Challenging +1.2 This requires implicit differentiation to find dy/dx, setting it to zero, and solving the resulting system. While it involves multiple steps and algebraic manipulation, the techniques are standard for A-level implicit differentiation questions. The 'show exactly one' aspect adds modest challenge beyond routine 'find stationary points' questions, but the algebra is manageable and the method is well-practiced.
10 In this question you must show detailed reasoning.
Show that the curve with equation \(x ^ { 2 } - 4 x y + 8 y ^ { 3 } - 4 = 0\) has exactly one stationary point.
DR: \(2x - 4y - 4x\frac{dy}{dx} + 24y\frac{dy}{dx} = 0\); \(\frac{dy}{dx} = 0\); \(2x - 4y = 0\); \(x^2 - 2x^2 - x^2 + 4 = 0\); \(x^3 - x^2 - 4 = 0\); \(f(2) = 0\); \((x-2)(x^2 + x + 2) = 0\); \(\Delta = -7 < 0\) so quadratic has no real roots, hence just one stationary point
M1*, A1, M1d*, A1, M1, B1, M1, A1, E1
Attempt implicit differentiation; Obtain correct derivative; Either rearrange and use, or substitute; Obtain \(2x - 4y = 0\), or equiv; Eliminate \(x\) or \(y\) from eqn of curve; Obtain correct cubic; Identify \(x = 2\) as root or \((x-2)\) as factor; Attempt to factorise cubic − any valid method; Correct quadratic quotient; Justify one stationary point; Deal with at least one \(y\) term correctly; OR \(4x^2 - 8x^2 + 8x^3 - 4 = 0\) OR \(2y^3 - y^2 - 1 = 0\); BC; OR \(f(1) = 0\); OR \((y-1)(2y^2 + y + 1) = 0\); Allow for dividing by root of their cubic
| DR: $2x - 4y - 4x\frac{dy}{dx} + 24y\frac{dy}{dx} = 0$; $\frac{dy}{dx} = 0$; $2x - 4y = 0$; $x^2 - 2x^2 - x^2 + 4 = 0$; $x^3 - x^2 - 4 = 0$; $f(2) = 0$; $(x-2)(x^2 + x + 2) = 0$; $\Delta = -7 < 0$ so quadratic has no real roots, hence just one stationary point | M1*, A1, M1d*, A1, M1, B1, M1, A1, E1 | Attempt implicit differentiation; Obtain correct derivative; Either rearrange and use, or substitute; Obtain $2x - 4y = 0$, or equiv; Eliminate $x$ or $y$ from eqn of curve; Obtain correct cubic; Identify $x = 2$ as root or $(x-2)$ as factor; Attempt to factorise cubic − any valid method; Correct quadratic quotient; Justify one stationary point; Deal with at least one $y$ term correctly; OR $4x^2 - 8x^2 + 8x^3 - 4 = 0$ OR $2y^3 - y^2 - 1 = 0$; BC; OR $f(1) = 0$; OR $(y-1)(2y^2 + y + 1) = 0$; Allow for dividing by root of their cubic |
10 In this question you must show detailed reasoning.\\
Show that the curve with equation $x ^ { 2 } - 4 x y + 8 y ^ { 3 } - 4 = 0$ has exactly one stationary point.
\hfill \mbox{\textit{OCR H240/01 2018 Q10 [10]}}