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2013
P2 Q3
Topical
Maclaurin
13 P2 Q3
2013 H2 Mathematics Paper 2 Question 3
Maclaurin Series
Answers
(i)
f
(
0
)
=
0
{f(0)=0}
f
(
0
)
=
0
f
′
(
0
)
=
2
{f'(0)=2}
f
′
(
0
)
=
2
f
′
′
(
0
)
=
−
4
{f''(0)=-4}
f
′′
(
0
)
=
−
4
f
′
′
′
(
0
)
=
14
{f'''(0)=14}
f
′′′
(
0
)
=
14
f
(
x
)
≈
2
x
−
2
x
2
+
7
3
x
3
{f(x)\approx2 x - 2 x^2 + \frac{7}{3} x^3}
f
(
x
)
≈
2
x
−
2
x
2
+
3
7
x
3
(ii)
a
=
−
1
,
{a=-1,\;}
a
=
−
1
,
n
=
2
{n=2}
n
=
2
Third non-zero term
=
−
1
3
x
3
{= - \frac{1}{3}x^3}
=
−
3
1
x
3
Full solutions
(i)
Let
y
=
f
(
x
)
=
ln
(
1
+
2
sin
x
)
{y=f(x)=\ln (1+2\sin x)}
y
=
f
(
x
)
=
ln
(
1
+
2
sin
x
)
e
y
=
1
+
2
sin
x
\mathrm{e}^y = 1 + 2 \sin x
e
y
=
1
+
2
sin
x
Differentiating implicitly w.r.t.
x
,
{x,}
x
,
e
y
d
y
d
x
=
2
cos
x
e
y
d
2
y
d
x
2
+
e
y
(
d
y
d
x
)
2
=
−
2
sin
x
e
y
d
3
y
d
x
3
+
e
y
d
y
d
x
d
2
y
d
x
2
+
2
e
y
d
y
d
x
d
2
y
d
x
2
+
e
y
(
d
y
d
x
)
3
=
−
2
cos
x
\begin{gather*} \mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cos x \\ \mathrm{e}^y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \mathrm{e}^y \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 =-2 \sin x \\ \mathrm{e}^y \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + \mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2 \mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \mathrm{e}^y \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^3 =-2 \cos x \\ \end{gather*}
e
y
d
x
d
y
=
2
cos
x
e
y
d
x
2
d
2
y
+
e
y
(
d
x
d
y
)
2
=
−
2
sin
x
e
y
d
x
3
d
3
y
+
e
y
d
x
d
y
d
x
2
d
2
y
+
2
e
y
d
x
d
y
d
x
2
d
2
y
+
e
y
(
d
x
d
y
)
3
=
−
2
cos
x
When
x
=
0
,
{x=0,}
x
=
0
,
y
=
ln
(
1
+
2
sin
0
)
=
0
e
0
d
y
d
x
=
2
cos
0
d
y
d
x
=
2
e
0
d
2
y
d
x
2
+
e
0
(
2
)
2
=
−
2
sin
0
d
2
y
d
x
2
=
−
4
e
0
d
3
y
d
x
3
+
3
e
0
(
2
)
(
−
4
)
+
e
0
(
2
)
3
=
−
2
cos
0
d
3
y
d
x
3
=
14
\begin{gather*} y = \ln(1+2\sin 0) = 0 \\ \mathrm{e}^0 \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cos 0 \\ \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \\ \mathrm{e}^0 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \mathrm{e}^0 (2)^2 =-2 \sin 0 \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -4 \\ \mathrm{e}^0 \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 3 \mathrm{e}^0 (2)(-4) + \mathrm{e}^0 (2)^3 =-2 \cos 0 \\ \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 14 \end{gather*}
y
=
ln
(
1
+
2
sin
0
)
=
0
e
0
d
x
d
y
=
2
cos
0
d
x
d
y
=
2
e
0
d
x
2
d
2
y
+
e
0
(
2
)
2
=
−
2
sin
0
d
x
2
d
2
y
=
−
4
e
0
d
x
3
d
3
y
+
3
e
0
(
2
)
(
−
4
)
+
e
0
(
2
)
3
=
−
2
cos
0
d
x
3
d
3
y
=
14
f
(
0
)
=
0
■
f
′
(
0
)
=
2
■
f
′
′
(
0
)
=
−
4
■
f
′
′
′
(
0
)
=
14
■
\begin{align*} f(0) &= 0 \; \blacksquare \\ f'(0) &= 2 \; \blacksquare \\ f''(0) &= -4 \; \blacksquare \\ f'''(0) &= 14 \; \blacksquare \\ \end{align*}
f
(
0
)
f
′
(
0
)
f
′′
(
0
)
f
′′′
(
0
)
=
0
■
=
2
■
=
−
4
■
=
14
■
Maclaurin series for
f
(
x
)
:
{f(x):}
f
(
x
)
:
f
(
x
)
=
0
+
2
x
−
4
2
x
2
+
14
3
!
x
3
+
…
=
2
x
−
2
x
2
+
7
3
x
3
+
…
■
\begin{align*} f(x) &= 0 + 2x - \frac{4}{2} x^2 + \frac{14}{3!}x^3 + \ldots \\ &= 2 x - 2 x^2 + \frac{7}{3} x^3 + \ldots \blacksquare \end{align*}
f
(
x
)
=
0
+
2
x
−
2
4
x
2
+
3
!
14
x
3
+
…
=
2
x
−
2
x
2
+
3
7
x
3
+
…
■
(ii)
e
a
x
sin
n
x
=
(
1
+
a
x
+
a
2
x
2
2
+
…
)
(
n
x
−
n
3
x
3
6
+
…
)
=
n
x
+
a
n
x
2
+
(
−
n
3
6
+
a
2
n
2
)
x
3
+
…
\begin{align*} & \mathrm{e}^{ax} \sin nx \\ & = \left( 1 + ax + \frac{a^2x^2}{2} + \ldots \right) \left( nx - \frac{n^3x^3}{6} + \ldots \right) \\ & = nx + anx^2 + \left( -\frac{n^3}{6}+\frac{a^2n}{2} \right)x^3 + \ldots \end{align*}
e
a
x
sin
n
x
=
(
1
+
a
x
+
2
a
2
x
2
+
…
)
(
n
x
−
6
n
3
x
3
+
…
)
=
n
x
+
an
x
2
+
(
−
6
n
3
+
2
a
2
n
)
x
3
+
…
Since the first two non-zero terms are equal,
n
=
2
■
a
n
=
−
2
a
=
−
1
■
\begin{align*} n &= 2 \; \blacksquare \\ an &= -2 \\ a &= -1 \; \blacksquare \\ \end{align*}
n
an
a
=
2
■
=
−
2
=
−
1
■
Third non-zero term
=
(
−
2
3
6
+
(
−
1
)
2
(
2
)
2
)
x
3
=
−
1
3
x
3
■
\begin{align*} & \textrm{Third non-zero term} \\ & = \left( -\frac{2^3}{6} + \frac{(-1)^2(2)}{2} \right)x^3 \\ &= - \frac{1}{3} x^3 \; \blacksquare \end{align*}
Third non-zero term
=
(
−
6
2
3
+
2
(
−
1
)
2
(
2
)
)
x
3
=
−
3
1
x
3
■
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