CAIE P1 2011 November — Question 8 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2011
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeDetermine constant from stationary point condition
DifficultyModerate -0.3 This is a straightforward multi-part calculus question testing standard techniques: using the stationary point condition to find a constant, finding the second derivative to classify stationary points, and integrating to recover the original function. All parts follow routine procedures with no problem-solving insight required, making it slightly easier than average but still requiring competent execution of multiple techniques.
Spec1.07e Second derivative: as rate of change of gradient1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation

A curve \(y = \mathrm{f}(x)\) has a stationary point at \(P(3, -10)\). It is given that \(\mathrm{f}'(x) = 2x^2 + kx - 12\), where \(k\) is a constant.
  1. Show that \(k = -2\) and hence find the \(x\)-coordinate of the other stationary point, \(Q\). [4]
  2. Find \(\mathrm{f}''(x)\) and determine the nature of each of the stationary points \(P\) and \(Q\). [2]
  3. Find \(\mathrm{f}(x)\). [4]

(i) \(f'(3) = 0 \Rightarrow 18 + 3k - 12 = 0\)
\(k = -2\)
\((x-3)(x+2) = 0\)
AnswerMarks Guidance
\(x = -2, (\) Allow also \(= 3)\)M1, A1, M1, A1 [4] AG
(ii) \(f''(x) = 4x - 2\)
\(f''(3) > 0\) hence min at \(P\)
AnswerMarks Guidance
\(f''(-2) < 0\) hence max at \(Q\)B1, B1 [2] 3 min, -2 max independent of \(f''(x)\)
(iii) \(f(x) = \frac{2}{3}x^3 - x^2 - 12x (+c)\)
Sub \((3, -10) \rightarrow -10 = 18 - 9 - 36 + c\)
AnswerMarks Guidance
\(c = 17\)B2,1,0, M1, A1 [4] Accept anywhere in question; Dependent on c present; Condone \(y =\) or equation \(=\)
**(i)** $f'(3) = 0 \Rightarrow 18 + 3k - 12 = 0$

$k = -2$

$(x-3)(x+2) = 0$

$x = -2, ($ Allow also $= 3)$ | M1, A1, M1, A1 [4] | AG

**(ii)** $f''(x) = 4x - 2$

$f''(3) > 0$ hence min at $P$

$f''(-2) < 0$ hence max at $Q$ | B1, B1 [2] | 3 min, -2 max independent of $f''(x)$

**(iii)** $f(x) = \frac{2}{3}x^3 - x^2 - 12x (+c)$

Sub $(3, -10) \rightarrow -10 = 18 - 9 - 36 + c$

$c = 17$ | B2,1,0, M1, A1 [4] | Accept anywhere in question; Dependent on c present; Condone $y =$ or equation $=$

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A curve $y = \mathrm{f}(x)$ has a stationary point at $P(3, -10)$. It is given that $\mathrm{f}'(x) = 2x^2 + kx - 12$, where $k$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Show that $k = -2$ and hence find the $x$-coordinate of the other stationary point, $Q$. [4]

\item Find $\mathrm{f}''(x)$ and determine the nature of each of the stationary points $P$ and $Q$. [2]

\item Find $\mathrm{f}(x)$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2011 Q8 [10]}}