The answer
(i) shown
(ii) \(\dfrac{1}{8}(x+1)(3x-5)^{\frac{5}{3}} + c\) (\(k = \tfrac18\))
(iii) \(0\), equal areas above and below the \(x\)-axis
O-Level A-Math 2020 Paper 2 Question 7 · Verified worked solution by the Genius Plus Academy teaching team
The question
(i) Show that \(\dfrac{d}{dx}\left\{x(3x-5)^{\frac{5}{3}}\right\} = (8x-5)(3x-5)^{\frac{2}{3}}\). [5]
(ii) Hence find \(\displaystyle\int x(3x-5)^{\frac{2}{3}}\,dx\), in the form \(k(x+1)(3x-5)^{\frac{5}{3}} + c\). [4]
(iii) Find \(\displaystyle\int_{-1}^{\frac{5}{3}} x(3x-5)^{\frac{2}{3}}\,dx\) and explain what this implies about the curve \(y = x(3x-5)^{\frac{2}{3}}\). [2]
(i) Product rule, with chain rule on \((3x-5)^{5/3}\): \[\dfrac{d}{dx}\left\{x(3x-5)^{5/3}\right\} = (3x-5)^{5/3} + x\cdot\tfrac{5}{3}(3x-5)^{2/3}\cdot 3 = (3x-5)^{5/3} + 5x(3x-5)^{2/3}.\] Factor out \((3x-5)^{2/3}\): \(= (3x-5)^{2/3}\big[(3x-5) + 5x\big] = (8x-5)(3x-5)^{2/3}\). (shown)
(ii) From (i), \(\displaystyle\int (8x-5)(3x-5)^{2/3}\,dx = x(3x-5)^{5/3}\). Split the left side and use \(\displaystyle\int (3x-5)^{2/3}dx = \dfrac{(3x-5)^{5/3}}{3 \cdot \tfrac53} = \tfrac15(3x-5)^{5/3}\): \[8\!\int\! x(3x-5)^{2/3}dx - 5\!\int\!(3x-5)^{2/3}dx = x(3x-5)^{5/3} \Rightarrow 8\!\int\! x(3x-5)^{2/3}dx = x(3x-5)^{5/3} + (3x-5)^{5/3}.\] So \(\displaystyle\int x(3x-5)^{2/3}\,dx = \dfrac{1}{8}(x+1)(3x-5)^{5/3} + c\), giving \(k = \dfrac18\).
(iii) \(\displaystyle\int_{-1}^{5/3} x(3x-5)^{2/3}dx = \dfrac{1}{8}\Big[(x+1)(3x-5)^{5/3}\Big]_{-1}^{5/3}\). At \(x = \tfrac53\) the factor \((3x-5)^{5/3} = 0\); at \(x = -1\) the factor \((x+1) = 0\). Both ends give \(0\), so the integral \(= 0\). Since \((3x-5)^{2/3} \geqslant 0\), the curve is below the \(x\)-axis for \(-1 < x < 0\) and above it for \(0 < x < \tfrac53\); the zero value means the **area below the \(x\)-axis (from \(-1\) to \(0\)) exactly equals the area above it (from \(0\) to \(\tfrac53\))**, so the signed area cancels to zero.
Answer: (i) shown
(ii) \(\dfrac{1}{8}(x+1)(3x-5)^{\frac{5}{3}} + c\) (\(k = \tfrac18\))
(iii) \(0\), equal areas above and below the \(x\)-axis
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 A-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 product rule + chain rule question from Differentiation & Integration, worth 11 marks: 5 + 4 + 2.
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