Question 2 - A-Level Maths

Edexcel C3 2010 June — Question 2 7 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2010
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Equation of normal line
Difficulty	Moderate -0.3 This is a straightforward C3 differentiation question requiring the chain rule to find dy/dx, evaluating at x=2 to find the gradient, then using the perpendicular gradient formula to find the normal. All steps are routine and well-practiced, making it slightly easier than average, though it does require careful algebraic manipulation to reach the integer form.
Spec	1.07m Tangents and normals: gradient and equations 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

2. A curve $C$ has equation $$y = \frac { 3 } { ( 5 - 3 x ) ^ { 2 } } , \quad x \neq \frac { 5 } { 3 }$$ The point $P$ on $C$ has $x$-coordinate 2. Find an equation of the normal to $C$ at $P$ in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

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Question 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
At $P$, $y = 3$	B1
$\frac{dy}{dx} = 3(-2)(5-3x)^{-3}(-3)$ or $\frac{18}{(5-3x)^3}$	M1A1	$\pm k(5-3x)^{-3}$ can be implied.
$\frac{dy}{dx} = \frac{18}{(5-3(2))^3} \{= -18\}$	M1	Substituting $x=2$ into their $\frac{dy}{dx}$
$m(\mathbf{N}) = \frac{-1}{-18}$ or $\frac{1}{18}$	M1	Uses $m(\mathbf{N}) = -\frac{1}{\text{their } m(\mathbf{T})}$
$\mathbf{N}: y - 3 = \frac{1}{18}(x-2)$	M1	$y - y_1 = m(x-2)$ with their normal gradient or changed tangent gradient and their $y_1$
$\mathbf{N}: x - 18y + 52 = 0$	A1	All previous 6 marks must be earned. Completely correct solution required.
[7]

# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| At $P$, $y = 3$ | B1 | |
| $\frac{dy}{dx} = 3(-2)(5-3x)^{-3}(-3)$ or $\frac{18}{(5-3x)^3}$ | M1A1 | $\pm k(5-3x)^{-3}$ can be implied. |
| $\frac{dy}{dx} = \frac{18}{(5-3(2))^3} \{= -18\}$ | M1 | Substituting $x=2$ into their $\frac{dy}{dx}$ |
| $m(\mathbf{N}) = \frac{-1}{-18}$ or $\frac{1}{18}$ | M1 | Uses $m(\mathbf{N}) = -\frac{1}{\text{their } m(\mathbf{T})}$ |
| $\mathbf{N}: y - 3 = \frac{1}{18}(x-2)$ | M1 | $y - y_1 = m(x-2)$ with their normal gradient or changed tangent gradient and their $y_1$ |
| $\mathbf{N}: x - 18y + 52 = 0$ | A1 | All previous 6 marks must be earned. Completely correct solution required. |
| | **[7]** | |

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2. A curve $C$ has equation

$$y = \frac { 3 } { ( 5 - 3 x ) ^ { 2 } } , \quad x \neq \frac { 5 } { 3 }$$

The point $P$ on $C$ has $x$-coordinate 2. Find an equation of the normal to $C$ at $P$ in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\

\hfill \mbox{\textit{Edexcel C3 2010 Q2 [7]}}

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