CAIE P2 2023 June — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind equation of tangent
DifficultyModerate -0.3 This is a straightforward application of the quotient rule to find dy/dx, followed by evaluating at x=1 to get the gradient, then using point-slope form to write the tangent equation. All steps are routine and mechanical with no conceptual challenges, making it slightly easier than average but not trivial due to the algebraic manipulation required.
Spec1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

2 A curve has equation \(y = \frac { 2 + 3 \ln x } { 1 + 2 x }\).
Find the equation of the tangent to the curve at the point \(\left( 1 , \frac { 2 } { 3 } \right)\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Use quotient rule (or equivalent) to find first derivativeM1
Obtain \(\dfrac{(1+2x)\frac{3}{x} - (2+3\ln x)2}{(1+2x)^2}\)A1 OE
Substitute \(x=1\) and obtain \(\dfrac{5}{9}\)A1
Attempt equation of tangent through \(\left(1,\ \dfrac{2}{3}\right)\) with *their* numerical gradientM1 Must have made an attempt at differentiation
Obtain \(5x - 9y + 1 = 0\) or equivalent of required formA1
5
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use quotient rule (or equivalent) to find first derivative | M1 | |
| Obtain $\dfrac{(1+2x)\frac{3}{x} - (2+3\ln x)2}{(1+2x)^2}$ | A1 | OE |
| Substitute $x=1$ and obtain $\dfrac{5}{9}$ | A1 | |
| Attempt equation of tangent through $\left(1,\ \dfrac{2}{3}\right)$ with *their* numerical gradient | M1 | Must have made an attempt at differentiation |
| Obtain $5x - 9y + 1 = 0$ or equivalent of required form | A1 | |
| | **5** | |
2 A curve has equation $y = \frac { 2 + 3 \ln x } { 1 + 2 x }$.\\
Find the equation of the tangent to the curve at the point $\left( 1 , \frac { 2 } { 3 } \right)$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\

\hfill \mbox{\textit{CAIE P2 2023 Q2 [5]}}