| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a straightforward application of the quotient rule followed by solving a simple factorized equation. Part (i) is routine differentiation with the answer given, and part (ii) requires only setting the numerator to zero and substituting back—standard textbook exercise with no novel insight required, slightly above average due to algebraic manipulation. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
The equation of a curve is $y = \frac{x^2}{2x - 1}$.
(i) Show that $\frac{dy}{dx} = \frac{2x(x - 1)}{(2x - 1)^2}$. [4]
M1: Use quotient rule: $\frac{dy}{dx} = \frac{(2x)(2x - 1) - x^2(2)}{(2x - 1)^2}$
M1: Expand numerator: $4x^2 - 2x - 2x^2 = 2x^2 - 2x$
M1: Factor numerator: $2x(x - 1)$
A1: $\frac{dy}{dx} = \frac{2x(x - 1)}{(2x - 1)^2}$
(ii) Find the coordinates of the stationary points of the curve. You need not determine their nature. [4]
M1: Set $\frac{dy}{dx} = 0$, giving $2x(x - 1) = 0$
A1: $x = 0$ or $x = 1$
M1: Find corresponding $y$-values: when $x = 0$, $y = 0$; when $x = 1$, $y = 1$
A1: Stationary points are $(0, 0)$ and $(1, 1)$5 The equation of a curve is $y = \frac { x ^ { 2 } } { 2 x + 1 }$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( x + 1 ) } { ( 2 x + 1 ) ^ { 2 } }$.\\
(ii) Find the coordinates of the stationary points of the curve. You need not determine their nature.
\hfill \mbox{\textit{OCR MEI C3 2007 Q5 [8]}}