac{64^{2/3}\,b^{8}}{a^{6}} = \dfrac{16b^8}{a^6}\) (since \(64^{1/3}=4\), \(4^2=16\)).
(d) Common denominator \((x+2)(4-x)\). Numerator \(= x(4-x) + 3(x+2) - (x+2)(4-x) = (4x-x^2) + (3x+6) - (2x - x^2 + 8) = 5x - 2\). So the fraction is \(\dfrac{5x-2}{(x+2)(4-x)}\).
Answer: (a) \(-3, -2, -1, 0\)
(b) \(\dfrac{15pq^5}{2r^3}\)
(c) \(\dfrac{16b^8}{a^6}\)
(d) \(\dfrac{5x-2}{(x+2)(4-x)}\)
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 laws of indices question from Algebraic manipulation / Equations & inequalities, worth 10 marks: 3 + 3 + 2 + 2 ....
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