| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of power rule, point verification, normal line equation (negative reciprocal of gradient), and distance formula. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(4x \to k\) or \(3x^{-\frac{1}{2}} \to kx^{\frac{1}{2}}\) or \(-2x^2 \to kx\) | M1 | |
| \(\frac{dy}{dx} = 4 + \frac{9}{2} x^{\frac{1}{2}} - 4x\) | A1, A1 | |
| (b) For \(x = 4\), \(y = (4 \times 4) + (3 \times 4\sqrt{4}) - (2 \times 16) = 16 + 24 - 32 = 8\) | (*) | B1 |
| (c) \(\frac{dy}{dx} = 4 + 9 - 16 = -3\) | M: Evaluate their \(\frac{dy}{dx}\) at \(x = 4\) | M1 |
| Gradient of normal \(= \frac{1}{3}\) | A1ft | |
| Equation of normal: \(y - 8 = \frac{1}{3}(x - 4)\), \(3y = x + 20\) | (*) | M1, A1 |
| (d) \(y = 0: x = \ldots\) (−20) and use \((x - x_1)^2 + (y - y_1)^2\) | M1 | |
| \(PQ = \sqrt{24^2 + 8^2}\) or \(PQ^2 = 24^2 + 8^2\) | Follow through from (k, 0) | A1ft |
| May also be scored with \((-24)^2\) and \((-8)^2\). | ||
| \(= 8\sqrt{10}\) | A1 |
**(a)** $4x \to k$ or $3x^{-\frac{1}{2}} \to kx^{\frac{1}{2}}$ or $-2x^2 \to kx$ | M1 |
$\frac{dy}{dx} = 4 + \frac{9}{2} x^{\frac{1}{2}} - 4x$ | | A1, A1 | (3 marks)
**(b)** For $x = 4$, $y = (4 \times 4) + (3 \times 4\sqrt{4}) - (2 \times 16) = 16 + 24 - 32 = 8$ | (*) | B1 | (1 mark)
**(c)** $\frac{dy}{dx} = 4 + 9 - 16 = -3$ | M: Evaluate their $\frac{dy}{dx}$ at $x = 4$ | M1 |
Gradient of normal $= \frac{1}{3}$ | | A1ft |
Equation of normal: $y - 8 = \frac{1}{3}(x - 4)$, $3y = x + 20$ | (*) | M1, A1 | (4 marks)
**(d)** $y = 0: x = \ldots$ (−20) and use $(x - x_1)^2 + (y - y_1)^2$ | | M1 |
$PQ = \sqrt{24^2 + 8^2}$ or $PQ^2 = 24^2 + 8^2$ | Follow through from (k, 0) | A1ft |
May also be scored with $(-24)^2$ and $(-8)^2$. | | |
$= 8\sqrt{10}$ | | A1 | (3 marks)
**Total: 11 marks**
**Guidance:**
- (a) For the 2 A marks: coefficients need not be simplified, but powers must be simplified. All 3 terms correct: A1 A1. Two terms correct: A1 A0. Only one term correct: A0 A0. Allow the M1 A1 for finding C to be scored either in part (a) or in part (b).
- (b) There must be some evidence of the "24" value.
- (c) In this part, beware 'working backwards' from the given answer. A1ft: Follow through is just from the candidate's value of $\frac{dy}{dx}$. 2nd M: Is not given if an m value appears "from nowhere". 2nd M: Must be an attempt at a normal equation, not a tangent. 2nd M: Alternative is to use (4, 8) or (1, 2) in $y = mx + c$ to find a value for c.
- If calculation for the gradient value is seen in part (a), it must be used in part (b) to score the first M1 A1 in (c).
- Using (1, 2) instead of (2, 1): Loses the 2nd method mark in (a). Gains the 2nd method mark in (c).
- (d) M: Using the normal equation to attempt coordinates of Q (even if using $x = 0$ instead of $y = 0$), and using Pythagoras to attempt $PQ$ or $PQ^2$. Follow through from (k, 0), but not from (0, k)…
- A common wrong answer is to use $x = 0$ to give $\frac{20}{3}$. This scores M1 A0 A0.
- For final answer, accept other simplifications of $\sqrt{640}$, e.g. $2\sqrt{160}$ or $4\sqrt{40}$.
---8. The curve $C$ has equation $y = 4 x + 3 x ^ { \frac { 3 } { 2 } } - 2 x ^ { 2 } , \quad x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the point $P ( 4,8 )$ lies on $C$.
\item Show that an equation of the normal to $C$ at the point $P$ is
$$3 y = x + 20 .$$
The normal to $C$ at $P$ cuts the $x$-axis at the point $Q$.
\item Find the length $P Q$, giving your answer in a simplified surd form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2007 Q8 [11]}}