| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring standard techniques (power rule for polynomials and fractional powers) followed by finding a constant using given coordinates, then calculating a normal gradient. All steps are routine C1 procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(4x \to kx^2\) or \(6\sqrt{x} \to kx^{\frac{3}{2}}\) or \(\frac{8}{x^2} \to kx^{-1}\) (where \(k\) is a non-zero constant) | M1 | |
| \(f(x) = 2x^3, -4x^{\frac{3}{2}}, -8x^{-1}\) | A1, A1, A1 | (+ \(C\)) (+ \(C\) not required) |
| At \(x = 4, y = 1\): | M1 | Must be in part (a) |
| \(1 = (2 \times 16) - \left(4 \times 4^{\frac{3}{2}}\right) - (8 \times 4^{-1}) + C\) | — | |
| \(C = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(f'(4) = 16 - (6 \times 2) + \frac{8}{16} = \frac{9}{2}\) (= \(m\)) | M1 | Attempt \(f'(4)\) with the given \(f'\). Must be in part (b) |
| Gradient of normal is \(-\frac{2}{9}\) (= \(-\frac{1}{m}\)) | M1 | Attempt perp. grad. rule. Dependent on the use of their \(f'(x)\) |
| Eqn of normal: \(y - 1 = -\frac{2}{9}(x-4)\) (or any equiv. form, e.g. \(\frac{y-1}{x-4} = -\frac{2}{9}\)) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Final answer: gradient \(-\frac{1}{(9/2)}\) or \(-\frac{1}{4.5}\) | A0 | (but all M marks are available). |
(a) $4x \to kx^2$ or $6\sqrt{x} \to kx^{\frac{3}{2}}$ or $\frac{8}{x^2} \to kx^{-1}$ (where $k$ is a non-zero constant) | M1 |
$f(x) = 2x^3, -4x^{\frac{3}{2}}, -8x^{-1}$ | A1, A1, A1 | (+ $C$) (+ $C$ not required)
At $x = 4, y = 1$: | M1 | Must be in part (a)
$1 = (2 \times 16) - \left(4 \times 4^{\frac{3}{2}}\right) - (8 \times 4^{-1}) + C$ | —
$C = 3$ | A1 |
**Total for (a): 6 marks**
(b) $f'(4) = 16 - (6 \times 2) + \frac{8}{16} = \frac{9}{2}$ (= $m$) | M1 | Attempt $f'(4)$ with the given $f'$. Must be in part (b)
Gradient of normal is $-\frac{2}{9}$ (= $-\frac{1}{m}$) | M1 | Attempt perp. grad. rule. Dependent on the use of their $f'(x)$
Eqn of normal: $y - 1 = -\frac{2}{9}(x-4)$ (or any equiv. form, e.g. $\frac{y-1}{x-4} = -\frac{2}{9}$) | M1, A1 |
Typical answers for A1: $y = -\frac{2}{9}x + \frac{17}{9}$ $(2x + 9y - 17 = 0)$ $(y = -0.2x + 1.8)$
Final answer: gradient $-\frac{1}{(9/2)}$ or $-\frac{1}{4.5}$ | A0 | (but all M marks are available).
**Total for (b): 4 marks**
**Total: 10 marks**
---9. The curve $C$ has equation $y = \mathrm { f } ( x ) , x > 0$, and $\mathrm { f } ^ { \prime } ( x ) = 4 x - 6 \sqrt { } x + \frac { 8 } { x ^ { 2 } }$.
Given that the point $P ( 4,1 )$ lies on $C$,
\begin{enumerate}[label=(\alph*)]
\item find $\mathrm { f } ( x )$ and simplify your answer.
\item Find an equation of the normal to $C$ at the point $P ( 4,1 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2008 Q9 [10]}}