CAIE P1 2013 November — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (straightforward integration + point)
DifficultyModerate -0.8 This is a straightforward integration question requiring only the power rule for two simple terms, followed by using a boundary condition to find the constant. It's slightly easier than average as it involves direct application of standard techniques with no problem-solving or manipulation required.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums
2 A curve has equation \(y = \mathrm { f } ( x )\). It is given that \(\mathrm { f } ^ { \prime } ( x ) = x ^ { - \frac { 3 } { 2 } } + 1\) and that \(\mathrm { f } ( 4 ) = 5\). Find \(\mathrm { f } ( x )\).
AnswerMarks Guidance
\(f(x) = 2x^{-\frac{1}{3}} + x (+c)\)M1A1 Attempt integ \(x^{-\frac{1}{3}}\) or \(+ x\) needed for M
\(5 = -2 \times \frac{1}{2} + 4 + c\)M1 Sub \((4, 5)\). \(c\) must be present
\(c = 2\)A1
[4] 
$f(x) = 2x^{-\frac{1}{3}} + x (+c)$ | M1A1 | Attempt integ $x^{-\frac{1}{3}}$ or $+ x$ needed for M
$5 = -2 \times \frac{1}{2} + 4 + c$ | M1 | Sub $(4, 5)$. $c$ must be present
$c = 2$ | A1 |
| | [4]
2 A curve has equation $y = \mathrm { f } ( x )$. It is given that $\mathrm { f } ^ { \prime } ( x ) = x ^ { - \frac { 3 } { 2 } } + 1$ and that $\mathrm { f } ( 4 ) = 5$. Find $\mathrm { f } ( x )$.

\hfill \mbox{\textit{CAIE P1 2013 Q2 [4]}}
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