Math Repository
about
topic
al
year
ly
Yearly
2009
P1 Q11
Topical
Maxima
09 P1 Q11
2009 H2 Mathematics Paper 1 Question 11
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
(ii)
(
1
2
,
1
2
e
)
,
{\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2\mathrm{e}}} \right), \;}
(
2
1
,
2
e
1
)
,
(
−
1
2
,
−
1
2
e
)
{\left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2\mathrm{e}}} \right)}
(
−
2
1
,
−
2
e
1
)
(iii)
1
2
(
1
−
e
−
n
2
)
{\frac{1}{2} \left(1 - \mathrm{e}^{-n^2}\right)}
2
1
(
1
−
e
−
n
2
)
Area of the region between the curve and the positive
x
-
{x\textrm{-}}
x
-
axis
=
1
2
units
2
{=\frac{1}{2} \textrm{ units}^2}
=
2
1
units
2
(iv)
1
−
e
−
4
{1 - \mathrm{e}^{-4}}
1
−
e
−
4
(v)
0.363
units
3
{0.363 \textrm{ units}^3}
0.363
units
3
Full solutions
(i)
(ii)
f
′
(
x
)
=
e
−
x
2
−
2
x
2
e
−
x
2
=
(
1
−
2
x
2
)
e
−
x
2
\begin{align*} f'(x) &= \mathrm{e}^{-x^2} - 2x^2 \mathrm{e}^{-x^2} \\ &= (1-2x^2) \mathrm{e}^{-x^2} \end{align*}
f
′
(
x
)
=
e
−
x
2
−
2
x
2
e
−
x
2
=
(
1
−
2
x
2
)
e
−
x
2
At turning points,
f
′
(
x
)
=
0
{f'(x) = 0}
f
′
(
x
)
=
0
(
1
−
2
x
2
)
e
−
x
2
=
0
1
−
2
x
2
=
0
\begin{gather*} (1-2x^2) \mathrm{e}^{-x^2} = 0 \\ 1-2x^2 = 0 \\ \end{gather*}
(
1
−
2
x
2
)
e
−
x
2
=
0
1
−
2
x
2
=
0
x
=
±
1
2
y
=
±
1
2
e
−
1
2
=
±
1
2
e
\begin{align*} x &= \pm \frac{1}{\sqrt{2}} \\ y &= \pm \frac{1}{\sqrt{2}} \mathrm{e}^{ - \frac{1}{2} } \\ &= \pm \frac{1}{\sqrt{2\mathrm{e}}} \\ \end{align*}
x
y
=
±
2
1
=
±
2
1
e
−
2
1
=
±
2
e
1
Coordinates of turning points:
(
1
2
,
1
2
e
)
■
(
−
1
2
,
−
1
2
e
)
■
\begin{align*} & \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2\mathrm{e}}} \right) \; \blacksquare \\ & \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2\mathrm{e}}} \right) \; \blacksquare \\ \end{align*}
(
2
1
,
2
e
1
)
■
(
−
2
1
,
−
2
e
1
)
■
(iii)
d
u
d
x
=
2
x
\frac{\mathrm{d}u}{\mathrm{d}x} = 2x
d
x
d
u
=
2
x
When
x
=
0
,
{x=0, \;}
x
=
0
,
u
=
0
{u=0}
u
=
0
When
x
=
n
,
{x=n, \;}
x
=
n
,
u
=
n
2
{u=n^2}
u
=
n
2
∫
0
n
f
(
x
)
d
x
=
∫
0
n
x
e
−
x
2
d
x
=
∫
0
n
2
x
e
−
u
1
2
x
d
u
=
1
2
∫
0
n
2
e
−
u
d
u
=
1
2
[
−
e
−
u
]
0
n
2
=
1
2
(
−
e
−
n
2
−
(
−
e
−
0
)
)
=
1
2
(
1
−
e
−
n
2
)
■
\begin{align*} & \int_0^n f(x) \; \mathrm{d}x \\ & = \int_0^n x \mathrm{e}^{-x^2} \; \mathrm{d}x \\ & = \int_0^{n^2} x \mathrm{e}^{-u} \frac{1}{2x} \; \mathrm{d}u \\ & = \frac{1}{2} \int_0^{n^2} \mathrm{e}^{-u} \; \mathrm{d}u \\ & = \frac{1}{2} \left[ - \mathrm{e}^{-u} \right]_0^{n^2} \\ & = \frac{1}{2} \left(- \mathrm{e}^{-n^2} - (-\mathrm{e}^{-0}) \right) \\ & = \frac{1}{2} \left(1 - \mathrm{e}^{-n^2}\right) \; \blacksquare \end{align*}
∫
0
n
f
(
x
)
d
x
=
∫
0
n
x
e
−
x
2
d
x
=
∫
0
n
2
x
e
−
u
2
x
1
d
u
=
2
1
∫
0
n
2
e
−
u
d
u
=
2
1
[
−
e
−
u
]
0
n
2
=
2
1
(
−
e
−
n
2
−
(
−
e
−
0
)
)
=
2
1
(
1
−
e
−
n
2
)
■
As
n
→
∞
,
e
−
n
2
→
0
{n \to \infty, \; \mathrm{e}^{-n^2} \to 0}
n
→
∞
,
e
−
n
2
→
0
1
2
(
1
−
e
−
n
2
)
→
1
2
\frac{1}{2} \left(1 - \mathrm{e}^{-n^2}\right) \to \frac{1}{2}
2
1
(
1
−
e
−
n
2
)
→
2
1
Hence the area of the region between the curve and the positive
x
-
{x\textrm{-}}
x
-
axis
=
1
2
units
2
■
{=\frac{1}{2} \textrm{ units}^2 \; \blacksquare}
=
2
1
units
2
■
(iv)
∫
−
2
2
∣
f
(
x
)
∣
d
x
=
2
∫
0
2
f
(
x
)
d
x
=
2
(
1
2
)
(
1
−
e
−
2
2
)
=
1
−
e
−
4
■
\begin{align*} & \int_{-2}^2 \left| f(x) \right| \; \mathrm{d}x \\ & = 2 \int_0^2 f(x) \; \mathrm{d}x \\ & = 2 \left( \frac{1}{2} \right) \left( 1 - \mathrm{e}^{-2^2} \right) \\ & = 1 - \mathrm{e}^{-4} \; \blacksquare \end{align*}
∫
−
2
2
∣
f
(
x
)
∣
d
x
=
2
∫
0
2
f
(
x
)
d
x
=
2
(
2
1
)
(
1
−
e
−
2
2
)
=
1
−
e
−
4
■
(v)
Volume of revolution
=
π
∫
0
1
(
x
e
−
x
2
)
2
d
x
=
0.363
(3 sf)
units
3
■
\begin{align*} & \textrm{Volume of revolution} \\ & = \pi \int_0^1 \left( x \mathrm{e}^{-x^2} \right)^2 \; \mathrm{d}x \\ & = 0.363 \textrm{ (3 sf)} \textrm{ units}^3 \; \blacksquare \end{align*}
Volume of revolution
=
π
∫
0
1
(
x
e
−
x
2
)
2
d
x
=
0.363
(3 sf)
units
3
■
Back to top ▲