| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | January |
| Marks | 11 |
| 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 C1 question testing standard differentiation applications: finding a normal (reciprocal of gradient), integration with a boundary condition, and analyzing when a derivative equals a given value. All parts use routine techniques with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Gradient of tangent at \(P\): \(m = 4\); Grad. of normal \(= -\frac{1}{m}\) (\(= -\frac{1}{4}\)) | B1, M1 | |
| Equation of normal: \(y - 4 = -\frac{1}{4}(x - 1)\) (\(4y = -x + 17\)) | M1 A1 | (4 marks) |
| (b) \((3x - 1)^2 = 9x^2 - 6x + 1\) | B1 | |
| Integrate: \(\frac{9x^3}{3} - \frac{6x^2}{2} + x\) \((+C)\) | M1 A1ft | |
| Substitute \((1, 4)\) to find \(c = \ldots\), \(c = 3\) | M1, A1 | (\(y = 3x^3 - 3x^2 + x + 3\)) |
| (c) Gradient of (tangent to) \(C\) is \(\geq 0\) | B1 | |
| Gradient of given line is \(< 0\) (\(-2\)) | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{(3x-1)^3}{9}\) \((+C)\) | M1 A1 (numerator) A1 (denominator) | |
| Substitute \((1, 4)\) to find \(c = \ldots\), \(c = \frac{28}{9}\) | M1, A1 | (\(y = \frac{(3x-1)^3}{9} + \frac{28}{9}\)) |
**(a)** Gradient of tangent at $P$: $m = 4$; Grad. of normal $= -\frac{1}{m}$ ($= -\frac{1}{4}$) | B1, M1 |
Equation of normal: $y - 4 = -\frac{1}{4}(x - 1)$ ($4y = -x + 17$) | M1 A1 | (4 marks)
**(b)** $(3x - 1)^2 = 9x^2 - 6x + 1$ | B1 |
Integrate: $\frac{9x^3}{3} - \frac{6x^2}{2} + x$ $(+C)$ | M1 A1ft |
Substitute $(1, 4)$ to find $c = \ldots$, $c = 3$ | M1, A1 | ($y = 3x^3 - 3x^2 + x + 3$) | (5 marks)
**(c)** Gradient of (tangent to) $C$ is $\geq 0$ | B1 |
Gradient of given line is $< 0$ ($-2$) | B1 | (2 marks)
**Total: 11 marks**
**Notes:**
(a) Using gradient of tangent is M0.
(b) **Alternative:**
$y = \frac{(3x-1)^3}{9}$ $(+C)$ | M1 A1 (numerator) A1 (denominator) |
Substitute $(1, 4)$ to find $c = \ldots$, $c = \frac{28}{9}$ | M1, A1 | ($y = \frac{(3x-1)^3}{9} + \frac{28}{9}$) |
---9. The gradient of the curve $C$ is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 3 x - 1 ) ^ { 2 } .$$
The point $P ( 1,4 )$ lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the normal to $C$ at $P$.
\item Find an equation for the curve $C$ in the form $y = \mathrm { f } ( x )$.
\item Using $\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 3 x - 1 ) ^ { 2 }$, show that there is no point on $C$ at which the tangent is parallel to the line $y = 1 - 2 x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2005 Q9 [11]}}