CAIE P2 2012 June — Question 5 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2012
SessionJune
Marks9
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TopicStandard Integrals and Reverse Chain Rule
TypeArea under curve using integration
DifficultyStandard +0.3 This is a straightforward two-part question requiring standard calculus techniques: (i) finding a stationary point by differentiation and solving an exponential equation, and (ii) computing a definite integral using reverse chain rule. Both parts are routine applications of core P2 methods with no novel problem-solving required, making it slightly easier than average.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits

5 \includegraphics[max width=\textwidth, alt={}, center]{0a45a806-007f-4840-85e7-16d4c1a2c599-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).
  1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
  2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).

AnswerMarks Guidance
(i) Differentiate to obtain expression of form \(ke^{3x} + m\)M1
Obtain correct \(2e^{3x} - 6\)A1
Equate attempt at first derivative to zero and attempt solutionDM1
Obtain \(\frac{1}{3} x = \ln 3\) or equivalentA1
Conclude \(x = \ln 9\) or \(a = 9\)A1 [5]
(ii) Integrate to obtain expression of form \(ae^{3x} + bx^2 + cx\)M1
Obtain correct \(8e^{3x} - 3x^2 + 3x\)A1
Substitute correct limits and attempt simplificationDM1
Obtain \(8e - 14\)A1 [4]
**(i)** Differentiate to obtain expression of form $ke^{3x} + m$ | M1 | |
| Obtain correct $2e^{3x} - 6$ | A1 | |
| Equate attempt at first derivative to zero and attempt solution | DM1 | |
| Obtain $\frac{1}{3} x = \ln 3$ or equivalent | A1 | |
| Conclude $x = \ln 9$ or $a = 9$ | A1 | [5] |

**(ii)** Integrate to obtain expression of form $ae^{3x} + bx^2 + cx$ | M1 | |
| Obtain correct $8e^{3x} - 3x^2 + 3x$ | A1 | |
| Substitute correct limits and attempt simplification | DM1 | |
| Obtain $8e - 14$ | A1 | [4] |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{0a45a806-007f-4840-85e7-16d4c1a2c599-3_528_757_251_694}

The diagram shows the curve $y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3$ and its minimum point $M$.\\
(i) Show that the $x$-coordinate of $M$ can be written in the form $\ln a$, where the value of $a$ is to be stated.\\
(ii) Find the exact value of the area of the region enclosed by the curve and the lines $x = 0 , x = 2$ and $y = 0$.

\hfill \mbox{\textit{CAIE P2 2012 Q5 [9]}}
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