CAIE P2 2011 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2011
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind gradient at a point - direct evaluation
DifficultyEasy -1.2 This is a straightforward application of the chain rule to differentiate ln(5x+1), giving dy/dx = 5/(5x+1), followed by direct substitution of x=4. It requires only one standard technique with no problem-solving or conceptual challenge, making it easier than average.
Spec1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

1 Find the gradient of the curve \(y = \ln ( 5 x + 1 )\) at the point where \(x = 4\).

AnswerMarks Guidance
Obtain derivative of the form \(\frac{k}{5x+1}\), where \(k = 1, 5\) or \(\frac{1}{5}\)M1
Obtain correct derivative \(\frac{5}{5x+1}\)A1
Substitute \(x = 4\) into expression for derivative and obtain \(\frac{5}{21}\)A1√ [3]
Obtain derivative of the form $\frac{k}{5x+1}$, where $k = 1, 5$ or $\frac{1}{5}$ | M1 |
Obtain correct derivative $\frac{5}{5x+1}$ | A1 |
Substitute $x = 4$ into expression for derivative and obtain $\frac{5}{21}$ | A1√ | [3]
1 Find the gradient of the curve $y = \ln ( 5 x + 1 )$ at the point where $x = 4$.

\hfill \mbox{\textit{CAIE P2 2011 Q1 [3]}}