The answer
(i) \(f'(x) = 16x^3 + e^{2x-1} - 3\)
(ii) \(f(x) = 4x^4 + \tfrac12 e^{2x-1} - 3x + 1\)
(iii) shown
O-Level A-Math 2020 Paper 2 Question 10 · Verified worked solution by the Genius Plus Academy teaching team
The question
The function \(f\) is defined for \(x \in \mathbb{R}\) with \(f''(x) = 48x^2 + 2e^{2x-1}\). The line \(x = \dfrac{1}{2}\) is the normal to \(y = f(x)\) at the point where \(y = \dfrac{1}{4}\).
(i) Find an expression for \(f'(x)\). [5]
(ii) Hence find an expression for \(f(x)\). [4]
(iii) Show that the tangent to \(y = f(x)\) where the curve meets the \(y\)-axis can be written as \(2e(y + 3x - 1) = 2x + 1\). [3]
(i) \(f'(x) = \displaystyle\int (48x^2 + 2e^{2x-1})\,dx = 16x^3 + e^{2x-1} + c\). The normal at the point is the vertical line \(x = \tfrac12\), so the tangent there is horizontal: \(f'\!\left(\tfrac12\right) = 0\). \[16\left(\tfrac18\right) + e^{0} + c = 2 + 1 + c = 0 \Rightarrow c = -3, \qquad f'(x) = 16x^3 + e^{2x-1} - 3.\]
(ii) \(f(x) = \displaystyle\int f'(x)\,dx = 4x^4 + \tfrac12 e^{2x-1} - 3x + c_2\). The point \(\left(\tfrac12, \tfrac14\right)\) lies on the curve: \[4\left(\tfrac{1}{16}\right) + \tfrac12 e^{0} - 3\left(\tfrac12\right) + c_2 = \tfrac14 + \tfrac12 - \tfrac32 + c_2 = \tfrac14 \Rightarrow c_2 = 1.\] So \(f(x) = 4x^4 + \dfrac12 e^{2x-1} - 3x + 1\).
(iii) The curve meets the \(y\)-axis at \(x = 0\): \(f(0) = \tfrac12 e^{-1} + 1 = 1 + \dfrac{1}{2e}\), and the gradient there is \(f'(0) = e^{-1} - 3 = \dfrac{1}{e} - 3\). The tangent is \[y - \left(1 + \tfrac{1}{2e}\right) = \left(\tfrac1e - 3\right)x \Rightarrow y + 3x - 1 = \tfrac1e x + \tfrac{1}{2e} = \tfrac1e\left(x + \tfrac12\right) = \dfrac{2x + 1}{2e}.\] Multiplying both sides by \(2e\): \(\;2e(y + 3x - 1) = 2x + 1\). (shown)
Answer: (i) \(f'(x) = 16x^3 + e^{2x-1} - 3\)
(ii) \(f(x) = 4x^4 + \tfrac12 e^{2x-1} - 3x + 1\)
(iii) shown
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 integrate f″ twice question from Integration / Applications of differentiation, worth 12 marks: 5 + 4 + 3.
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