| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Curve above or below axis |
| Difficulty | Standard +0.3 This is a straightforward multi-part question covering standard AS-level techniques: completing the square or using the discriminant to show a quadratic is always positive, finding where a function decreases using differentiation, solving simultaneous equations, and finding tangency conditions. All parts are routine applications of core methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.07m Tangents and normals: gradient and equations1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 2x - 3\) \(= 0\) when \(x = 1.5, y = 1.75\) This is a minimum point, \(1.75 > 0\) Curve lies above the \(x\)-axis. | M1 A1 A1 \(\checkmark\) [3] | Completing square or using calculus. Correct 1.75 from some method. Correct deduction for candidate's \(+ve\) \(y\). |
| (ii) Decreasing function for \(x < 1.5\). | A1 \(\checkmark\) [1] | Correct deduction for candidate's value of \(x\). Allow \(\leq\). |
| (iii) \(y = x^2 - 3x + 4\) with \(y + 2x = 6\) Eliminate \(y\) to give \(x^2 - x - 2 = 0\) or eliminate \(x\) to give \(y^2 - 10y + 16 = 0\) | M1 DM1 | Attempt at eqn in \(x\) or \(y\) and set to 0. |
| \(\Rightarrow (-1, 8)\) and \((2, 2)\) | A1 [3] | All values. |
| (iv) \(x^2 - 3x + 4 = k - 2x\) \(\Rightarrow x^2 - x + 4 - k = 0\) or \(2x - 3 = -2\) Use of \(b^2 - 4ac = 0\) or \(x = \frac{1}{2} \Rightarrow y = 2\frac{3}{4}\) | M1 M1 A1 [3] | Equates and sets to 0. Uses \(b^2 - 4ac\) on eqn \(= 0\). CAO |
$y = x^2 - 3x + 4$
(i) $\frac{dy}{dx} = 2x - 3$ $= 0$ when $x = 1.5, y = 1.75$ This is a minimum point, $1.75 > 0$ Curve lies above the $x$-axis. | M1 A1 A1 $\checkmark$ [3] | Completing square or using calculus. Correct 1.75 from some method. Correct deduction for candidate's $+ve$ $y$.
(ii) Decreasing function for $x < 1.5$. | A1 $\checkmark$ [1] | Correct deduction for candidate's value of $x$. Allow $\leq$.
(iii) $y = x^2 - 3x + 4$ with $y + 2x = 6$ Eliminate $y$ to give $x^2 - x - 2 = 0$ or eliminate $x$ to give $y^2 - 10y + 16 = 0$ | M1 DM1 | Attempt at eqn in $x$ or $y$ and set to 0.
$\Rightarrow (-1, 8)$ and $(2, 2)$ | A1 [3] | All values.
(iv) $x^2 - 3x + 4 = k - 2x$ $\Rightarrow x^2 - x + 4 - k = 0$ or $2x - 3 = -2$ Use of $b^2 - 4ac = 0$ or $x = \frac{1}{2} \Rightarrow y = 2\frac{3}{4}$ | M1 M1 A1 [3] | Equates and sets to 0. Uses $b^2 - 4ac$ on eqn $= 0$. CAO
$k = 3\frac{3}{4}$10 The equation of a curve is $y = x ^ { 2 } - 3 x + 4$.\\
(i) Show that the whole of the curve lies above the $x$-axis.\\
(ii) Find the set of values of $x$ for which $x ^ { 2 } - 3 x + 4$ is a decreasing function of $x$.
The equation of a line is $y + 2 x = k$, where $k$ is a constant.\\
(iii) In the case where $k = 6$, find the coordinates of the points of intersection of the line and the curve.\\
(iv) Find the value of $k$ for which the line is a tangent to the curve.
\hfill \mbox{\textit{CAIE P1 2005 Q10 [10]}}