CAIE P1 2014 June — Question 4 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeEquation of tangent line
DifficultyModerate -0.3 This is a straightforward application of the chain rule to find dy/dx, followed by routine tangent line calculation. It requires finding the y-coordinate at x=-1, computing the derivative using chain rule (which is standard for this form), and applying y-y₁=m(x-x₁). Slightly easier than average due to being a direct single-method question with no conceptual challenges, though the chain rule application and algebraic manipulation keep it near average difficulty.
Spec1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations
4 A curve has equation \(y = \frac { 4 } { ( 3 x + 1 ) ^ { 2 } }\). Find the equation of the tangent to the curve at the point where the line \(x = - 1\) intersects the curve.
AnswerMarks Guidance
\(\frac{dy}{dx} = [-2 \times 4(3x + 1)^3] \times [3]\)B1B1 [5] \([-2 \times 4u^3] \times [3]\) is B0B1 unless resolved
When \(x = -1\), \(\frac{dy}{dx} = 3\)B1
When \(x = -1, y = 1\) soiB1
\(y - 1 = 3(x + 1)\) (\(\to y = 3x + 4\))B1 \(\checkmark\) [5] Ft on their '3' only (not \(-\frac{1}{3}\)). Dep on diffn 
$\frac{dy}{dx} = [-2 \times 4(3x + 1)^3] \times [3]$ | B1B1 [5] | $[-2 \times 4u^3] \times [3]$ is B0B1 unless resolved

When $x = -1$, $\frac{dy}{dx} = 3$ | B1 | 

When $x = -1, y = 1$ soi | B1 | 
$y - 1 = 3(x + 1)$ ($\to y = 3x + 4$) | B1 $\checkmark$ [5] | Ft on their '3' only (not $-\frac{1}{3}$). Dep on diffn
4 A curve has equation $y = \frac { 4 } { ( 3 x + 1 ) ^ { 2 } }$. Find the equation of the tangent to the curve at the point where the line $x = - 1$ intersects the curve.

\hfill \mbox{\textit{CAIE P1 2014 Q4 [5]}}
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