CAIE Further Paper 2 2024 November — Question 2 7 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyStandard +0.3 This is a straightforward implicit differentiation question with standard techniques. Part (a) requires applying the chain rule and product rule to differentiate ln(xy), then substituting a point - routine for Further Maths students. Part (b) extends to finding the second derivative, which adds one more layer but follows a mechanical process. The algebra is manageable and the question type is a textbook exercise rather than requiring any novel insight.
Spec1.07s Parametric and implicit differentiation
The curve \(C\) has equation $$4y^2 + 4\ln(xy) = 1.$$
  1. Show that, at the point \(\left(2, \frac{1}{2}\right)\) on \(C\), \(\frac{dy}{dx} = -\frac{1}{6}\). [3]
  2. Find the value of \(\frac{d^2y}{dx^2}\) at the point \(\left(2, \frac{1}{2}\right)\). [4]
Question 2:
AnswerMarks
2(a)d ( )
4y2 =8yy'
AnswerMarks Guidance
dxB1 Differentiates y2 correctly.
d xy'+ y
(4ln(xy))=4 =4y'y−1+4x−1
AnswerMarks Guidance
dx xyB1 Differentiates ln(xy) correctly.
8 (1) y'+4(2)y'+4 (1) =0y'=−1
AnswerMarks Guidance
2 2 6B1 Substitutes (2,1) into
2
8y'y+4y'y−1+4x−1=0, AG.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
2(b)EITHER Solution 1
8(y')2+8y''y−4(y')2 y−2+4y−1y''−4x−2 =0(B1B1 Differentiates 8y'y+4y'y−1+4x−1=0.
B1 for 8(y')2 +8y''y and B1 for other terms.
8 (−1)2 +8y'' (1)−16 (−1)2 +8y''−1=0
AnswerMarks Guidance
6 2 6M1) ( 2,1)
Substitutes .
2
OR Solution 2
y'=−x−1 ( 2y+ y−1 )−1
AnswerMarks Guidance
y''=x−1 ( 2y+ y−1 )−2( 2y'−y−2y' ) +x−2 ( 2y+ y−1 )−1(B1B1 B1 for x−1 ( 2y+ y−1 )−2( 2y'−y−2y' ) and B1
for x−2 ( 2y+ y−1 )−1 .
y''=1(1+2)−2(−1)(2−4)+1(1+2)−1
AnswerMarks Guidance
2 6 4M1) ( 2,1)
Substitutes and y'=−1.
2 6
12y''−11=0 y''= 11
AnswerMarks
9 108A1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | d ( )
4y2 =8yy'
dx | B1 | Differentiates y2 correctly.
d xy'+ y
(4ln(xy))=4 =4y'y−1+4x−1
dx xy | B1 | Differentiates ln(xy) correctly.
8 (1) y'+4(2)y'+4 (1) =0y'=−1
2 2 6 | B1 | Substitutes (2,1) into
2
8y'y+4y'y−1+4x−1=0, AG.
3
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | EITHER Solution 1
8(y')2+8y''y−4(y')2 y−2+4y−1y''−4x−2 =0 | (B1B1 | Differentiates 8y'y+4y'y−1+4x−1=0.
B1 for 8(y')2 +8y''y and B1 for other terms.
8 (−1)2 +8y'' (1)−16 (−1)2 +8y''−1=0
6 2 6 | M1) | ( 2,1)
Substitutes .
2
OR Solution 2
y'=−x−1 ( 2y+ y−1 )−1
y''=x−1 ( 2y+ y−1 )−2( 2y'−y−2y' ) +x−2 ( 2y+ y−1 )−1 | (B1B1 | B1 for x−1 ( 2y+ y−1 )−2( 2y'−y−2y' ) and B1
for x−2 ( 2y+ y−1 )−1 .
y''=1(1+2)−2(−1)(2−4)+1(1+2)−1
2 6 4 | M1) | ( 2,1)
Substitutes and y'=−1.
2 6
12y''−11=0 y''= 11
9 108 | A1
4
Question | Answer | Marks | Guidance
The curve $C$ has equation
$$4y^2 + 4\ln(xy) = 1.$$

\begin{enumerate}[label=(\alph*)]
\item Show that, at the point $\left(2, \frac{1}{2}\right)$ on $C$, $\frac{dy}{dx} = -\frac{1}{6}$. [3]
\item Find the value of $\frac{d^2y}{dx^2}$ at the point $\left(2, \frac{1}{2}\right)$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q2 [7]}}
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