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Specimen 2017 H2 Mathematics Paper 2 Question 4
Maclaurin Series
Answers
(i)
k
=
2.
{k=2.}
k
=
2.
(ii)
2
x
+
2
x
2
+
4
x
3
{2 x + 2 x^2 + 4 x^3}
2
x
+
2
x
2
+
4
x
3
.
(iii)
n
=
2
,
a
=
2.
{n=2, a=2.}
n
=
2
,
a
=
2.
Third non-zero term of the series expansion
=
8
3
x
3
.
{=\frac{8}{3}x^3.}
=
3
8
x
3
.
Full solutions
(i)
d
y
d
x
=
sec
2
(
e
2
x
−
1
)
⋅
(
2
e
2
x
)
=
2
e
2
x
(
tan
2
(
e
2
x
−
1
)
+
1
)
=
2
e
2
x
(
1
+
y
2
)
■
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \sec^2 (\mathrm{e}^{2x}-1) \cdot (2 \mathrm{e}^{2x}) \\ &= 2\mathrm{e}^{2x} \left( \tan^2(\mathrm{e}^{2x}-1)+1 \right) \\ &= 2\mathrm{e}^{2x} (1+y^2) \; \blacksquare \end{align*}
d
x
d
y
=
sec
2
(
e
2
x
−
1
)
⋅
(
2
e
2
x
)
=
2
e
2
x
(
tan
2
(
e
2
x
−
1
)
+
1
)
=
2
e
2
x
(
1
+
y
2
)
■
d
2
y
d
x
2
=
4
e
2
x
(
1
+
y
2
)
+
2
e
2
x
(
2
y
d
y
d
x
)
=
2
d
y
d
x
+
4
e
2
x
y
d
y
d
x
=
2
d
y
d
x
(
1
+
2
e
2
x
y
)
\begin{align*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= 4\mathrm{e}^{2x}(1+y^2) + 2\mathrm{e}^{2x}\left( 2y \frac{\mathrm{d}y}{\mathrm{d}x} \right) \\ &= 2\frac{\mathrm{d}y}{\mathrm{d}x} + 4\mathrm{e}^{2x}y\frac{\mathrm{d}y}{\mathrm{d}x} \\ &= 2\frac{\mathrm{d}y}{\mathrm{d}x} \left( 1 + 2\mathrm{e}^{2x}y \right) \end{align*}
d
x
2
d
2
y
=
4
e
2
x
(
1
+
y
2
)
+
2
e
2
x
(
2
y
d
x
d
y
)
=
2
d
x
d
y
+
4
e
2
x
y
d
x
d
y
=
2
d
x
d
y
(
1
+
2
e
2
x
y
)
d
3
y
d
x
3
=
2
d
2
y
d
x
2
(
1
+
2
e
2
x
y
)
+
2
d
y
d
x
(
4
e
2
x
y
+
2
e
2
x
d
y
d
x
)
\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 2\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}(1+2\mathrm{e}^{2x}y) + 2\frac{\mathrm{d}y}{\mathrm{d}x}\left(4\mathrm{e}^{2x}y + 2\mathrm{e}^{2x}\frac{\mathrm{d}y}{\mathrm{d}x}\right)
d
x
3
d
3
y
=
2
d
x
2
d
2
y
(
1
+
2
e
2
x
y
)
+
2
d
x
d
y
(
4
e
2
x
y
+
2
e
2
x
d
x
d
y
)
When
x
=
0
,
{x=0,}
x
=
0
,
y
=
tan
(
e
2
(
0
)
−
1
)
=
0
d
y
d
x
=
2
e
2
(
0
)
(
1
+
0
2
)
=
2
d
2
y
d
x
2
=
2
(
2
)
(
1
+
2
e
2
(
0
)
(
0
)
)
=
4
d
3
y
d
x
3
=
2
(
4
)
(
1
+
2
e
2
(
0
)
(
0
)
)
+
2
(
2
)
(
4
e
2
(
0
)
(
0
)
+
2
e
2
(
0
)
(
2
)
)
=
24
\begin{align*} y &= \tan(\mathrm{e}^{2(0)}-1) \\ &= 0 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 2 \mathrm{e}^{2(0)} (1 + 0^2) \\ &= 2 \\ \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= 2 (2) (1+2\mathrm{e}^{2(0)}(0)) \\ &= 4 \\ \frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} &= 2(4)(1+2\mathrm{e}^{2(0)}(0)) + 2(2)(4\mathrm{e}^{2(0)}(0)+2\mathrm{e}^{2(0)}(2)) \\ &=24 \end{align*}
y
d
x
d
y
d
x
2
d
2
y
d
x
3
d
3
y
=
tan
(
e
2
(
0
)
−
1
)
=
0
=
2
e
2
(
0
)
(
1
+
0
2
)
=
2
=
2
(
2
)
(
1
+
2
e
2
(
0
)
(
0
))
=
4
=
2
(
4
)
(
1
+
2
e
2
(
0
)
(
0
))
+
2
(
2
)
(
4
e
2
(
0
)
(
0
)
+
2
e
2
(
0
)
(
2
))
=
24
(ii)
Maclaurin series for
tan
(
e
2
x
−
1
)
=
0
+
2
x
+
4
2
!
x
2
+
24
3
!
x
3
+
…
=
2
x
+
2
x
2
+
4
x
3
+
…
■
\begin{align*} & \textrm{Maclaurin series for } \tan(\mathrm{e}^{2x}-1) \\ & = 0 + 2x + \frac{4}{2!}x^2 + \frac{24}{3!}x^3 + \ldots \\ & = 2x + 2x^2 + 4x^3 + \ldots \; \blacksquare \end{align*}
Maclaurin series for
tan
(
e
2
x
−
1
)
=
0
+
2
x
+
2
!
4
x
2
+
3
!
24
x
3
+
…
=
2
x
+
2
x
2
+
4
x
3
+
…
■
(iii)
e
a
x
ln
(
1
+
n
x
)
=
(
1
+
a
x
+
1
2
a
2
x
2
+
1
6
a
3
x
3
)
(
n
x
−
1
2
n
2
x
2
+
1
3
n
3
x
3
)
+
…
=
n
x
+
(
a
n
−
1
2
n
2
)
x
2
+
(
1
3
n
3
−
1
2
a
n
2
+
1
2
a
2
n
)
x
3
+
…
\begin{align*} & \mathrm{e}^{ax} \ln (1+nx) \\ & = (1 + a x + \frac{1}{2} a^2 x^2 + \frac{1}{6} a^3 x^3) (n x - \frac{1}{2} n^2 x^2 + \frac{1}{3} n^3 x^3) + \ldots \\ & = nx + \left(an-\frac{1}{2}n^2 \right)x^2 + \left(\frac{1}{3}n^3 - \frac{1}{2}an^2 + \frac{1}{2}a^2n\right) x^3 + \ldots \\ \end{align*}
e
a
x
ln
(
1
+
n
x
)
=
(
1
+
a
x
+
2
1
a
2
x
2
+
6
1
a
3
x
3
)
(
n
x
−
2
1
n
2
x
2
+
3
1
n
3
x
3
)
+
…
=
n
x
+
(
an
−
2
1
n
2
)
x
2
+
(
3
1
n
3
−
2
1
a
n
2
+
2
1
a
2
n
)
x
3
+
…
Since the first two non-zero terms are equal,
n
=
2
■
n = 2 \; \blacksquare
n
=
2
■
a
n
−
1
2
n
2
=
2
a
=
2
■
\begin{gather*} an - \frac{1}{2}n^2 = 2 \\ a = 2 \; \blacksquare \end{gather*}
an
−
2
1
n
2
=
2
a
=
2
■
Third non-zero term of the series expansion
=
(
1
3
n
3
−
1
2
a
n
2
+
1
2
a
2
n
)
x
3
=
8
3
x
3
■
\begin{align*} & \textrm{Third non-zero term of the series expansion} \\ &= \left(\frac{1}{3}n^3 -\frac{1}{2}an^2 + \frac{1}{2}a^2n\right) x^3 \\ &= \frac{8}{3}x^3 \; \blacksquare \end{align*}
Third non-zero term of the series expansion
=
(
3
1
n
3
−
2
1
a
n
2
+
2
1
a
2
n
)
x
3
=
3
8
x
3
■
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