Question 7
Solution
(a) Expand $ \left(1 + \frac{x}{3}\right)^{-3} $ up to $ x^3 $
We use the Binomial Expansion for $ (1 + u)^n $, valid for $ |u| \lt 1 $:
$ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots $
Here, $ u = \frac{x}{3} $ and $ n = -3 $. Substituting, we get:
$ \left(1 + \frac{x}{3}\right)^{-3} = 1 + (-3)\left(\frac{x}{3}\right) + \frac{(-3)(-4)}{2!}\left(\frac{x}{3}\right)^2 + \frac{(-3)(-4)(-5)}{3!}\left(\frac{x}{3}\right)^3 + \dots $
Simplify term by term:
$\text{First term: } 1.$
$\text{Second term: } -3 \cdot \frac{x}{3} = -x.$
$\text{Third term: } \frac{(-3)(-4)}{2} \cdot \frac{x^2}{9} = \frac{12}{2} \cdot \frac{x^2}{9} = \frac{6x^2}{9} = \frac{2x^2}{3}.$
$\text{Fourth term: } \frac{(-3)(-4)(-5)}{6} \cdot \frac{x^3}{27} = \frac{-60}{6} \cdot \frac{x^3}{27} = -10 \cdot \frac{x^3}{27} = -\frac{10x^3}{27}.$
Thus, the expansion is:
$ \left(1 + \frac{x}{3}\right)^{-3} = 1 - x + \frac{2x^2}{3} - \frac{10x^3}{27}. $
(b) Range of validity
The Binomial Expansion is valid when $ \left| \frac{x}{3} \right| \lt 1 $, i.e.:
$ |x| \lt 3. $
Thus, the range of validity is:
$ \boxed{|x| \lt 3}. $
(c) Express $ (3 + x)^{-3} $ in the form $ P(1 + Qx)^{-3} $
Factor $ 3 $ from $ (3 + x) $:
$ (3 + x)^{-3} = 3^{-3} \left(1 + \frac{x}{3}\right)^{-3}. $
Here, $ 3^{-3} = \frac{1}{27} $, so:
$ (3 + x)^{-3} = \frac{1}{27} \left(1 + \frac{x}{3}\right)^{-3}. $
Thus, $ P = \frac{1}{27} $ and $ Q = \frac{1}{3} $, and the expression is:
$ \boxed{(3 + x)^{-3} = \frac{1}{27}(1 + \frac{x}{3})^{-3}.}$
(d) Obtain the series expansion for $ f(x) $ up to $ x^2 $
The function is:
$ f(x) = \frac{1 + 4x}{(3 + x)^3}. $
Substitute the expansion of $ (3 + x)^{-3} $ from part (c):
$ (3 + x)^{-3} = \frac{1}{27}\left(1 - x + \frac{2x^2}{3}\right). $
Thus:
$ f(x) = (1 + 4x) \cdot \frac{1}{27}\left(1 - x + \frac{2x^2}{3}\right). $
Expand:
$ f(x) = \frac{1}{27} \left[ (1)(1 - x + \frac{2x^2}{3}) + (4x)(1 - x + \frac{2x^2}{3}) \right]. $
Simplify:
$ f(x) = \frac{1}{27} \left[ 1 - x + \frac{2x^2}{3} + 4x - 4x^2 + \frac{8x^3}{3} \right]. $
$ f(x) = \frac{1}{27} \left[ 1 + 3x - \frac{10x^2}{3} \right]. $
Thus:
$ f(x) = \frac{1}{27} + \frac{x}{9} - \frac{10x^2}{81}. $
(e) Estimate $ \int_0^{0.2} f(x)\,dx $
Using the expansion for $ f(x) $:
$ f(x) = \frac{1}{27} + \frac{x}{9} - \frac{10x^2}{81}. $
Integrate term by term:
$ \int f(x)\,dx = \frac{x}{27} + \frac{x^2}{18} - \frac{10x^3}{243}. $
Evaluate from $ 0 $ to $ 0.2 $:
$ \int_0^{0.2} f(x)\,dx = \left[\frac{x}{27} + \frac{x^2}{18} - \frac{10x^3}{243}\right]_0^{0.2}. $
Substitute $ x = 0.2 $:
$ = \frac{0.2}{27} + \frac{0.04}{18} - \frac{0.08}{243}. $
Simplify:
$ = 0.0074074 + 0.0022222 - 0.0003292. $
$ \approx 0.0093004 \text{ (to 5 significant figures)}. $
Final answer:
$ \boxed{0.0093004}. $
Question 8
![The points $A$ and $B$ have coordinates $(-6,8)$ and $(12,2)$ respectively. (a) Find an equation of the straight line passing through $A$ and $B$ in the form $a x+b y+c=0$, where $a, b$ and $c$ are integers to be found. [3 Marks] (b) Find the exact length of $A B$ The point $X$ with coordinates $(m, n)$ lies on $A B$ such that $A X: X B=1: [2 Marks]$ (c) Find the value of $m$ and the value of $n$ [2 Marks] The line $L$ passes through the point $X$ and is perpendicular to $A B$ The point $C$ with coordinates $(p, q)$ lies on $L$ where $p>0$ and $q>0$ Given that $A B$ is a diameter of a circle and $C$ also lies on the circumference of the circle, (d) find (i) the exact value of $p$ (ii) the exact value of $q$ [7 Marks] (e) Find the exact area of triangle $A B C$ [3 Marks]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqQyN2YdYWoXDiO_SCRUZXZvEMqaEHxvlSlp4ZLjD9BG3SHDrREK-xcerbigisZ0s-AMif5k_IAQmZvlygtWAJntvFnfRNWgXY0OfTnDxlwaOXsET0JsykiMiFbH5bcM2yUEGEoo0y4HYpoKIeTTM4TXDQWMr9bLjjHkbm_L7Mk3mpDRfYTsfiq20jAsi_/s1600/q8a.png){kind=link}
Solution
The points $ A $ and $ B $ have coordinates $ (-6, 8) $ and $ (12, 2) $ respectively.
(a) Equation of the line passing through $ A $ and $ B $
The slope of the line $ AB $ is:
$ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 8}{12 - (-6)} = \frac{-6}{18} = -\frac{1}{3}. $
The equation of a line in slope-intercept form is:
$ y - y_1 = m(x - x_1), $
Using point $ A(-6, 8) $:
$ y - 8 = -\frac{1}{3}(x + 6). $
Simplify:
$ y - 8 = -\frac{1}{3}x - 2 \implies y = -\frac{1}{3}x + 6. $
Rearranging to the form $ ax + by + c = 0 $:
$ x + 3y - 18 = 0. $
Thus, the equation of the line is:
$ \boxed{x + 3y - 18 = 0}. $
(b) Length of $ AB $
The length of a line segment between two points is:
$ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. $
Substitute $ A(-6, 8) $ and $ B(12, 2) $:
$ AB = \sqrt{(12 - (-6))^2 + (2 - 8)^2} = \sqrt{18^2 + (-6)^2} = \sqrt{324 + 36} = \sqrt{360}. $
Simplify:
$ AB = 6\sqrt{10}. $
Thus, the length of $ AB $ is:
$ \boxed{6\sqrt{10}}. $
(c) Coordinates of $ X $ dividing $ AB $ in the ratio $ 1:2 $
The coordinates of $ X $ are given by:
$ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}. $
Substitute $ m_1 = 1, m_2 = 2 $:
$ x = \frac{(1)(12) + (2)(-6)}{3} = 0, \quad y = \frac{(1)(2) + (2)(8)}{3} = 6. $
Thus, the coordinates of $ X $ are:
$ \boxed{(0, 6)}. $
(d) Line $ L $ perpendicular to $ AB $ passing through $ X(0, 6) $
The slope of a perpendicular line is $ 3 $.
The equation of $ L $ is:
$ y - 6 = 3(x - 0) \implies y = 3x + 6. $
$ \boxed{y = 3x + 6}. $
(d)(i) Coordinates of $ C $
The midpoint of $ AB $ is:
$ \left(\frac{-6 + 12}{2}, \frac{8 + 2}{2}\right) = (3, 5). $
The radius is $ 3\sqrt{10} $.
The equation of the circle is:
$ (x - 3)^2 + (y - 5)^2 = 90. $
Substitute $ y = 3x + 6 $:
$ (x - 3)^2 + (3x + 1)^2 = 90. $
Simplify:
$ (x^2 - 6x + 9) + (9x^2 + 6x + 1) = 90 \implies 10x^2 + 10 = 90 \implies x^2 = 8. $
$ x = 2\sqrt{2}. $
Substitute into $ y = 3x + 6 $:
$ y = 6\sqrt{2} + 6. $
Thus:
$ \boxed{(2\sqrt{2}, 6 + 6\sqrt{2})}. $
(e) Area of triangle $ ABC $
The area formula is:
$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. $
Substitute values:
$ \frac{1}{2} \left| -6(2 - (6 + 6\sqrt{2})) + 12((6 + 6\sqrt{2}) - 8) + 2\sqrt{2}(8 - 2) \right|. $
Simplify:
$ \frac{1}{2} \left| 24 + 36\sqrt{2} - 24 + 72\sqrt{2} + 12\sqrt{2} \right|. $
$ = \frac{1}{2} \cdot 120\sqrt{2} = 60\sqrt{2}. $
Final answer:
$ \boxed{60\sqrt{2}}. $
Question 9
Figure 4 shows a right pyramid with vertex $V$ and base $A B C D E$ which is a 
regular pentagon. $$ \begin{aligned} A B & =B C=C D=D E=E A=2 x \mathrm{~cm} \\ V A & =V B=V C=V D=V E=3 x \mathrm{~cm} \end{aligned} $$ Find, in degrees to one decimal place, the size of the angle between the plane $V B C$ and the base $A B C D E$ [6 Marks]
Solution
First Method
Let $M$ be the midpoint of $BC$, and $O$ be the centre of the regular pentagon.
$ \angle BOC = \frac{360^\circ}{5} = 72^\circ $
Thus,
$ \angle OBM = 180^\circ - 90^\circ - 36^\circ = 54^\circ $
Since
$ BM = \frac{1}{2}BC = \frac{1}{2}(2x) = x, $
$ OM = BM \tan 54^\circ = x \tan 54^\circ $
In $ \triangle VMC $,
$ VM^2 = VC^2 - MC^2 = (3x)^2 - x^2 = 8x^2 $
$ VM = \sqrt{8}\,x $
In $ \triangle VMO $,
$ \cos \angle VMO = \frac{OM}{VM} = \frac{x \tan 54^\circ}{\sqrt{8}\,x} = \frac{\tan 54^\circ}{\sqrt{8}} $
$ = 0.4866 $
$ \angle VMO = 60.88^\circ \approx 60.9^\circ $
Second Method
Let $M$ be midpoint of $BC$.
Step 1: Find $ BE $ in $ \triangle ABE $
Using the cosine rule in $ \triangle ABE $, we have:
$ BE^2 = AB^2 + AE^2 - 2 \cdot AB \cdot AE \cdot \cos(\angle BAE). $
Substituting $ AB = AE = 2x $ and $ \cos(108^\circ) = -\cos(72^\circ) $ where $ \cos(72^\circ) = \frac{\sqrt{5} - 1}{4} $:
$ BE^2 = (2x)^2 + (2x)^2 - 2 \cdot (2x) \cdot (2x) \cdot \left(-\frac{\sqrt{5} - 1}{4}\right) $
$ BE^2 = 4x^2 + 4x^2 + 8x^2 \cdot \frac{\sqrt{5} - 1}{4} $
$ BE^2 = 8x^2 + 2x^2(\sqrt{5} - 1) $
$ BE^2 = 2x^2(3 + \sqrt{5}) $
Thus:
$ BE = x \sqrt{2(3 + \sqrt{5})} $
Step 2: Find $ ME $ in $ \triangle BEM $
Since $ M $ is the midpoint of $ BC $, $ BM = x $. Using Pythagoras:
$ ME^2 = BE^2 - BM^2 $
$ ME^2 = 2x^2(3 + \sqrt{5}) - x^2 $
$ ME^2 = x^2(2(3 + \sqrt{5}) - 1) $
$ ME^2 = x^2(6 + 2\sqrt{5} - 1) $
$ ME^2 = x^2(5 + 2\sqrt{5}) $
Thus:
$ ME = x \sqrt{5 + 2\sqrt{5}} $
Step 3: Find $ VM $ in $ \triangle VCM $
Using $ VM^2 = (3x)^2 - x^2 $:
$ VM^2 = 9x^2 - x^2 $
$ VM^2 = 8x^2 $
$ VM = 2\sqrt{2}x $
Step 4: Find $ \cos(\angle VME) $ Using Cosine Rule
$ \cos(\angle VME) = \frac{ME^2 + VM^2 - VE^2}{2 \cdot ME \cdot VM} $
Substitute values:
$ \cos(\angle VME) = \frac{x^2(5 + 2\sqrt{5}) + 8x^2 - 9x^2} {2 \cdot x \sqrt{5 + 2\sqrt{5}} \cdot 2\sqrt{2}x} $
Simplify the numerator:
$ \text{Numerator} = x^2(5 + 2\sqrt{5}) + 8x^2 - 9x^2 $
$ = x^2(4 + 2\sqrt{5}) $
Simplify the denominator:
$ \text{Denominator} = 4x^2 \sqrt{2} \sqrt{5 + 2\sqrt{5}} $
Combine:
$ \cos(\angle VME) = \frac{4 + 2\sqrt{5}}{4 \sqrt{2} \sqrt{5 + 2\sqrt{5}}} $
$ = \frac{2 + \sqrt{5}}{2 \sqrt{2} \sqrt{5 + 2\sqrt{5}}} = 60.88 $
Final Expression for $ \angle VME $:
The angle is $ \boxed{60.9^\circ} $
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