| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This question requires finding a tangent equation (standard differentiation and point-slope form) and then computing an area using integration. Both are routine A-level techniques with straightforward algebra. The multi-step nature and combination of calculus skills makes it slightly above average difficulty, but no novel insight is required. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = -4x^{-\frac{3}{2}} - 1\) | B1 | Needs both terms |
| \(= -\frac{3}{2}\) when \(x = 4\) | M1 | Subs \(x = 4\) into \(dy/dx\) |
| Eqn of \(BC\): \(y - 0 = -\frac{3}{2}(x - 4)\) | M1 | Must be using differential + correct form of line at \(B(4,0)\) |
| \(\rightarrow C\left(1, 4\frac{1}{2}\right)\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area under curve \(= \int\left(\frac{8}{\sqrt{x}} - x\right)\) | ||
| \(= \frac{8x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{1}{2}x^2\) | B1 B1 | Both unsimplified |
| Limits 1 to 4 \(\rightarrow 8\frac{1}{2}\) | M1 | Using correct limits |
| Area under tangent \(= \frac{1}{2} \times 4\frac{1}{2} \times 3 = 6\frac{3}{4}\) | M1 | Or could use calculus |
| Shaded area \(= 1\frac{3}{4}\) | A1 | |
| [5] |
## Question 11:
**Part (i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -4x^{-\frac{3}{2}} - 1$ | B1 | Needs both terms |
| $= -\frac{3}{2}$ when $x = 4$ | M1 | Subs $x = 4$ into $dy/dx$ |
| Eqn of $BC$: $y - 0 = -\frac{3}{2}(x - 4)$ | M1 | Must be using differential + correct form of line at $B(4,0)$ |
| $\rightarrow C\left(1, 4\frac{1}{2}\right)$ | A1 | |
| | [4] | |
**Part (ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area under curve $= \int\left(\frac{8}{\sqrt{x}} - x\right)$ | | |
| $= \frac{8x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{1}{2}x^2$ | B1 B1 | Both unsimplified |
| Limits 1 to 4 $\rightarrow 8\frac{1}{2}$ | M1 | Using correct limits |
| Area under tangent $= \frac{1}{2} \times 4\frac{1}{2} \times 3 = 6\frac{3}{4}$ | M1 | Or could use calculus |
| Shaded area $= 1\frac{3}{4}$ | A1 | |
| | [5] | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{13cfb59a-7781-4786-a625-919b01a2a4f0-4_643_570_849_790}
The diagram shows part of the curve $y = \frac { 8 } { \sqrt { } x } - x$ and points $A ( 1,7 )$ and $B ( 4,0 )$ which lie on the curve. The tangent to the curve at $B$ intersects the line $x = 1$ at the point $C$.\\
(i) Find the coordinates of $C$.\\
(ii) Find the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2013 Q11 [9]}}