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2020
P1 Q11
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Maxima
20 P1 Q11
2020 H2 Mathematics Paper 1 Question 11
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
tan
θ
=
4
x
x
2
+
4
a
+
a
2
{\tan \theta = \frac{4x}{x^2 + 4a + a^2}}
tan
θ
=
x
2
+
4
a
+
a
2
4
x
(ii)
x
=
4
a
+
a
2
{x = \sqrt{4a+a^2}}
x
=
4
a
+
a
2
tan
θ
=
2
4
a
+
a
2
{\tan \theta = \frac{2}{\sqrt{4a+a^2}}}
tan
θ
=
4
a
+
a
2
2
(iii)
The optimal point and optimal angle may not correspond to the highest chance of scoring
(iv)
∠
K
D
A
≈
π
4
rad
{\angle KDA \approx \frac{\pi}{4} \textrm{ rad}}
∠
KD
A
≈
4
π
rad
(v)
0.0801
≤
θ
<
π
2
{0.0801 \leq \theta < \frac{\pi}{2}}
0.0801
≤
θ
<
2
π
Full solutions
(i)
tan
θ
=
tan
(
∠
A
K
D
−
∠
A
K
C
)
=
tan
∠
A
K
D
−
tan
∠
A
K
C
1
+
tan
∠
A
K
D
⋅
tan
A
K
C
=
a
+
4
x
−
a
x
1
+
a
+
4
x
⋅
a
x
=
4
x
1
+
a
2
+
4
a
x
2
=
4
x
x
2
+
4
a
+
a
2
■
\begin{align*} \tan \theta &= \tan( \angle AKD - \angle AKC ) \\ &= \frac{\tan \angle AKD - \tan \angle AKC}{1 + \tan \angle AKD \cdot \tan AKC} \\ &= \frac{\frac{a+4}{x} - \frac{a}{x} }{ 1 + \frac{a+4}{x} \cdot \frac{a}{x} } \\ &= \frac{ \frac{4}{x} }{ 1 + \frac{a^2 + 4a }{x^2} } \\ &= \frac{4x}{x^2 + 4a + a^2} \; \blacksquare \end{align*}
tan
θ
=
tan
(
∠
A
KD
−
∠
A
K
C
)
=
1
+
tan
∠
A
KD
⋅
tan
A
K
C
tan
∠
A
KD
−
tan
∠
A
K
C
=
1
+
x
a
+
4
⋅
x
a
x
a
+
4
−
x
a
=
1
+
x
2
a
2
+
4
a
x
4
=
x
2
+
4
a
+
a
2
4
x
■
(ii)
Differentiating w.r.t.
x
,
{x,}
x
,
d
d
x
(
tan
θ
)
=
4
(
x
2
+
4
a
+
a
2
)
−
(
4
x
)
(
2
x
)
(
x
2
+
4
a
+
a
2
)
2
\frac{\mathrm{d}}{\mathrm{d}x} \left( \tan \theta \right) = \frac{4(x^2+4a+a^2) - (4x)(2x)}{(x^2+4a+a^2)^2}
d
x
d
(
tan
θ
)
=
(
x
2
+
4
a
+
a
2
)
2
4
(
x
2
+
4
a
+
a
2
)
−
(
4
x
)
(
2
x
)
At maximum
tan
θ
,
d
d
x
(
tan
θ
)
=
0
{\tan \theta, \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left( \tan \theta \right) = 0}
tan
θ
,
d
x
d
(
tan
θ
)
=
0
4
x
2
+
16
a
+
4
a
2
−
8
x
2
=
0
x
2
=
4
a
+
a
2
x
=
4
a
+
a
2
■
\begin{gather*} 4x^2 + 16 a + 4a^2 - 8x^2 = 0 \\ x^2 = 4a + a^2 \\ x = \sqrt{4a+a^2} \; \blacksquare \end{gather*}
4
x
2
+
16
a
+
4
a
2
−
8
x
2
=
0
x
2
=
4
a
+
a
2
x
=
4
a
+
a
2
■
tan
θ
=
4
4
a
+
a
2
4
a
+
a
2
+
4
a
+
a
2
=
2
4
a
+
a
2
4
a
+
a
2
=
2
4
a
+
a
2
■
\begin{align*} \tan \theta &= \frac{4 \sqrt{4 a + a^2} }{4a + a^2 + 4a + a^2} \\ &= \frac{2\sqrt{4a+a^2}}{4a+a^2} \\ &= \frac{2}{\sqrt{4a+a^2}} \; \blacksquare \end{align*}
tan
θ
=
4
a
+
a
2
+
4
a
+
a
2
4
4
a
+
a
2
=
4
a
+
a
2
2
4
a
+
a
2
=
4
a
+
a
2
2
■
(iii)
The optimal point and optimal angle may not correspond to the highest chance of scoring
■
{\blacksquare}
■
(iv)
tan
K
D
A
=
x
a
+
4
=
4
a
+
a
2
a
+
4
=
a
a
+
4
\begin{align*} \tan KDA &= \frac{x}{a+4} \\ &= \frac{\sqrt{4a+a^2}}{a+4} \\ &= \sqrt{\frac{a}{a+4}} \end{align*}
tan
KD
A
=
a
+
4
x
=
a
+
4
4
a
+
a
2
=
a
+
4
a
As
a
≫
4
,
a
4
+
a
≈
1
{a \gg 4, \frac{a}{4+a} \approx 1}
a
≫
4
,
4
+
a
a
≈
1
tan
∠
K
D
A
≈
1
∠
K
D
A
≈
π
4
rad
■
\begin{align*} \tan \angle KDA &\approx 1 \\ \angle KDA &\approx \frac{\pi}{4} \textrm{ rad} \; \blacksquare \end{align*}
tan
∠
KD
A
∠
KD
A
≈
1
≈
4
π
rad
■
(v)
Since
X
Y
=
50
,
{XY = 50, \;}
X
Y
=
50
,
C
D
=
4
{CD = 4}
C
D
=
4
and
X
C
=
D
Y
,
{XC = DY,}
XC
=
D
Y
,
0
<
a
≤
23
0 < a \leq 23
0
<
a
≤
23
0
<
4
a
≤
92
and
0
<
a
2
≤
529
\begin{align*} &0 < 4a \leq 92 \quad \textrm{ and} \\ &0 < a^2 \leq 529 \end{align*}
0
<
4
a
≤
92
and
0
<
a
2
≤
529
0
<
4
a
+
a
2
≤
621
0
<
4
a
+
a
2
≤
621
\begin{align*} & 0 < 4a + a^2 \leq 621 \\ & 0 < \sqrt{4a + a^2} \leq \sqrt{621} \end{align*}
0
<
4
a
+
a
2
≤
621
0
<
4
a
+
a
2
≤
621
1
4
a
+
a
2
≥
1
621
2
4
a
+
a
2
≥
2
621
tan
θ
≥
2
621
\begin{align*} \frac{1}{\sqrt{4a + a^2}} &\geq \frac{1}{\sqrt{621}} \\ \frac{2}{\sqrt{4a + a^2}} &\geq \frac{2}{\sqrt{621}} \\ \tan \theta &\geq \frac{2}{\sqrt{621}} \\ \end{align*}
4
a
+
a
2
1
4
a
+
a
2
2
tan
θ
≥
621
1
≥
621
2
≥
621
2
tan
−
1
2
621
≤
θ
<
π
2
0.0801
≤
θ
<
π
2
■
\begin{gather*} \tan^{-1} \frac{2}{\sqrt{621}} \leq \theta < \frac{\pi}{2} \\ 0.0801 \leq \theta < \frac{\pi}{2} \; \blacksquare \end{gather*}
tan
−
1
621
2
≤
θ
<
2
π
0.0801
≤
θ
<
2
π
■
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