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Specimen 2017 H2 Mathematics Paper 1 Question 5
Differentiation II: Maxima, Minima, Rates of Change
Definite Integrals: Areas and Volumes
Answers
(a)
Maximum point at
x
=
e
3
.
{x=\mathrm{e}^3.}
x
=
e
3
.
(b)
20
units
2
.
{20 \textrm{ units}^2.}
20
units
2
.
Full solutions
(a)
d
y
d
x
=
3
(
ln
x
)
2
−
(
ln
x
)
3
x
2
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 \left( \ln x \right)^2 - \left( \ln x \right)^3}{x^2}
d
x
d
y
=
x
2
3
(
ln
x
)
2
−
(
ln
x
)
3
At turning point,
d
y
d
x
=
0
{\frac{\mathrm{d}y}{\mathrm{d}x} = 0}
d
x
d
y
=
0
3
(
ln
x
)
2
−
(
ln
x
)
3
=
0
(
ln
x
)
2
(
3
−
ln
x
)
=
0
\begin{gather*} 3 \left( \ln x \right)^2 - \left( \ln x \right)^3 = 0 \\ \left( \ln x \right)^2 ( 3 - \ln x ) = 0 \end{gather*}
3
(
ln
x
)
2
−
(
ln
x
)
3
=
0
(
ln
x
)
2
(
3
−
ln
x
)
=
0
Since
x
>
1
,
{x > 1,}
x
>
1
,
3
−
ln
x
=
0
x
=
e
3
■
\begin{gather*} 3 - \ln x = 0 \\ x = \mathrm{e}^3 \; \blacksquare \end{gather*}
3
−
ln
x
=
0
x
=
e
3
■
When
x
<
e
3
,
{x < \mathrm{e}^3, }
x
<
e
3
,
d
y
d
x
>
0
{\frac{\mathrm{d}y}{\mathrm{d}x} > 0}
d
x
d
y
>
0
When
x
>
e
3
,
{x > \mathrm{e}^3, }
x
>
e
3
,
d
y
d
x
<
0
{\frac{\mathrm{d}y}{\mathrm{d}x} < 0}
d
x
d
y
<
0
Hence it is a maximum point when
x
=
e
3
■
{x=\mathrm{e}^3 \; \blacksquare}
x
=
e
3
■
(b)
Exact area of region
=
∫
e
e
3
1
x
(
ln
x
)
3
d
x
=
[
1
4
(
ln
x
)
4
]
e
e
3
=
(
ln
e
3
)
4
−
(
ln
e
)
4
4
=
20
units
2
■
\begin{align*} & \textrm{Exact area of region} \\ & = \int_{\mathrm{e}}^{\mathrm{e}^3} \frac{1}{x} \left( \ln x \right)^3 \; \mathrm{d}x \\ & = \left[ \frac{1}{4} \left( \ln x \right)^4 \right]_{\mathrm{e}}^{\mathrm{e}^3} \\ & = \frac{(\ln \mathrm{e}^3)^4 - (\ln \mathrm{e})^4}{4} \\ & = 20 \textrm{ units}^2 \; \blacksquare \end{align*}
Exact area of region
=
∫
e
e
3
x
1
(
ln
x
)
3
d
x
=
[
4
1
(
ln
x
)
4
]
e
e
3
=
4
(
ln
e
3
)
4
−
(
ln
e
)
4
=
20
units
2
■
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