AQA C2 — Question 8

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent meets curve/axis — further geometry
DifficultyStandard +0.3 This is a structured multi-part C2 question involving standard differentiation of powers, finding tangent equations, and basic integration. While part (d) requires combining several results to find an area, each individual step follows routine procedures with clear guidance. The question is slightly easier than average due to its scaffolded nature and use of only basic techniques.
Spec1.07a Derivative as gradient: of tangent to curve1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals

8 A curve, drawn from the origin \(O\), crosses the \(x\)-axis at the point \(A ( 9,0 )\). Tangents to the curve at \(O\) and \(A\) meet at the point \(P\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{48c5470e-6489-4b25-98a6-1b4e101ab01c-006_763_879_466_577} The curve, defined for \(x \geqslant 0\), has equation $$y = x ^ { \frac { 3 } { 2 } } - 3 x$$
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
    1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(O\) and hence write down an equation of the tangent at \(O\).
    2. Show that the equation of the tangent at \(A ( 9,0 )\) is \(2 y = 3 x - 27\).
    3. Hence find the coordinates of the point \(P\) where the two tangents meet.
  2. Find \(\int \left( x ^ { \frac { 3 } { 2 } } - 3 x \right) \mathrm { d } x\).
  3. Calculate the area of the shaded region bounded by the curve and the tangents \(O P\) and \(A P\).

Question 8:
(a)
AnswerMarks
\(b = a^c\)B1
(b)
AnswerMarks Guidance
\(\log_2(x+7)^2 - \log_2(x+5) = 3\)M1
\(\log_2\frac{(x+7)^2}{x+5} = 3\)M1
\((x+7)^2 = 8(x+5)\)A1
\(x^2 + 14x + 49 = 8x + 40\)
\(x^2 + 6x + 9 = 0\)A1
\((x+3)^2 = 0 \Rightarrow x = -3\) onlyA1 one solution shown, check valid: \(x+7=4>0\), \(x+5=2>0\) ✓
# Question 8:

**(a)**
$b = a^c$ | B1 |

**(b)**
$\log_2(x+7)^2 - \log_2(x+5) = 3$ | M1 |
$\log_2\frac{(x+7)^2}{x+5} = 3$ | M1 |
$(x+7)^2 = 8(x+5)$ | A1 |
$x^2 + 14x + 49 = 8x + 40$ | |
$x^2 + 6x + 9 = 0$ | A1 |
$(x+3)^2 = 0 \Rightarrow x = -3$ only | A1 | one solution shown, check valid: $x+7=4>0$, $x+5=2>0$ ✓ | A1 |

---
8 A curve, drawn from the origin $O$, crosses the $x$-axis at the point $A ( 9,0 )$. Tangents to the curve at $O$ and $A$ meet at the point $P$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{48c5470e-6489-4b25-98a6-1b4e101ab01c-006_763_879_466_577}

The curve, defined for $x \geqslant 0$, has equation

$$y = x ^ { \frac { 3 } { 2 } } - 3 x$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $O$ and hence write down an equation of the tangent at $O$.
\item Show that the equation of the tangent at $A ( 9,0 )$ is $2 y = 3 x - 27$.
\item Hence find the coordinates of the point $P$ where the two tangents meet.
\end{enumerate}\item Find $\int \left( x ^ { \frac { 3 } { 2 } } - 3 x \right) \mathrm { d } x$.
\item Calculate the area of the shaded region bounded by the curve and the tangents $O P$ and $A P$.
\end{enumerate}

\hfill \mbox{\textit{AQA C2  Q8}}