CAIE P2 2024 June — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionJune
Marks5
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TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule application, followed by substitution of given coordinates. While it involves multiple differentiation techniques, it's a standard textbook exercise with no novel problem-solving required, making it slightly easier than average.
Spec1.07s Parametric and implicit differentiation
2 A curve has equation \(x ^ { 2 } \ln y + y ^ { 2 } + 4 x = 9\).
Find the gradient of the curve at the point \(( 2,1 )\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt use of product rule for differentiating \(x^2 \ln y\)M1 Allow if \(\frac{dy}{dx}\) missing
Obtain \(2x\ln y + \frac{x^2}{y}\frac{dy}{dx}\)A1
Obtain \(+2y\frac{dy}{dx}\)B1
Substitute \(x=2\), \(y=1\) in equation equal to 0 involving at least one \(\frac{dy}{dx}\) and solve for \(\frac{dy}{dx}\)M1 Correct equation: \(2x\ln y + \frac{x^2}{y}\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0\)
Obtain \(-\frac{2}{3}\)A1 Or exact equivalent. Not from wrong working.
Total5 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of product rule for differentiating $x^2 \ln y$ | M1 | Allow if $\frac{dy}{dx}$ missing |
| Obtain $2x\ln y + \frac{x^2}{y}\frac{dy}{dx}$ | A1 | |
| Obtain $+2y\frac{dy}{dx}$ | B1 | |
| Substitute $x=2$, $y=1$ in equation equal to 0 involving at least one $\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$ | M1 | Correct equation: $2x\ln y + \frac{x^2}{y}\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0$ |
| Obtain $-\frac{2}{3}$ | A1 | Or exact equivalent. Not from wrong working. |
| **Total** | **5** | |
2 A curve has equation $x ^ { 2 } \ln y + y ^ { 2 } + 4 x = 9$.\\
Find the gradient of the curve at the point $( 2,1 )$.\\

\hfill \mbox{\textit{CAIE P2 2024 Q2 [5]}}
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Q1 3 Q2 5 Q3 8 Q4 7 Q5 9 Q6 9 Q7 9