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Question 0821
Maclaurin Series
2021 Paper 1 Question 7 Variant
Question
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It is given that
y
=
e
2
tan
−
1
x
,
{y=\mathrm{e}^{2 \tan^{-1} x},}
y
=
e
2
t
a
n
−
1
x
,
for
x
∈
R
.
{x \in \mathbb{R}.}
x
∈
R
.
(a)
Show that
(
1
+
x
2
)
d
2
y
d
x
2
=
(
2
−
2
x
)
d
y
d
x
.
(1+x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=(2-2x)\frac{\mathrm{d}y}{\mathrm{d}x}.
(
1
+
x
2
)
d
x
2
d
2
y
=
(
2
−
2
x
)
d
x
d
y
.
[4]
(b)
Find the first 4 terms of the Maclaurin expansion of
e
2
tan
−
1
x
.
{\mathrm{e}^{2 \tan^{-1} x}.}
e
2
t
a
n
−
1
x
.
[5]
Answer
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