CAIE P2 2022 March — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind tangent line equation
DifficultyModerate -0.8 This is a straightforward differentiation and tangent line question. It requires applying the chain rule to ln(2x+5) to get dy/dx = 8/(2x+5), evaluating at x=-2 to find the gradient m=8, then using point-slope form with the given point. All steps are routine with no problem-solving or insight required, making it easier than average.
Spec1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations
2 A curve has equation \(y = 7 + 4 \ln ( 2 x + 5 )\).
Find the equation of the tangent to the curve at the point ( \(- 2,7\) ), giving your answer in the form \(y = m x + c\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Differentiate to obtain form \(\frac{k}{2x+5}\)M1 Any non-zero constant \(k\)
Obtain correct \(\frac{8}{2x+5}\)A1
Substitute \(x = -2\) to obtain gradient \(8\)A1
Attempt equation of tangent through \((-2, 7)\) with numerical gradientM1
Obtain \(y = 8x + 23\)A1
Total5 
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain form $\frac{k}{2x+5}$ | M1 | Any non-zero constant $k$ |
| Obtain correct $\frac{8}{2x+5}$ | A1 | |
| Substitute $x = -2$ to obtain gradient $8$ | A1 | |
| Attempt equation of tangent through $(-2, 7)$ with numerical gradient | M1 | |
| Obtain $y = 8x + 23$ | A1 | |
| **Total** | **5** | |
2 A curve has equation $y = 7 + 4 \ln ( 2 x + 5 )$.\\
Find the equation of the tangent to the curve at the point ( $- 2,7$ ), giving your answer in the form $y = m x + c$.\\

\hfill \mbox{\textit{CAIE P2 2022 Q2 [5]}}
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Q1 3 Q2 5 Q3 5 Q4 7 Q5 8 Q6 10 Q7 12