CAIE P3 2021 November — Question 7 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring standard application of the chain rule and algebraic manipulation to find dy/dx, followed by setting the numerator to zero to find stationary points. The algebra is routine and the method is a direct textbook exercise, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation

7 The equation of a curve is \(\ln ( x + y ) = x - 2 y\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x + y - 1 } { 2 ( x + y ) + 1 }\).
  2. Find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis. \(\quad\) [3]

Question 7(a):
AnswerMarks Guidance
AnswerMark Guidance
Use chain rule to differentiate LHS\*M1
Obtain \(\frac{1}{x+y}\left(1 + \frac{dy}{dx}\right)\)A1
Equate derivative of LHS to \(1 - 2\frac{dy}{dx}\) and solve for \(\frac{dy}{dx}\)DM1
Obtain the given answer correctlyA1
Question 7(b):
AnswerMarks Guidance
AnswerMark Guidance
State \(x + y = 1\)B1
Obtain or imply \(x - 2y = 0\)B1
Obtain coordinates \(x = \frac{2}{3}\) and \(y = \frac{1}{3}\)B1
## Question 7(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use chain rule to differentiate LHS | \*M1 | |
| Obtain $\frac{1}{x+y}\left(1 + \frac{dy}{dx}\right)$ | A1 | |
| Equate derivative of LHS to $1 - 2\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$ | DM1 | |
| Obtain the given answer correctly | A1 | |

## Question 7(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| State $x + y = 1$ | B1 | |
| Obtain or imply $x - 2y = 0$ | B1 | |
| Obtain coordinates $x = \frac{2}{3}$ and $y = \frac{1}{3}$ | B1 | |

---
7 The equation of a curve is $\ln ( x + y ) = x - 2 y$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x + y - 1 } { 2 ( x + y ) + 1 }$.
\item Find the coordinates of the point on the curve where the tangent is parallel to the $x$-axis. $\quad$ [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2021 Q7 [7]}}