CAIE P1 2014 November — Question 10 9 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2014
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeClassify nature of stationary points
DifficultyModerate -0.3 This is a straightforward integration problem with standard techniques. Part (i) requires simple substitution into the second derivative to determine concavity. Parts (ii) and (iii) involve routine integration of power functions and using given conditions to find constants—all standard AS-level calculus with no novel problem-solving required. Slightly easier than average due to the mechanical nature of the steps.
Spec1.07e Second derivative: as rate of change of gradient1.07f Convexity/concavity: points of inflection1.08a Fundamental theorem of calculus: integration as reverse of differentiation

A curve is such that \(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\). The curve has a stationary point at \(P\) where \(x = 2\).
  1. State, with a reason, the nature of this stationary point. [1]
  2. Find an expression for \(\frac{dy}{dx}\). [4]
  3. Given that the curve passes through the point \((1, 13)\), find the coordinates of the stationary point \(P\). [4]

\(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\)
AnswerMarks Guidance
(i) (If \(x = 2\)) it's negative \(\rightarrow\) MaxB1 [1] www
(ii) \(\left(\frac{dy}{dx}\right) = -12x^{-2} - 4x + (A)\)
\(= 0\) when \(x = 2\)
AnswerMarks Guidance
\(\rightarrow A = 11\)B2,1,0, M1, A1 [4] oe one per term; Attempt at the constant \(A\) after \(\int n\); co
(iii) \((y =) 12x^{-1} - 2x^2 + Ax + (c)\)
\(y = 13\) when \(x = 1 \rightarrow c = -8\)
AnswerMarks Guidance
(If \(x = 2\)) \(y = 12\)B2,1,0, ✓, M1, A1 [4] oe Doesn't need '+c', but does need a term \(A\) to give "\(4x\)".; Attempt at \(c\) after \(\int n\); co
$\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4$

**(i)** (If $x = 2$) it's negative $\rightarrow$ Max | B1 [1] | www

**(ii)** $\left(\frac{dy}{dx}\right) = -12x^{-2} - 4x + (A)$

$= 0$ when $x = 2$

$\rightarrow A = 11$ | B2,1,0, M1, A1 [4] | oe one per term; Attempt at the constant $A$ after $\int n$; co

**(iii)** $(y =) 12x^{-1} - 2x^2 + Ax + (c)$

$y = 13$ when $x = 1 \rightarrow c = -8$

(If $x = 2$) $y = 12$ | B2,1,0, ✓, M1, A1 [4] | oe Doesn't need '+c', but does need a term $A$ to give "$4x$".; Attempt at $c$ after $\int n$; co
A curve is such that $\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4$. The curve has a stationary point at $P$ where $x = 2$.

\begin{enumerate}[label=(\roman*)]
\item State, with a reason, the nature of this stationary point. [1]
\item Find an expression for $\frac{dy}{dx}$. [4]
\item Given that the curve passes through the point $(1, 13)$, find the coordinates of the stationary point $P$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2014 Q10 [9]}}