| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Standard +0.3 This is a standard two-part question combining tangency conditions (discriminant = 0) and area between curves. Part (i) requires setting up a quadratic and using b²-4ac=0, while part (ii) involves solving a quadratic and integrating a difference of functions. Both are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 + 4x + c - 8 (= 0)\) | M1 | Attempt to simplify to 3-term quadratic |
| \(16 - 4(c - 8) = 0\) | M1 | Apply \(b^2 - 4ac = 0, '= 0'\) soi |
| \(c = 12\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2 - 2x = 2 \to x = (-2)\) | M1 | Equate derivs of curve and line. Expect \(x = -2\) |
| \(-4 + c = 8 + 4 - 4\) | M1 | Sub their \(x = -2\) into line and curve, and equate |
| \(c = 12\) | A1 [3] | |
| (ii) \(x^2 + 4x + 3 \to (x + 1)(x + 3) (= 0) \to x = -1\) or \(-3\) | B1 | |
| \(\int[8 - 2x - x^2] - [(2x + 11) \text{ or area of trapezium}]\) | M1M1 | Attempt to integrate. At some stage subtract |
| \(\left[8x - x^2 - \frac{x^3}{3}\right] - [x^2 + 11x \text{ or } 8x - x^2 - \frac{x^3}{3} - \frac{1}{2}(5 + 9) \times 2\right]\) | A1B1 | A1 for curve, B1 for line OR \(-3x - 2x^2 - \frac{x^3}{3}\) A2,1,0 |
| Apply their limits to at least integral for curve \(1\frac{1}{3}\) oe | M1 A1 [7] | For M marks allow reversed limits and/or subtraction of areas but then final A0 |
(i) $x^2 + 4x + c - 8 (= 0)$ | M1 | Attempt to simplify to 3-term quadratic
$16 - 4(c - 8) = 0$ | M1 | Apply $b^2 - 4ac = 0, '= 0'$ soi
$c = 12$ | A1 [3] |
OR
$-2 - 2x = 2 \to x = (-2)$ | M1 | Equate derivs of curve and line. Expect $x = -2$
$-4 + c = 8 + 4 - 4$ | M1 | Sub their $x = -2$ into line and curve, and equate
$c = 12$ | A1 [3] |
(ii) $x^2 + 4x + 3 \to (x + 1)(x + 3) (= 0) \to x = -1$ or $-3$ | B1 |
$\int[8 - 2x - x^2] - [(2x + 11) \text{ or area of trapezium}]$ | M1M1 | Attempt to integrate. At some stage subtract
$\left[8x - x^2 - \frac{x^3}{3}\right] - [x^2 + 11x \text{ or } 8x - x^2 - \frac{x^3}{3} - \frac{1}{2}(5 + 9) \times 2\right]$ | A1B1 | A1 for curve, B1 for line OR $-3x - 2x^2 - \frac{x^3}{3}$ A2,1,0
Apply their limits to at least integral for curve $1\frac{1}{3}$ oe | M1 A1 [7] | For M marks allow reversed limits and/or subtraction of areas but then final A011 A line has equation $y = 2 x + c$ and a curve has equation $y = 8 - 2 x - x ^ { 2 }$.\\
(i) For the case where the line is a tangent to the curve, find the value of the constant $c$.\\
(ii) For the case where $c = 11$, find the $x$-coordinates of the points of intersection of the line and the curve. Find also, by integration, the area of the region between the line and the curve.
\hfill \mbox{\textit{CAIE P1 2014 Q11 [10]}}