AQA C2 2009 June — Question 5 13 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2009
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward C2 calculus question involving standard differentiation of fractional powers, finding stationary points by setting dy/dx=0, and working with tangent lines. All techniques are routine for this level, though part (d) requires combining multiple steps. The fractional indices are standard C2 fare and no novel problem-solving is needed.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives

The diagram shows part of a curve with a maximum point \(M\). \includegraphics{figure_5} The equation of the curve is $$y = 15x^{\frac{3}{2}} - x^{\frac{5}{2}}$$
  1. Find \(\frac{dy}{dx}\). [3]
  2. Hence find the coordinates of the maximum point \(M\). [4]
  3. The point \(P(1, 14)\) lies on the curve. Show that the equation of the tangent to the curve at \(P\) is \(y = 20x - 6\). [3]
  4. The tangents to the curve at the points \(P\) and \(M\) intersect at the point \(R\). Find the length of \(RM\). [3]

Question 5:
5
Question 5:
5
The diagram shows part of a curve with a maximum point $M$.

\includegraphics{figure_5}

The equation of the curve is
$$y = 15x^{\frac{3}{2}} - x^{\frac{5}{2}}$$

\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$. [3]
\item Hence find the coordinates of the maximum point $M$. [4]
\item The point $P(1, 14)$ lies on the curve. Show that the equation of the tangent to the curve at $P$ is $y = 20x - 6$. [3]
\item The tangents to the curve at the points $P$ and $M$ intersect at the point $R$. Find the length of $RM$. [3]
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2009 Q5 [13]}}