| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question covering standard P1 techniques: solving a quadratic inequality (routine factorization), completing the square (direct application of formula), and finding a discriminant condition for equal roots. All parts are textbook exercises requiring no problem-solving insight, though the composite function in part (iii) adds minor complexity beyond the most basic questions. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2x^2 - 3x - 9 > 0 \to x = 3\) or \(1\frac{1}{2}\). Set of \(x > 3\), or \(x < -1\frac{1}{2}\) | M1 A1, A1 [3] | For solving quadratic. Ignore \(>\) or \(\ge\) condone \(\ge\) or \(\le\) |
| (ii) \(2y^2 - 3x = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}\). Vertex \(\left(\frac{3}{4}, -\frac{9}{8}\right)\) | B3, 2, 1, B1✓ [4] | – \(x^2\) in bracket is an error. ✓ on 'c' and 'b'. |
| (iii) \(gf(x) = 6x^2 - 9x + k = 0\). Use of \(b^2 - 4ac \to k = \frac{27}{8}\) oe. | B1, M1 A1 [3] | Used on a quadratic (even fg). |
$f : x \mapsto 2x^2 - 3x$, $g : x \mapsto 3x + k$
(i) $2x^2 - 3x - 9 > 0 \to x = 3$ or $1\frac{1}{2}$. Set of $x > 3$, or $x < -1\frac{1}{2}$ | M1 A1, A1 [3] | For solving quadratic. Ignore $>$ or $\ge$ condone $\ge$ or $\le$
(ii) $2y^2 - 3x = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}$. Vertex $\left(\frac{3}{4}, -\frac{9}{8}\right)$ | B3, 2, 1, B1✓ [4] | – $x^2$ in bracket is an error. ✓ on 'c' and 'b'.
(iii) $gf(x) = 6x^2 - 9x + k = 0$. Use of $b^2 - 4ac \to k = \frac{27}{8}$ oe. | B1, M1 A1 [3] | Used on a quadratic (even fg).10 A curve has equation $y = 2 x ^ { 2 } - 3 x$.\\
(i) Find the set of values of $x$ for which $y > 9$.\\
(ii) Express $2 x ^ { 2 } - 3 x$ in the form $a ( x + b ) ^ { 2 } + c$, where $a , b$ and $c$ are constants, and state the coordinates of the vertex of the curve.
The functions f and g are defined for all real values of $x$ by
$$\mathrm { f } ( x ) = 2 x ^ { 2 } - 3 x \quad \text { and } \quad \mathrm { g } ( x ) = 3 x + k$$
where $k$ is a constant.\\
(iii) Find the value of $k$ for which the equation $\mathrm { gf } ( x ) = 0$ has equal roots.
\hfill \mbox{\textit{CAIE P1 2013 Q10 [10]}}