| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Curve Sketching |
| Type | Rational curve intersections |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (i) requires sketching two standard curves (reciprocal squared and quadratic), which is routine. Part (ii) involves setting the equations equal, leading to a quartic in x² that factors nicely to give x⁴ - 2x² - 3 = 0, solvable by substitution. While it requires algebraic manipulation and exact form answers, the steps are straightforward with no novel problem-solving required. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of \(y = \frac{3}{x^2}\); Sketch of \(y = x^2 - 2\); Intercepts of \((0, -2)\), \((\sqrt{2}, 0)\), \((-\sqrt{2}, 0)\) | B1, M1, A1 | Must be in both quadrants, with axes intended as asymptotes; Positive quadratic, symmetrical in \(y\)-axis; All 3 intercepts correct; Allow just eg \(\sqrt{2}\) if marked on relevant axis |
| Answer | Marks | Guidance |
|---|---|---|
| DR: \(3 = x^4 - 2x^3\); \((x^2 - 3)(x^2 + 1) = 0\); \(x^2 = 3\); \(x = \pm\sqrt{3}\); A1; points are \((\sqrt{3}, 1)\) and \((-\sqrt{3}, 1)\); \(x^2 = -1\) has no solutions as \(x^2 \geq 0\) | M1, M1, M1, A1, A1, B1 | Equate and remove fractions; Attempt to solve disguised quadratic; Attempt to find \(x\) from their roots; Correct \(x\) values; Both correct coordinates; Justify no roots for \(x^2 = -1\); Could use substitution; At least one value of \(x\); A0 if extra roots, or if decimals |
**(i)**
| Sketch of $y = \frac{3}{x^2}$; Sketch of $y = x^2 - 2$; Intercepts of $(0, -2)$, $(\sqrt{2}, 0)$, $(-\sqrt{2}, 0)$ | B1, M1, A1 | Must be in both quadrants, with axes intended as asymptotes; Positive quadratic, symmetrical in $y$-axis; All 3 intercepts correct; Allow just eg $\sqrt{2}$ if marked on relevant axis |
**(ii)**
| DR: $3 = x^4 - 2x^3$; $(x^2 - 3)(x^2 + 1) = 0$; $x^2 = 3$; $x = \pm\sqrt{3}$; A1; points are $(\sqrt{3}, 1)$ and $(-\sqrt{3}, 1)$; $x^2 = -1$ has no solutions as $x^2 \geq 0$ | M1, M1, M1, A1, A1, B1 | Equate and remove fractions; Attempt to solve disguised quadratic; Attempt to find $x$ from their roots; Correct $x$ values; Both correct coordinates; Justify no roots for $x^2 = -1$; Could use substitution; At least one value of $x$; A0 if extra roots, or if decimals |4 (i) Sketch the curves $y = \frac { 3 } { x ^ { 2 } }$ and $y = x ^ { 2 } - 2$ on the axes provided in the Printed Answer Booklet.\\
(ii) In this question you must show detailed reasoning.
Find the exact coordinates of the points of intersection of the curves $y = \frac { 3 } { x ^ { 2 } }$ and $y = x ^ { 2 } - 2$.
\hfill \mbox{\textit{OCR H240/01 2018 Q4 [9]}}