The answer
\((\sqrt{2},\,4)\) and \((-\sqrt{2},\,4)\), both minimum points
O-Level A-Math 2020 Paper 1 Question 6 · Verified worked solution by the Genius Plus Academy teaching team
The question
Find the coordinates of the stationary points of the curve \(y = x^2 + \dfrac{4}{x^2}\) and determine the nature of each stationary point. [6]
Write \(y = x^2 + 4x^{-2}\), so \(\dfrac{dy}{dx} = 2x - 8x^{-3} = 2x - \dfrac{8}{x^3}\).
At a stationary point \(\dfrac{dy}{dx} = 0\): \(2x = \dfrac{8}{x^3} \Rightarrow 2x^4 = 8 \Rightarrow x^4 = 4 \Rightarrow x^2 = 2\) (rejecting \(x^2 = -2\)), so \(x = \pm\sqrt{2}\).
At \(x = \pm\sqrt{2}\): \(y = (\pm\sqrt2)^2 + \dfrac{4}{(\pm\sqrt2)^2} = 2 + \dfrac{4}{2} = 4\). The stationary points are \((\sqrt{2},\,4)\) and \((-\sqrt{2},\,4)\).
Second derivative: \(\dfrac{d^2y}{dx^2} = 2 + 24x^{-4} = 2 + \dfrac{24}{x^4}\), which is \(> 0\) for all \(x \neq 0\). Hence both stationary points are minimum points.
Answer: \((\sqrt{2},\,4)\) and \((-\sqrt{2},\,4)\), both minimum points
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 stationary points question from Applications of differentiation, worth 6.
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