| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of power rule for part (a), simple algebraic manipulation for part (b), integration with boundary condition for part (c), and standard normal line calculation for part (d). All techniques are standard textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07e Second derivative: as rate of change of gradient1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 3x^2 + 2\) | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(x^2\) is always positive, \(\frac{d^2y}{dx^2} \geq 2\) for all \(x\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{x^4}{4} + x^2 - 7x + (k)\) [\(k\) not required here] | M1 A2(1,0) | |
| \(4 = \frac{2^4}{4} + 2^2 - 14 + k \Rightarrow k = 10 \Rightarrow y = \frac{x^4}{4} + x^2 - 7x + 10\) | M1 A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=2\): \(\frac{dy}{dx} = 8 + 4 - 7 = 5\) | M1 A1 | |
| Gradient of normal \(= -\frac{1}{5}\) | M1 | |
| \(y - 4 = -\frac{1}{5}(x-2) \Rightarrow x + 5y - 22 = 0\) | M1 A1 | 5 marks |
## Question 10:
### Part (a):
$\frac{d^2y}{dx^2} = 3x^2 + 2$ | M1 A1 | **2 marks**
### Part (b):
Since $x^2$ is always positive, $\frac{d^2y}{dx^2} \geq 2$ for all $x$ | B1 | **1 mark**
### Part (c):
$y = \frac{x^4}{4} + x^2 - 7x + (k)$ [$k$ not required here] | M1 A2(1,0) |
$4 = \frac{2^4}{4} + 2^2 - 14 + k \Rightarrow k = 10 \Rightarrow y = \frac{x^4}{4} + x^2 - 7x + 10$ | M1 A1 | **5 marks**
### Part (d):
$x=2$: $\frac{dy}{dx} = 8 + 4 - 7 = 5$ | M1 A1 |
Gradient of normal $= -\frac{1}{5}$ | M1 |
$y - 4 = -\frac{1}{5}(x-2) \Rightarrow x + 5y - 22 = 0$ | M1 A1 | **5 marks**
**Total: 13 marks**10. For the curve $C$ with equation $y = \mathrm { f } ( x )$, \(\frac { d y } { d x } = x ^ { 3 } + 2 x - 7 .\) \\
(a) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$. \\
(2) \\
(b) Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \geq 2$ for all values of $x$. \\
(1) \\
Given that the point $P ( 2,4 )$ lies on $C$, \\
(c) find $y$ in terms of $x$, \\
(5) \\
(d) find an equation for the normal to $C$ at $P$ in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers. \\
(5) \\
\begin{enumerate}
\item continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [13]}}