| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 7 |
| 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 P1 differentiation question requiring standard techniques: differentiate a polynomial and square root term, evaluate the derivative at a point, find the negative reciprocal for the normal gradient, and write the line equation in the required form. All steps are routine with no problem-solving insight needed, making it easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| VIIIV SIHI NI III M I I N OC | VIIV SIHI NI IM IMM ION OC | VI4V SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2}x^2 + 2x^{-\frac{1}{2}}\) | M1A1 A1 | M1: reducing power by one on any \(x\) term. A1: two correct terms (unsimplified, ignore spurious extra terms like \(-15\)). A1: fully correct and simplified. Accept \(\frac{x^2}{2}+\frac{2}{\sqrt{x}}\). Withhold final A if "+c" included |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left.\frac{dy}{dx}\right | _{x=4} = \frac{1}{2}\times4^2 + 2\times\frac{1}{\sqrt{4}} = (9)\) | M1 |
| Gradient of normal is \(-\frac{1}{9}\) | dM1 | Correct method of using negative reciprocal to find gradient of normal |
| \(y - \frac{11}{3} = -\frac{1}{9}(x-4) \Rightarrow x + 9y - 37 = 0\) | M1 A1 | M1: attempt at finding equation of normal using changed gradient and point \(\left(4,\frac{11}{3}\right)\). If \(y=mx+c\) used must proceed to \(c=...\). A1: \(x+9y-37=0\) or any (positive or negative) integer multiple. Accept different order, must include "=0". ISW after correct answer |
# Question 5:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = \frac{1}{2}x^2 + 2x^{-\frac{1}{2}}$ | M1A1 A1 | M1: reducing power by one on any $x$ term. A1: two correct terms (unsimplified, ignore spurious extra terms like $-15$). A1: fully correct and simplified. Accept $\frac{x^2}{2}+\frac{2}{\sqrt{x}}$. Withhold final A if "+c" included |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $\left.\frac{dy}{dx}\right|_{x=4} = \frac{1}{2}\times4^2 + 2\times\frac{1}{\sqrt{4}} = (9)$ | M1 | For substituting $x=4$ into their $\frac{dy}{dx}$ |
| Gradient of normal is $-\frac{1}{9}$ | dM1 | Correct method of using negative reciprocal to find gradient of normal |
| $y - \frac{11}{3} = -\frac{1}{9}(x-4) \Rightarrow x + 9y - 37 = 0$ | M1 A1 | M1: attempt at finding equation of normal using changed gradient and point $\left(4,\frac{11}{3}\right)$. If $y=mx+c$ used must proceed to $c=...$. A1: $x+9y-37=0$ or any (positive or negative) integer multiple. Accept different order, must include "=0". ISW after correct answer |
---5. A curve has equation
$$y = \frac { x ^ { 3 } } { 6 } + 4 \sqrt { x } - 15 \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving the answer in simplest form.
The point $P \left( 4 , \frac { 11 } { 3 } \right)$ lies on the curve.
\item Find the equation of the normal to the curve at $P$. Write your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
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VIIIV SIHI NI III M I I N OC & VIIV SIHI NI IM IMM ION OC & VI4V SIHI NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel P1 2019 Q5 [7]}}