CAIE P1 2006 June — Question 10 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2006
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeFind stationary points and nature
DifficultyStandard +0.3 This is a standard stationary points question requiring differentiation, solving cubic equations, and basic integration. While it has multiple parts, each step follows routine procedures: find dy/dx, set to zero, use the condition that minimum is on x-axis to find k, then integrate. Slightly easier than average due to straightforward algebra and no novel problem-solving required.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08e Area between curve and x-axis: using definite integrals

10 \includegraphics[max width=\textwidth, alt={}, center]{cbcb15b4-1870-4dfd-b6e9-839aa4601511-4_515_885_662_630} The diagram shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 9 x + k\), where \(k\) is a constant. The curve has a minimum point on the \(x\)-axis.
  1. Find the value of \(k\).
  2. Find the coordinates of the maximum point of the curve.
  3. State the set of values of \(x\) for which \(x ^ { 3 } - 3 x ^ { 2 } - 9 x + k\) is a decreasing function of \(x\).
  4. Find the area of the shaded region.

\(y = x^3 - 3x^2 - 9x + k\)
AnswerMarks Guidance
(i) \(\frac{dy}{dx} = 3x^2 - 6x - 9\)M1 A1 Attempt to differentiate. All correct.
\(= 0\) when \(x = 3\) or \(x = -1\)DM1 Sets a differential to 0.
→ \(x = 3, y = 0\) → \(k = 27\)A1 co.
(ii) \(x = -1 \to y = 32\)B1 For his second value.
(iii) \(-1 < x < 3\)B1 Realises the need to look at -ve m. (accept \(\leq\))
(iv) Integrate \(y\) to get area. → \(\left[\frac{x^4}{4} - x^3 - \frac{9x^2}{2} + kx\right]\)M1 A2,1 Attempt at integration. -1 each error.
→ \(33.75\) when \(x = 3\)A1 co.
Total: [4] [1] [1] [4]
$y = x^3 - 3x^2 - 9x + k$

**(i)** $\frac{dy}{dx} = 3x^2 - 6x - 9$ | M1 A1 | Attempt to differentiate. All correct.
$= 0$ when $x = 3$ or $x = -1$ | DM1 | Sets a differential to 0.
→ $x = 3, y = 0$ → $k = 27$ | A1 | co.

**(ii)** $x = -1 \to y = 32$ | B1 | For his second value.

**(iii)** $-1 < x < 3$ | B1 | Realises the need to look at -ve m. (accept $\leq$)

**(iv)** Integrate $y$ to get area. → $\left[\frac{x^4}{4} - x^3 - \frac{9x^2}{2} + kx\right]$ | M1 A2,1 | Attempt at integration. -1 each error.
→ $33.75$ when $x = 3$ | A1 | co.

**Total: [4] [1] [1] [4]**

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10\\
\includegraphics[max width=\textwidth, alt={}, center]{cbcb15b4-1870-4dfd-b6e9-839aa4601511-4_515_885_662_630}

The diagram shows the curve $y = x ^ { 3 } - 3 x ^ { 2 } - 9 x + k$, where $k$ is a constant. The curve has a minimum point on the $x$-axis.\\
(i) Find the value of $k$.\\
(ii) Find the coordinates of the maximum point of the curve.\\
(iii) State the set of values of $x$ for which $x ^ { 3 } - 3 x ^ { 2 } - 9 x + k$ is a decreasing function of $x$.\\
(iv) Find the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2006 Q10 [10]}}