CAIE P2 2019 June — Question 3 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind normal equation at point
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate using product and chain rules, substitute a point to find the gradient, then find the perpendicular gradient and write the normal equation. While it involves multiple techniques, these are standard procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation
3 Find the equation of the normal to the curve $$x ^ { 2 } \ln y + 2 x + 5 y = 11$$ at the point \(( 3,1 )\).
Question 3:
Normal to \(x^2\ln y + 2x + 5y = 11\) at \((3, 1)\)
AnswerMarks Guidance
\(2x\ln y + x^2 \cdot \dfrac{1}{y}\dfrac{dy}{dx} + 2 + 5\dfrac{dy}{dx} = 0\)M1 A1 Implicit differentiation: M1 for attempt, A1 fully correct
At \((3,1)\): \(2(3)\ln 1 + 9\dfrac{dy}{dx} + 2 + 5\dfrac{dy}{dx} = 0\)M1 Substituting \((3,1)\)
\(0 + 14\dfrac{dy}{dx} + 2 = 0\)
\(\dfrac{dy}{dx} = -\dfrac{1}{7}\)A1 Correct gradient of tangent
Gradient of normal \(= 7\)M1 Using \(m_1 m_2 = -1\)
\(y - 1 = 7(x-3)\)M1 Using point and normal gradient
\(y = 7x - 20\)A1 
# Question 3:

Normal to $x^2\ln y + 2x + 5y = 11$ at $(3, 1)$

| $2x\ln y + x^2 \cdot \dfrac{1}{y}\dfrac{dy}{dx} + 2 + 5\dfrac{dy}{dx} = 0$ | M1 A1 | Implicit differentiation: M1 for attempt, A1 fully correct |
|---|---|---|
| At $(3,1)$: $2(3)\ln 1 + 9\dfrac{dy}{dx} + 2 + 5\dfrac{dy}{dx} = 0$ | M1 | Substituting $(3,1)$ |
| $0 + 14\dfrac{dy}{dx} + 2 = 0$ | | |
| $\dfrac{dy}{dx} = -\dfrac{1}{7}$ | A1 | Correct gradient of tangent |
| Gradient of normal $= 7$ | M1 | Using $m_1 m_2 = -1$ |
| $y - 1 = 7(x-3)$ | M1 | Using point and normal gradient |
| $y = 7x - 20$ | A1 | |

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3 Find the equation of the normal to the curve

$$x ^ { 2 } \ln y + 2 x + 5 y = 11$$

at the point $( 3,1 )$.\\

\hfill \mbox{\textit{CAIE P2 2019 Q3 [7]}}
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