Question 4 - A-Level Maths

CAIE P2 2020 March — Question 4 6 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2020
Session	March
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Find tangent equation at point
Difficulty	Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate both sides with respect to x (including chain rule for the ln term), substitute the given point to find dy/dx, then use point-slope form for the tangent. While it involves multiple techniques (implicit differentiation, chain rule, logarithmic differentiation), these are standard P2 skills with no novel problem-solving required, making it slightly easier than average.
Spec	1.07l Derivative of ln(x): and related functions 1.07m Tangents and normals: gradient and equations 1.07s Parametric and implicit differentiation

4 A curve has equation $$3 x ^ { 2 } - y ^ { 2 } - 4 \ln ( 2 y + 3 ) = 26$$ Find the equation of the tangent to the curve at the point $( 3 , - 1 )$.

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
Answer	Marks	Guidance
Differentiate $-y^2$ to obtain $-2y\frac{dy}{dx}$	B1
Differentiate $-4\ln(2y+3)$ to obtain $\frac{-8}{2y+3}\frac{dy}{dx}$	B1
Attempt differentiation of all terms	M1	Dependent on appearance of at least one $\frac{dy}{dx}$
Substitute $x=3$, $y=-1$ to find numerical value of $\frac{dy}{dx}$	M1
Obtain $\frac{dy}{dx} = 3$	A1
Obtain equation $y = 3x - 10$	A1	OE

## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate $-y^2$ to obtain $-2y\frac{dy}{dx}$ | B1 | |
| Differentiate $-4\ln(2y+3)$ to obtain $\frac{-8}{2y+3}\frac{dy}{dx}$ | B1 | |
| Attempt differentiation of all terms | M1 | Dependent on appearance of at least one $\frac{dy}{dx}$ |
| Substitute $x=3$, $y=-1$ to find numerical value of $\frac{dy}{dx}$ | M1 | |
| Obtain $\frac{dy}{dx} = 3$ | A1 | |
| Obtain equation $y = 3x - 10$ | A1 | OE |

---

Show LaTeX source

4 A curve has equation

$$3 x ^ { 2 } - y ^ { 2 } - 4 \ln ( 2 y + 3 ) = 26$$

Find the equation of the tangent to the curve at the point $( 3 , - 1 )$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q4 [6]}}

This paper (7 questions)

View full paper

Q1 4 Q2 6 Q3 6 Q4 6 Q5 9 Q6 9 Q7 10