CAIE P1 2020 November — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionNovember
Marks4
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TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (reverse chain rule / composite functions)
DifficultyModerate -0.8 This is a straightforward integration question requiring only standard techniques: integrating (x-3)^{-2} using the reverse chain rule and integrating x using the power rule, then applying the boundary condition to find the constant. It's simpler than average A-level questions as it involves direct application of basic integration rules with no problem-solving or manipulation required.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits
2 The equation of a curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( x - 3 ) ^ { 2 } } + x\). It is given that the curve passes through the point (2, 7). Find the equation of the curve.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\((y =)\left[-(x-3)^{-1}\right]\left[+\frac{1}{2}x^2\right](+c)\)B1 B1
\(7 = 1 + 2 + c\)M1 Substitute \(x = 2\), \(y = 7\) into an integrated expression (\(c\) present). Expect \(c = 4\)
\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + 4\)A1 OE
4 
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(y =)\left[-(x-3)^{-1}\right]\left[+\frac{1}{2}x^2\right](+c)$ | B1 B1 | |
| $7 = 1 + 2 + c$ | M1 | Substitute $x = 2$, $y = 7$ into an integrated expression ($c$ present). Expect $c = 4$ |
| $y = -(x-3)^{-1} + \frac{1}{2}x^2 + 4$ | A1 | OE |
| | **4** | |
2 The equation of a curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( x - 3 ) ^ { 2 } } + x$. It is given that the curve passes through the point (2, 7).

Find the equation of the curve.\\

\hfill \mbox{\textit{CAIE P1 2020 Q2 [4]}}
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