Question 1 - A-Level Maths

Edexcel C3 2013 January — Question 1 7 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2013
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Tangents, normals and gradients
Type	Find tangent at given point (polynomial/algebraic)
Difficulty	Moderate -0.8 This is a straightforward two-part question requiring basic substitution to find a coordinate, then applying the chain rule and point-slope form. Both parts are routine C3 techniques with no problem-solving insight needed, making it easier than average.
Spec	1.07i Differentiate x^n: for rational n and sums 1.07m Tangents and normals: gradient and equations 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

The curve $C$ has equation

$$y = ( 2 x - 3 ) ^ { 5 }$$ The point $P$ lies on $C$ and has coordinates $( w , - 32 )$.
Find

the value of $w$,
the equation of the tangent to $C$ at the point $P$ in the form $y = m x + c$, where $m$ and $c$ are constants.

Show mark scheme Show mark scheme source

(a)

Answer	Marks	Guidance
$-32 = (2w-3)^5 \Rightarrow w = \frac{1}{2}$ oe	M1A1	Accept positive sign used of $y$, ie $y=+32$. Sight of just the answer would score both marks as long as no incorrect working is seen.

(b)

Answer	Marks	Guidance
$\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2$ or $10(2x-3)^4$	M1A1	A correct (un simplified) form of the differential. Accept $\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2$ or $\frac{dy}{dx} = 10(2x-3)^4$
When $x = \frac{1}{2}$, Gradient $= 160$	M1	This is awarded for an attempt to find the gradient of the tangent to the curve at $P$. Award for substituting their numerical value to part (a) into their differential to find the numerical gradient of the tangent.
Equation of tangent is $'160' = \frac{y-(-32)}{x - \frac{1}{2}}$ oe	dM1	Award for a correct method to find an equation of the tangent to the curve at $P$. It is dependent upon the previous M mark being awarded. Award for 'their 160' $= \frac{y-(-32)}{x-\text{their } \frac{1}{2}}$. If they use $y = mx + c$ it must be a full method, using $m=$ 'their 160', their $\cdot \frac{1}{2}$ ' and -32. An attempt must be seen to find $c=...$
$y = 160x - 112$	cso A1	The question is specific and requires the answer in this form. You may isw in this question after a correct answer.
(7 marks)

**(a)**

$-32 = (2w-3)^5 \Rightarrow w = \frac{1}{2}$ oe | M1A1 | Accept positive sign used of $y$, ie $y=+32$. Sight of just the answer would score both marks as long as no incorrect working is seen. | (2 marks)

**(b)**

$\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2$ or $10(2x-3)^4$ | M1A1 | A correct (un simplified) form of the differential. Accept $\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2$ or $\frac{dy}{dx} = 10(2x-3)^4$ | (2 marks)

When $x = \frac{1}{2}$, Gradient $= 160$ | M1 | This is awarded for an attempt to find the gradient of the tangent to the curve at $P$. Award for substituting their numerical value to part (a) into their differential to find the numerical gradient of the tangent. |

Equation of tangent is $'160' = \frac{y-(-32)}{x - \frac{1}{2}}$ oe | dM1 | Award for a correct method to find an equation of the tangent to the curve at $P$. It is dependent upon the previous M mark being awarded. Award for 'their 160' $= \frac{y-(-32)}{x-\text{their } \frac{1}{2}}$. If they use $y = mx + c$ it must be a full method, using $m=$ 'their 160', their $\cdot \frac{1}{2}$ ' and -32. An attempt must be seen to find $c=...$ |

$y = 160x - 112$ | cso A1 | The question is specific and requires the answer in this form. You may isw in this question after a correct answer. | (5 marks)

| (7 marks) |

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Show LaTeX source

\begin{enumerate}
  \item The curve $C$ has equation
\end{enumerate}

$$y = ( 2 x - 3 ) ^ { 5 }$$

The point $P$ lies on $C$ and has coordinates $( w , - 32 )$.\\
Find\\
(a) the value of $w$,\\
(b) the equation of the tangent to $C$ at the point $P$ in the form $y = m x + c$, where $m$ and $c$ are constants.\\

\hfill \mbox{\textit{Edexcel C3 2013 Q1 [7]}}

This paper (8 questions)

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Q1 7 Q2 8 Q3 9 Q4 8 Q5 11 Q6 11 Q7 12 Q8 9

Answer	Marks	Guidance
\(\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2\) or \(10(2x-3)^4\)	M1A1	A correct (un simplified) form of the differential. Accept \(\frac{dy}{dx} = 5 \times (2x-3)^4 \times 2\) or \(\frac{dy}{dx} = 10(2x-3)^4\)
When \(x = \frac{1}{2}\), Gradient \(= 160\)	M1	This is awarded for an attempt to find the gradient of the tangent to the curve at \(P\). Award for substituting their numerical value to part (a) into their differential to find the numerical gradient of the tangent.
Equation of tangent is \('160' = \frac{y-(-32)}{x - \frac{1}{2}}\) oe	dM1	Award for a correct method to find an equation of the tangent to the curve at \(P\). It is dependent upon the previous M mark being awarded. Award for 'their 160' \(= \frac{y-(-32)}{x-\text{their } \frac{1}{2}}\). If they use \(y = mx + c\) it must be a full method, using \(m=\) 'their 160', their \(\cdot \frac{1}{2}\) ' and -32. An attempt must be seen to find \(c=...\)
\(y = 160x - 112\)	cso A1	The question is specific and requires the answer in this form. You may isw in this question after a correct answer.
(7 marks)