Standard +0.8 This question requires finding a Maclaurin series by squaring ln(1+x), then using it to solve an intersection problem. Part (a) is standard Further Maths technique (squaring a known series), but part (b) requires setting up and solving a polynomial equation from the approximation, which adds problem-solving depth. The multi-step nature and need to connect series expansion to a geometric application elevates this above routine exercises, though it remains within expected Further Maths scope.
Find the Maclaurin series of \(( \ln ( 1 + x ) ) ^ { 2 }\) up to and including the term in \(x ^ { 4 }\).
The diagram below shows parts of the graphs of the curves with equations \(y = ( \ln ( 1 + x ) ) ^ { 2 }\) and \(y = 2 x ^ { 3 }\).
The curves intersect at the origin, \(O\), and at the point \(A\).
\includegraphics[max width=\textwidth, alt={}, center]{fbb82fa2-b316-44ae-a19e-197b45f51c87-4_663_906_831_248}
\section*{(b) In this question you must show detailed reasoning.}
Use your answer to part (a) to determine an approximation for the value of the \(x\)-coordinate of \(A\). Give your answer to \(\mathbf { 2 }\) decimal places.
Squaring the correct result for l n ( 1 + x ) with at least the
first three terms.
First two terms correct.
Third term correct allow any equivalent fraction or exact
decimal equivalent. ISW and ignore higher power terms.
Alternative method
2 l n ( x + 1 )
f(0)=0 and f ( x ) =
Answer
Marks
Guidance
x + 1
M1
M1
second derivative is zero too).
For f(0) = 0 and correct first derivative (or stating that the
second derivative is zero too).
Answer
Marks
Guidance
f(0)=0, f(0)=2, f(0)=−6
A1
Correct values for first, second and third derivatives
2−2ln(x+1)
For reference: f(x)= and
(x+1)2
2 ( − 3 + 2 l n ( x + 1 ) )
f ( x ) = and
( x + 1 ) 3
22−12ln(x+1)
f(4)(x)= and f (4 ) ( 0 ) = 2 2 .
(x+1)4
1 1
( l n (1 + x ) ) 2 = x 2 − x 3 + x 4 + ...
Answer
Marks
Guidance
1 2
A1
All terms correct (allow simplified or exact decimal
equivalents for coefficients). ISW and ignore higher
power terms.
[3]
f(0)=0, f(0)=2, f(0)=−6
A1
Correct values for first, second and third derivatives
2−2ln(x+1)
For reference: f(x)= and
(x+1)2
2 ( − 3 + 2 l n ( x + 1 ) )
f ( x ) = and
( x + 1 ) 3
22−12ln(x+1)
f(4)(x)= and f (4 ) ( 0 ) = 2 2 .
(x+1)4
1 1
( l n (1 + x ) ) 2 = x 2 − x 3 + x 4 + ...
1 2
A1
Answer
Marks
Guidance
9
(b)
DR
11
x2 −x3+ x4 +...=2x3
12
11
x2 1−3x+ x2 =0
12
x=0.37(6690...) or 2.89(6036...)
Answer
Marks
but x = 0 .3 8 as expansion is only valid for − 1 x 1 .
M1*
M1dep*
A1
Answer
Marks
[3]
3.1a
1.1
Answer
Marks
3.2a
Setting their expansion from part (a) equal to 2 x 3 . Their
expansion must have at least three terms with non-zero
quadratic, cubic and quartic terms.
Rearranging and factorising (allow sign slips only).
Ignore higher order terms if the factor of x2 is seen – their
expansion must have contained no constant or linear
terms. Stating their quadratic 1 − 3 x + 11 12 x 2 = 0 is fine
for this mark. Stating either non-zero root of the correct
(or their) quadratic can imply this mark.
Selects the smallest non-zero root (so must see both
1 8 8 3
correct roots either exact or to at least 2
1 1
decimal places) and justifies with interval for validity
1 8 − 8 3
being − 1 x 1 . Allow awrt 0.38 or .
1 1
Condone any indication that expansion is only valid
between values of − 1 and 1 or for values less than, or
less than or equal, to 1.
Note that including x5 terms in expansion gives
x = 0 .3 5 9 9 7 and including x5 and x6 terms in expansion
give x = 0 .3 6 5 0 4 . Ignore any attempt to work out
corresponding y-coordinate.
Question 9:
9 | (a) | 2
x 2 x 3
( l n ( 1 + x ) ) 2 = x − + + ...
2 3
= x 2 − x 3 + ...
1 1
. . . + x 4 + . . .
1 2 | M1
B1
A1
[3] | 3.1a
1.1
2.2a | Squaring the correct result for l n ( 1 + x ) with at least the
first three terms.
First two terms correct.
Third term correct allow any equivalent fraction or exact
decimal equivalent. ISW and ignore higher power terms.
Alternative method
2 l n ( x + 1 )
f(0)=0 and f ( x ) =
x + 1 | M1 | M1 | For f(0) = 0 and correct first derivative (or stating that the
second derivative is zero too). | For f(0) = 0 and correct first derivative (or stating that the
second derivative is zero too).
f(0)=0, f(0)=2, f(0)=−6 | A1 | Correct values for first, second and third derivatives
2−2ln(x+1)
For reference: f(x)= and
(x+1)2
2 ( − 3 + 2 l n ( x + 1 ) )
f ( x ) = and
( x + 1 ) 3
22−12ln(x+1)
f(4)(x)= and f (4 ) ( 0 ) = 2 2 .
(x+1)4
1 1
( l n (1 + x ) ) 2 = x 2 − x 3 + x 4 + ...
1 2 | A1 | All terms correct (allow simplified or exact decimal
equivalents for coefficients). ISW and ignore higher
power terms.
[3]
f(0)=0, f(0)=2, f(0)=−6
A1
Correct values for first, second and third derivatives
2−2ln(x+1)
For reference: f(x)= and
(x+1)2
2 ( − 3 + 2 l n ( x + 1 ) )
f ( x ) = and
( x + 1 ) 3
22−12ln(x+1)
f(4)(x)= and f (4 ) ( 0 ) = 2 2 .
(x+1)4
1 1
( l n (1 + x ) ) 2 = x 2 − x 3 + x 4 + ...
1 2
A1
9 | (b) | DR
11
x2 −x3+ x4 +...=2x3
12
11
x2 1−3x+ x2 =0
12
x=0.37(6690...) or 2.89(6036...)
but x = 0 .3 8 as expansion is only valid for − 1 x 1 . | M1*
M1dep*
A1
[3] | 3.1a
1.1
3.2a | Setting their expansion from part (a) equal to 2 x 3 . Their
expansion must have at least three terms with non-zero
quadratic, cubic and quartic terms.
Rearranging and factorising (allow sign slips only).
Ignore higher order terms if the factor of x2 is seen – their
expansion must have contained no constant or linear
terms. Stating their quadratic 1 − 3 x + 11 12 x 2 = 0 is fine
for this mark. Stating either non-zero root of the correct
(or their) quadratic can imply this mark.
Selects the smallest non-zero root (so must see both
1 8 8 3
correct roots either exact or to at least 2
1 1
decimal places) and justifies with interval for validity
1 8 − 8 3
being − 1 x 1 . Allow awrt 0.38 or .
1 1
Condone any indication that expansion is only valid
between values of − 1 and 1 or for values less than, or
less than or equal, to 1.
Note that including x5 terms in expansion gives
x = 0 .3 5 9 9 7 and including x5 and x6 terms in expansion
give x = 0 .3 6 5 0 4 . Ignore any attempt to work out
corresponding y-coordinate.
9 (a) Find the Maclaurin series of $( \ln ( 1 + x ) ) ^ { 2 }$ up to and including the term in $x ^ { 4 }$.
The diagram below shows parts of the graphs of the curves with equations $y = ( \ln ( 1 + x ) ) ^ { 2 }$ and $y = 2 x ^ { 3 }$.
The curves intersect at the origin, $O$, and at the point $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{fbb82fa2-b316-44ae-a19e-197b45f51c87-4_663_906_831_248}
\section*{(b) In this question you must show detailed reasoning.}
Use your answer to part (a) to determine an approximation for the value of the $x$-coordinate of $A$. Give your answer to $\mathbf { 2 }$ decimal places.
\hfill \mbox{\textit{OCR Further Pure Core 1 2024 Q9 [6]}}