| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Area under curve using substitution |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard differentiation of a square root function (chain rule), basic integration with substitution, finding a tangent line equation, and calculating an area between a curve and tangent. All techniques are routine P1/AS-level material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = -\frac{1}{2}(4-x)^{\frac{1}{2}}\times-1\) | B1 B1 | Without (−1). For (×−1). |
| \(\int y\,dx = 8x - \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} \div -1\) | 3 × B1 [5] | B1 for "8x" and "+c". B1 for all except −(−1). B1 for +(−1). (n.b. these 5 marks can be gained in(ii) or (iii)) |
| (ii) Eqn \(y-7 = \frac{1}{2}(x-3)\) → \(y = \frac{1}{2}x + 5\frac{1}{2}\) | M1A1 [2] | M1 unsimplified. A1 as y=mx+c |
| (iii) Area under curve = \(\int\) from 0 to 3 (58/3) Area under line = \(\frac{1}{2}(5\frac{1}{2}+7)\times3\) | M1 M1 | Use of limits – needs use of "0" Correct method |
| Or \(\left[\frac{1}{4}x^2+\frac{11x}{2}\right]\) from 0 to 3 | M1 A1 [4] | M1 Subtraction. A1 co |
| → \(\frac{58}{3}-\frac{75}{4}=\frac{7}{12}\) |
$y = 8-\sqrt{4-x}$
**(i)** $\frac{dy}{dx} = -\frac{1}{2}(4-x)^{\frac{1}{2}}\times-1$ | B1 B1 | Without (−1). For (×−1).
$\int y\,dx = 8x - \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} \div -1$ | 3 × B1 [5] | B1 for "8x" and "+c". B1 for all except −(−1). B1 for +(−1). (n.b. these 5 marks can be gained in(ii) or (iii))
**(ii)** Eqn $y-7 = \frac{1}{2}(x-3)$ → $y = \frac{1}{2}x + 5\frac{1}{2}$ | M1A1 [2] | M1 unsimplified. A1 as y=mx+c
**(iii)** Area under curve = $\int$ from 0 to 3 (58/3) Area under line = $\frac{1}{2}(5\frac{1}{2}+7)\times3$ | M1 M1 | Use of limits – needs use of "0" Correct method
Or $\left[\frac{1}{4}x^2+\frac{11x}{2}\right]$ from 0 to 3 | M1 A1 [4] | M1 Subtraction. A1 co
→ $\frac{58}{3}-\frac{75}{4}=\frac{7}{12}$ | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{1a4ddaa9-1ec2-4138-bfcb-a482fe6c942f-3_849_565_1466_790}
The diagram shows part of the curve $y = 8 - \sqrt { } ( 4 - x )$ and the tangent to the curve at $P ( 3,7 )$.\\
(i) Find expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\int y \mathrm {~d} x$.\\
(ii) Find the equation of the tangent to the curve at $P$ in the form $y = m x + c$.\\
(iii) Find, showing all necessary working, the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2014 Q9 [11]}}