| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Standard +0.3 This is a standard tangent/normal question requiring chain rule differentiation of a simple composite function, finding gradient at a point, writing line equations, and basic coordinate geometry. While multi-part with several steps, each component is routine A-level technique with no novel insight required. Slightly above average due to the normal line calculation and area integration, but still straightforward. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 4(x-2)^3\) | B1 | Or \(4x^3 - 24x^2 + 48x - 32\) |
| Grad of tangent \(= -4\) | M1 | Sub \(x = 1\) into their derivative |
| Eq. of tangent is \(y - 1 = -4(x-1)\) | M1 | Line thru \((1, 1)\) and with \(m\) from deriv |
| \(\rightarrow B\left(\frac{5}{4}, 0\right)\) | A1 | |
| Grad of normal \(= \frac{1}{4}\) | M1 | Use of \(m_1m_2 = -1\) |
| Eq. of normal is \(y - 1 = \frac{1}{4}(x-1) \rightarrow C\left(0, \frac{3}{4}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(AC^2 = 1^2 + \left(\frac{1}{4}\right)^2\) | M1 | |
| \(\frac{\sqrt{17}}{4}\) | A1 | Allow \(\sqrt{\frac{17}{16}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(\int(x-2)^4dx = \frac{(x-2)^5}{5}\) | B1 | Or \(\frac{x^5}{5} - 2x^4 + 8x^3 - 16x^2 + 16x\) |
| \(\left[0 - \left(-\frac{1}{5}\right)\right] = \frac{1}{5}\) | M1 | Apply limits \(1 \rightarrow 2\) for curve Or \(\int(-4x+5)dx = \frac{1}{8}\) |
| \(\Delta = \frac{1}{2} \times 1 \times\) (their \(\frac{5}{4} - 1\)) \(= \frac{1}{8}\) | M1 | |
| \(\frac{1}{5} - \frac{1}{8} = \frac{3}{40}\) or \(0.075\) | A1 |
(i) $\frac{dy}{dx} = 4(x-2)^3$ | B1 | Or $4x^3 - 24x^2 + 48x - 32$
Grad of tangent $= -4$ | M1 | Sub $x = 1$ into their derivative
Eq. of tangent is $y - 1 = -4(x-1)$ | M1 | Line thru $(1, 1)$ and with $m$ from deriv
$\rightarrow B\left(\frac{5}{4}, 0\right)$ | A1 |
Grad of normal $= \frac{1}{4}$ | M1 | Use of $m_1m_2 = -1$
Eq. of normal is $y - 1 = \frac{1}{4}(x-1) \rightarrow C\left(0, \frac{3}{4}\right)$ | A1 |
[6]
(ii) $AC^2 = 1^2 + \left(\frac{1}{4}\right)^2$ | M1 |
$\frac{\sqrt{17}}{4}$ | A1 | Allow $\sqrt{\frac{17}{16}}$
[2]
(iii) $\int(x-2)^4dx = \frac{(x-2)^5}{5}$ | B1 | Or $\frac{x^5}{5} - 2x^4 + 8x^3 - 16x^2 + 16x$
$\left[0 - \left(-\frac{1}{5}\right)\right] = \frac{1}{5}$ | M1 | Apply limits $1 \rightarrow 2$ for curve Or $\int(-4x+5)dx = \frac{1}{8}$
$\Delta = \frac{1}{2} \times 1 \times$ (their $\frac{5}{4} - 1$) $= \frac{1}{8}$ | M1 |
$\frac{1}{5} - \frac{1}{8} = \frac{3}{40}$ or $0.075$ | A1 |
[4]10\\
\includegraphics[max width=\textwidth, alt={}, center]{d0074ac8-42d2-49f4-a417-4a348537bccc-4_521_809_258_669}
The diagram shows part of the curve $y = ( x - 2 ) ^ { 4 }$ and the point $A ( 1,1 )$ on the curve. The tangent at $A$ cuts the $x$-axis at $B$ and the normal at $A$ cuts the $y$-axis at $C$.\\
(i) Find the coordinates of $B$ and $C$.\\
(ii) Find the distance $A C$, giving your answer in the form $\frac { \sqrt { } a } { b }$, where $a$ and $b$ are integers.\\
(iii) Find the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2013 Q10 [12]}}