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2016
P1 Q8
Topical
Maclaurin
16 P1 Q8
2016 H2 Mathematics Paper 1 Question 8
Maclaurin Series
Answers
(i)
f
′
′
(
x
)
=
2
a
2
y
+
2
a
3
y
{f''(x)=2a^2y+2a^3y}
f
′′
(
x
)
=
2
a
2
y
+
2
a
3
y
f
′
′
′
(
x
)
=
2
a
3
(
1
+
4
y
2
+
3
y
4
)
{f'''(x)}\allowbreak{=2a^3(1+4y^2+3y^4)}
f
′′′
(
x
)
=
2
a
3
(
1
+
4
y
2
+
3
y
4
)
(ii)
1
+
2
a
x
+
2
a
2
x
2
+
8
a
3
3
x
3
{1 + 2ax + 2a^2x^2 + \frac{8a^3}{3}x^3}
1
+
2
a
x
+
2
a
2
x
2
+
3
8
a
3
x
3
(iii)
tan
2
x
=
2
x
+
8
3
x
3
+
…
{\tan 2x = 2 x + \frac{8}{3} x^3 + \ldots}
tan
2
x
=
2
x
+
3
8
x
3
+
…
Full solutions
(i)
f
(
x
)
=
tan
(
a
x
+
b
)
f
′
(
x
)
=
a
sec
2
(
a
x
+
b
)
=
a
(
1
+
tan
2
(
a
x
+
b
)
)
=
a
(
1
+
y
2
)
=
a
+
a
y
2
■
\begin{align*} f(x) &= \tan (ax+b) \\ f'(x) &= a \sec^2 (ax+b) \\ &= a \left( 1 + \tan^2 (ax+b) \right) \\ &= a (1+y^2) \\ &= a + ay^2 \; \blacksquare \end{align*}
f
(
x
)
f
′
(
x
)
=
tan
(
a
x
+
b
)
=
a
sec
2
(
a
x
+
b
)
=
a
(
1
+
tan
2
(
a
x
+
b
)
)
=
a
(
1
+
y
2
)
=
a
+
a
y
2
■
Differentiating implicitly w.r.t.
x
,
{x,}
x
,
f
′
′
(
x
)
=
2
a
y
d
y
d
x
=
2
a
y
(
a
+
a
y
2
)
=
2
a
2
y
+
2
a
2
y
3
■
\begin{align*} f''(x) &= 2ay \frac{\mathrm{d}y}{\mathrm{d}x} \\ &= 2ay (a+ay^2) \\ & = 2a^2y + 2a^2y^3 \; \blacksquare \end{align*}
f
′′
(
x
)
=
2
a
y
d
x
d
y
=
2
a
y
(
a
+
a
y
2
)
=
2
a
2
y
+
2
a
2
y
3
■
f
′
′
′
(
x
)
=
2
a
2
d
y
d
x
+
6
a
2
y
2
d
y
d
x
=
2
a
2
(
a
+
a
y
2
)
+
6
a
2
y
2
(
a
+
a
y
2
)
=
2
a
3
+
2
a
3
y
2
+
6
a
3
y
2
+
6
a
3
y
4
=
2
a
3
(
1
+
4
y
2
+
3
y
4
)
■
\begin{align*} f'''(x) &= 2a^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 6a^2 y^2 \frac{\mathrm{d}y}{\mathrm{d}x} \\ &= 2a^2 (a+ay^2) + 6a^2 y^2 (a+ay^2) \\ &= 2a^3 + 2a^3y^2 + 6a^3y^2 + 6a^3y^4 \\ &= 2a^3 \left( 1 + 4y^2+3y^4 \right) \; \blacksquare \end{align*}
f
′′′
(
x
)
=
2
a
2
d
x
d
y
+
6
a
2
y
2
d
x
d
y
=
2
a
2
(
a
+
a
y
2
)
+
6
a
2
y
2
(
a
+
a
y
2
)
=
2
a
3
+
2
a
3
y
2
+
6
a
3
y
2
+
6
a
3
y
4
=
2
a
3
(
1
+
4
y
2
+
3
y
4
)
■
(ii)
When b=
1
4
π
,
{\frac{1}{4}\pi,}
4
1
π
,
f
(
0
)
=
tan
1
4
π
=
1
f
′
(
0
)
=
a
+
a
(
1
)
2
=
2
a
f
′
′
(
0
)
=
2
a
2
(
1
)
+
2
a
2
(
1
)
3
=
4
a
2
f
′
′
′
(
0
)
=
2
a
3
(
1
+
4
(
1
)
2
+
3
(
1
)
4
)
=
16
a
3
\begin{align*} f(0) &= \tan \frac{1}{4}\pi \\ &= 1 \\ f'(0) &= a + a(1)^2 \\ &= 2a \\ f''(0) &= 2a^2(1)+2a^2(1)^3 \\ &= 4a^2 \\ f'''(0) &= 2a^3 \left(1+4(1)^2+3(1)^4\right) \\ &= 16a^3 \end{align*}
f
(
0
)
f
′
(
0
)
f
′′
(
0
)
f
′′′
(
0
)
=
tan
4
1
π
=
1
=
a
+
a
(
1
)
2
=
2
a
=
2
a
2
(
1
)
+
2
a
2
(
1
)
3
=
4
a
2
=
2
a
3
(
1
+
4
(
1
)
2
+
3
(
1
)
4
)
=
16
a
3
Maclaurin series for
f
(
x
)
:
{f(x):}
f
(
x
)
:
f
(
x
)
=
1
+
2
a
x
+
4
a
2
2
x
2
+
16
a
3
3
!
x
3
+
…
=
1
+
2
a
x
+
2
a
2
x
2
+
8
a
3
3
x
3
+
…
■
\begin{align*} f(x) &= 1 + 2ax + \frac{4a^2}{2}x^2 + \frac{16a^3}{3!}x^3 + \ldots \\ &= 1 + 2ax + 2a^2x^2 + \frac{8a^3}{3}x^3 + \ldots \; \blacksquare \end{align*}
f
(
x
)
=
1
+
2
a
x
+
2
4
a
2
x
2
+
3
!
16
a
3
x
3
+
…
=
1
+
2
a
x
+
2
a
2
x
2
+
3
8
a
3
x
3
+
…
■
(iii)
When
a
=
2
{a=2}
a
=
2
and
b
=
0
,
{b=0,}
b
=
0
,
f
(
0
)
=
tan
0
=
0
f
′
(
0
)
=
2
+
2
(
0
)
2
=
2
f
′
′
(
0
)
=
2
a
2
(
0
)
+
2
a
2
(
0
)
3
=
0
f
′
′
′
(
0
)
=
2
(
2
)
3
(
1
+
4
(
0
)
2
+
3
(
0
)
4
)
=
16
\begin{align*} f(0) &= \tan 0 \\ &= 0 \\ f'(0) &= 2+2(0)^2 \\ &= 2 \\ f''(0) &= 2a^2(0) + 2a^2(0)^3 \\ &= 0 \\ f'''(0) &= 2(2)^3(1+4(0)^2+3(0)^4) \\ &= 16 \end{align*}
f
(
0
)
f
′
(
0
)
f
′′
(
0
)
f
′′′
(
0
)
=
tan
0
=
0
=
2
+
2
(
0
)
2
=
2
=
2
a
2
(
0
)
+
2
a
2
(
0
)
3
=
0
=
2
(
2
)
3
(
1
+
4
(
0
)
2
+
3
(
0
)
4
)
=
16
Maclaurin series for
tan
2
x
:
{\tan 2x:}
tan
2
x
:
f
(
x
)
=
0
+
2
x
+
0
x
2
+
16
3
!
x
3
+
…
=
2
x
+
8
3
x
3
+
…
■
\begin{align*} f(x) &= 0 + 2x + 0x^2 + \frac{16}{3!}x^3 + \ldots \\ &= 2x + \frac{8}{3}x^3 + \ldots \; \blacksquare \end{align*}
f
(
x
)
=
0
+
2
x
+
0
x
2
+
3
!
16
x
3
+
…
=
2
x
+
3
8
x
3
+
…
■
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