Question 1
Given that $\frac{a+2 \sqrt{5}}{3-\sqrt{5}}=\frac{11+b \sqrt{5}}{2}$ where $a$ is an integer and $b$ is prime, find the value of $a$ and the value of $b$Show your working clearly.
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Solution:
$\text{Given that } \frac{a+2 \sqrt{5}}{3-\sqrt{5}} = \frac{11+b\sqrt{5}}{2},$
$\text{where } a \text{ is an integer and } b \text{ is a prime, we solve as follows:}$
Step 1: Rationalize the denominator on the left-hand side
$ \frac{a+2\sqrt{5}}{3-\sqrt{5}} = \frac{(a+2\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}. $
$ (3-\sqrt{5})(3+\sqrt{5}) = 9-5 = 4, $
$ (a+2\sqrt{5})(3+\sqrt{5}) = 3a + a\sqrt{5} + 6\sqrt{5} + 10 = 3a+10+(a+6)\sqrt{5}. $
$ \text{Therefore } \frac{a+2\sqrt{5}}{3-\sqrt{5}} = \frac{3a+10+(a+6)\sqrt{5}}{4}. $
Step 2: Equate the two sides
$ \frac{3a+10+(a+6)\sqrt{5}}{4} = \frac{11+b\sqrt{5}}{2}. $
$ 3a+10+(a+6)\sqrt{5} = 22+2b\sqrt{5}. $
Step 3: Separate into rational and irrational parts
$ \text{Rational part: } 3a+10 = 22, \quad \text{Irrational part: } (a+6)\sqrt{5} = 2b\sqrt{5}. $
Step 4: Solve the rational part
$ 3a+10 = 22 \implies 3a = 12 \implies a = 4. $
Step 5: Solve the irrational part
$ a+6 = 2b \implies 4+6 = 2b \implies 10 = 2b \implies b = 5. $
Final Answer:
$ a = 4, \quad b = 5. $
Question 2
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Solution:
$\text{The } n\text{th term of a geometric series is given by } a_n = 8^{(1-2n)}.$
Step 1: General form of a geometric series
$ a_n = a r^{n-1}, $
$\text{where } a \text{ is the first term and } r \text{ is the common ratio.}$
Step 2: Find $a$ and $r$
$\text{For } n = 1, \text{ the first term is:}$
$ a_1 = 8^{(1-2\cdot 1)} = 8^{-1} = \frac{1}{8}. $
$\text{Thus, } a = \frac{1}{8}.$
$\text{The common ratio is:}$
$ r = \frac{a_{n+1}}{a_n} = \frac{8^{(1-2(n+1))}}{8^{(1-2n)}} = 8^{-2}. $
$ r = \frac{1}{64}. $
Step 3: Sum to infinity formula
$\text{The sum to infinity of a geometric series is:}$
$ S_\infty = \frac{a}{1-r}, \quad \text{where } |r| \lt 1. $
$\text{Substitute } a = \frac{1}{8} \text{ and } r = \frac{1}{64}:$
$ S_\infty = \frac{\frac{1}{8}}{1 - \frac{1}{64}}. $
Step 4: Simplify
$\text{The denominator is:}$
$ 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64}. $
$\text{Thus:}$
$ S_\infty = \frac{\frac{1}{8}}{\frac{63}{64}} = \frac{1}{8} \cdot \frac{64}{63} = \frac{64}{504}. $
$\text{Simplify:}$
$ S_\infty = \frac{8}{63}. $
Final Answer:
$ S_\infty = \frac{8}{63}. $
Question 3
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Solution:
Step 1: Formula for the length of an arc
$\text{Let } \theta_1 \text{ be the central angle of major sector } APB.$
$ \text{Arc length} = r \theta_1. $
$\text{Substitute the known values:}$
$ 53.2 = r \theta_1. \quad \text{(1)} $
Step 2: Formula for the area of a sector
$ \text{Sector area} = \frac{1}{2} r^2 \theta_1. $
$\text{Substitute the known values:}$
$ 372.4 = \frac{1}{2} r^2 \theta_1. \quad \text{(2)} $
Step 3: Solve for $r$ and $\theta$
$\text{From equation (1), express } \theta_1 \text{ in terms of } r:$
$ \theta_1 = \frac{53.2}{r}. \quad \text{(3)} $
$\text{Substitute equation (3) into equation (2):}$
$ 372.4 = \frac{1}{2} r^2 \left( \frac{53.2}{r} \right). $
$\text{Simplify:}$
$ 372.4 = \frac{1}{2} r \cdot 53.2. $
$ 372.4 = 26.6r. $
$\text{Solve for } r:$
$ r = \frac{372.4}{26.6} \approx 14.0 \,\mathrm{cm}. $
Step 4: Find $\theta_1$
$\text{Substitute } r = 14.0 \text{ into equation (3):}$
$ \theta_1 = \frac{53.2}{14.0} \approx 3.80 \,\mathrm{radians}. $
Final Answer
- $(i)\; r = 14.0 \,\mathrm{cm},$
- $(ii)\; \theta = 2\pi - 3.80 = 2.48 \,\mathrm{radians}.$
Question 4
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Solution:
$\text{The curve } S \text{ is given by the equation } y = \frac{x^2}{4} + 2 \text{ where } x \geq 0, \text{ and the line } l \text{ is given by } 2y - x - 4 = 0 \text{ where } x \geq 0. \text{ These intersect at points } A \text{ and } B.$
(a) (i) Coordinates of Point $A$
$\text{To find the intersection points of } S \text{ and } l, \text{ solve their equations simultaneously:}$
$y = \frac{x^2}{4} + 2 \quad \text{and} \quad 2y - x - 4 = 0.$
$\text{Substitute } y = \frac{x^2}{4} + 2 \text{ into } 2y - x - 4 = 0:$
$2\left(\frac{x^2}{4} + 2\right) - x - 4 = 0.$
$\text{Simplify:}$
$\frac{x^2}{2} + 4 - x - 4 = 0 \implies \frac{x^2}{2} - x = 0.$
$\text{Factorise:}$
$\frac{x}{2}(x - 2) = 0.$
$\text{This gives } x = 0 \text{ or } x = 2.$
$\text{For } x = 0, \text{ substitute into } y = \frac{x^2}{4} + 2:$
$y = \frac{0^2}{4} + 2 = 2.$
$\text{Thus, } A = (0, 2).$
(a) (ii) Coordinates of Point $B$
$\text{For } x = 2, \text{ substitute into } y = \frac{x^2}{4} + 2:$
$y = \frac{2^2}{4} + 2 = \frac{4}{4} + 2 = 1 + 2 = 3.$
$\text{Thus, } B = (2, 3).$
(b) Volume of the Solid Generated
$\text{The volume of the solid generated when the region between } S \text{ and } l \text{ is rotated about the } y\text{-axis is given by the formula:}$
$V = \pi \int_{y_1}^{y_2} \left( R(y)^2 - r(y)^2 \right) \, dy,$
$\text{where } R(y) \text{ and } r(y) \text{ are the outer and inner radii as functions of } y, \text{ and } y_1 \text{ and } y_2 \text{ are the bounds of } y.$
$\text{From the equations:}$
- $\text{The line } l \text{ can be rearranged as } x = 2y - 4, \text{ giving } r(y) = 2y - 4.$
- $\text{The curve } S \text{ is rearranged as } x = \sqrt{4(y - 2)}, \text{ giving } R(y) = \sqrt{4(y - 2)}.$
- $y_1 = 2 \text{ and } y_2 = 3.$
$\text{Substitute into the volume formula:}$
$V = \pi \int_{2}^{3} \left[ \left(\sqrt{4(y - 2)}\right)^2 - (2y - 4)^2 \right] \, dy.$
$\text{Simplify:}$
$V = \pi \int_{2}^{3} \left[ 4(y - 2) - (2y - 4)^2 \right] \, dy.$
$\text{Expand } (2y - 4)^2:$
$(2y - 4)^2 = 4y^2 - 16y + 16.$
$\text{Thus:}$
$V = -\pi \int_{2}^{3} \left( 4y^2 - 16y + 16 - 4y + 8 \right) \, dy.$
$\text{Combine terms:}$
$V = -\pi \int_{2}^{3} \left( 4y^2 - 20y + 24 \right) \, dy.$
$\text{Integrate term by term:}$
$\int 4y^2 \, dy = \frac{4y^3}{3}, \quad \int -20y \, dy = -10y^2, \quad \int 24 \, dy = 24y.$
$\text{Thus:}$
$V = -\pi \left[ \frac{4y^3}{3} - 10y^2 + 24y \right]_{2}^{3}.$
$\text{Evaluate at the bounds:}$
$\text{At } y = 3: \quad \frac{4(3)^3}{3} - 10(3)^2 + 24(3) = \frac{108}{3} - 90 + 72 = 36 - 90 + 72 = 18.$
$\text{At } y = 2: \quad \frac{4(2)^3}{3} - 10(2)^2 + 24(2) = \frac{32}{3} - 40 + 48 = \frac{32}{3} + 8 = \frac{56}{3}.$
$\text{Subtract:}$
$V = -\pi \left( 18 - \frac{56}{3} \right) = \pi \left( \frac{54}{3} - \frac{56}{3} \right) = \pi \left( \frac{2}{3} \right).$
$\text{Thus:}$
$V = \frac{2\pi}{3}.$
Question 5
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Solution:
(a) Draw the lines
$\text{The equations of the lines to be drawn are:}$
- $y = 2x + 5$
- $4y = x - 8,\; y = \frac{x - 8}{4}$
- $5y + 3x = 30,\; y = \frac{30 - 3x}{5}$
(b) Shade the region $R$
$y \leq 2x + 5, \quad 4y \geq x - 8, \quad 5y + 3x \leq 30.$
$\text{Shade below } y = 2x + 5,\; \text{above } y = \frac{x - 8}{4},\; \text{and below } y = \frac{30 - 3x}{5}.$
(c) Find the least value of $P = 2x - 5y$
$\text{Intersection of } y = 2x + 5 \text{ and } y = \frac{x - 8}{4}:$
$2x + 5 = \frac{x - 8}{4} \implies 8x + 20 = x - 8 \implies 7x = -28 \implies x = -4,\; y = -3.$
$\text{Intersection of } y = 2x + 5 \text{ and } y = \frac{30 - 3x}{5}:$
$2x + 5 = \frac{30 - 3x}{5} \implies 10x + 25 = 30 - 3x \implies 13x = 5$
$\implies x = \frac{5}{13},\; y = 2\left(\frac{5}{13}\right) + 5 = \frac{75}{13}.$
$\text{Intersection of } y = \frac{x - 8}{4} \text{ and } y = \frac{30 - 3x}{5}:$
$\frac{x - 8}{4} = \frac{30 - 3x}{5} \implies 5(x - 8) = 4(30 - 3x) \implies 5x - 40 = 120 - 12x$
$\implies 17x = 160 \implies x = \frac{160}{17},\; y = \frac{6}{17}.$
Evaluate $P = 2x - 5y$:
$P = 2(-4) - 5(-3) = 7.$
$P = 2\left(\frac{5}{13}\right) - 5\left(\frac{75}{13}\right) = \frac{-365}{13} \approx -28.07.$
$P = 2\left(\frac{160}{17}\right) - 5\left(\frac{6}{17}\right) = \frac{290}{17} \approx 17.05.$
Final Answer:
$P_{\min} = -28.1 \text{ at } \left(\frac{5}{13}, \frac{75}{13}\right).$
Question 6
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Solution:
(a) Show that the volume of the prism is $ \frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3 $
$\text{The cross-section } ABC \text{ is an isosceles triangle with:}$
$AB = AC = r \, \mathrm{cm}, \quad \angle CAB = \frac{\pi}{3} \, \text{radians}.$
$\text{The area of triangle } ABC \text{ is given by:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle CAB).$
$\text{Substitute the values:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot r \cdot r \cdot \sin\left(\frac{\pi}{3}\right).$
$\text{Since } \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \text{ the area becomes:}$
$\text{Area of } \triangle ABC = \frac{1}{2} \cdot r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^2 \, \mathrm{cm}^2.$
$\text{The volume of the prism is:}$
$\text{Volume} = \text{Area of cross-section } \times \text{height}.$
$\text{The height of the prism is } AE = 5 \, \mathrm{cm}, \text{ so:}$
$\text{Volume} = \frac{\sqrt{3}}{4} r^2 \cdot 5 = \frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3.$
$\text{Thus, the volume of the prism is:}$
$\boxed{\frac{5\sqrt{3}}{4} r^2 \, \mathrm{cm}^3}.$
(b) Rate of increase of the volume of the prism
$\text{The volume of the prism is:}$
$V = \frac{5\sqrt{3}}{4} r^2.$
$\text{Differentiate } V \text{ with respect to time } t:$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{4} \cdot 2r \cdot \frac{dr}{dt}.$
$\text{Simplify:}$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} r \cdot \frac{dr}{dt}.$
$\text{We are given that:}$
$\frac{dr}{dt} = 0.2 \, \mathrm{cm/s}.$
$\text{When the area of the rectangular face } BCDF \text{ is } 60 \, \mathrm{cm}^2,$
$\text{Area of } BCDF = BC \cdot AE = r \cdot 5.$
$\text{Substitute } 60:$
$r \cdot 5 = 60 \implies r = 12 \, \mathrm{cm}.$
$\text{Substitute into } \frac{dV}{dt}:$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 12 \cdot 0.2.$
$\text{Simplify:}$
$\frac{dV}{dt} = \frac{5\sqrt{3}}{2} \cdot 2.4 = 6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}.$
$\text{Thus, the rate of increase is:}$
$\boxed{6\sqrt{3} \, \mathrm{cm}^3/\mathrm{s}}.$
Question 7
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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.Click to view the solution
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)
regular pentagon. $$ \begin{aligned} A B & =B C=C D=D E=E A=2 x \mathrm{~cm} \\ V A & =V B=V C=V D=V E=3 x \mathrm{~cm} \end{aligned} $$
Click to view the solution
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} $
Question 10
The curve $C$ with equation $y=\frac{6-3x}{x-4}$ where $x \neq 4$, crosses the $x$-axis at the point $P$ and the $y$-axis at the point $Q$(a) Find the coordinates of
(i) $P$
(ii) $Q$
(b) Write down an equation of the asymptote to $C$ which is
(i) parallel to the $y$-axis
(ii) parallel to the $x$-axis
(c) Sketch $C$ showing clearly the asymptotes and the coordinates of the points $P$ and $Q$
The line $L$ is the normal to $C$ at the point on $C$ where $x=2$
(d) Find an equation of $L$
The line $L$ intersects $C$ again at the point $R$
(e) Find the $x$ coordinate of $R$
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Solution:
The given curve $C$ has the equation: $$ y=\frac{6-3x}{x-4}, \quad x \neq 4. $$(a) Coordinates of Points
(i) Point $P$ (Where $C$ crosses the $x$-axis)
At the $x$-axis, $y=0$. Substituting $y=0$ into the equation: $$ 0=\frac{6-3x}{x-4}. $$ For the fraction to be zero, the numerator must be zero: $$ 6-3x=0 \implies x=2. $$ Thus, the coordinates of $P$ are: $$ P=(2,0). $$(ii) Point $Q$ (Where $C$ crosses the $y$-axis)
At the $y$-axis, $x=0$. Substituting $x=0$ into the equation: $$ y=\frac{6-3(0)}{0-4}=\frac{6}{-4}=-\frac{3}{2}. $$ Thus, the coordinates of $Q$ are: $$ Q=\left(0,-\frac{3}{2}\right). $$(b) Asymptotes of $C$
(i) Asymptote Parallel to the $y$-axis
The denominator of the equation $$ y=\frac{6-3x}{x-4} $$ becomes zero when $$ x-4=0, $$ i.e., $$ x=4. $$ Thus, the vertical asymptote is: $$ x=4. $$(ii) Asymptote Parallel to the $x$-axis
As $x \to \infty$ or $x \to -\infty$, the dominant term in the numerator and denominator is $-3x$ and $x$, respectively. Simplifying: $$ y \to -3. $$ Thus, the horizontal asymptote is: $$ y=-3. $$(c) Sketch of the Curve $C$
The curve passes through $P(2,0)$ and $Q\left(0,-\frac{3}{2}\right)$, has a vertical asymptote at $x=4$, and a horizontal asymptote at $y=-3$. The sketch should clearly show these features.
(d) Equation of the Normal Line $L$ at $x=2$
First, find the slope of the tangent to $C$ at $x=2$. The derivative of $$ y=\frac{6-3x}{x-4} $$ is computed using the quotient rule: $$ y'=\frac{(x-4)(-3)-(6-3x)(1)}{(x-4)^2}. $$ Simplify: $$ y'=\frac{-3(x-4)-(6-3x)}{(x-4)^2} =\frac{-3x+12-6+3x}{(x-4)^2} =\frac{6}{(x-4)^2}. $$ At $x=2$, the slope of the tangent is: $$ y'(2)=\frac{6}{(2-4)^2} =\frac{6}{4} =\frac{3}{2}. $$ The slope of the normal is the negative reciprocal: $$ \text{slope of normal}=-\frac{2}{3}. $$ The point on $C$ at $x=2$ is $P(2,0)$. Using the point-slope form of a line: $$ y-0=-\frac{2}{3}(x-2). $$ Simplify: $$ y=-\frac{2}{3}x+\frac{4}{3}. $$ Thus, the equation of the normal line is: $$ L:y=-\frac{2}{3}x+\frac{4}{3}. $$(e) Intersection of $L$ and $C$
Substitute $$ y=-\frac{2}{3}x+\frac{4}{3} $$ into $$ y=\frac{6-3x}{x-4}: $$ $$ -\frac{2}{3}x+\frac{4}{3} = \frac{6-3x}{x-4}. $$ Clear the fractions by multiplying through by $3(x-4)$: $$ -2x(x-4)+4(x-4)=3(6-3x). $$ Expand: $$ -2x^2+8x+4x-16=18-9x. $$ Simplify: $$ -2x^2+12x-16=18-9x. $$ Rearrange to form a quadratic equation: $$ -2x^2+21x-34=0. $$ Divide through by $-1$: $$ 2x^2-21x+34=0. $$ Solve this quadratic using the quadratic formula: $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, \quad a=2,\ b=-21,\ c=34. $$ $$ x=\frac{-(-21)\pm\sqrt{(-21)^2-4(2)(34)}}{2(2)}. $$ $$ x=\frac{21\pm\sqrt{441-272}}{4}. $$ $$ x=\frac{21\pm\sqrt{169}}{4}. $$ $$ x=\frac{21\pm13}{4}. $$ $$ x=\frac{34}{4}=8.5 \quad \text{or} \quad x=\frac{8}{4}=2. $$ Since $x=2$ corresponds to the point $P$, the other intersection point is: $$ x=8.5. $$Question 11
The roots of a quadratic equation $E$ are $\alpha$ and $\beta$ where $\alpha>\beta>0$. Given that $\alpha-\beta=2\sqrt{6}$ and $\alpha^2+\beta^2=30$(a) show that
(i) $\alpha\beta=3$
(ii) $\alpha+\beta=6$
(b) Without solving $E$
(i) find the value of $\alpha^4+\beta^4$
(ii) find the exact value of $\alpha^4-\beta^4$
Given that $\alpha^4=P+Q\sqrt{6}$ where $P$ and $Q$ are positive integers,
(c) find the value of $P$ and the value of $Q$
Click to view the solution
Solution:
The roots of the quadratic equation $E$ are $\alpha$ and $\beta$, where $\alpha>\beta>0$.We are given: $$ \alpha-\beta=2\sqrt{6}, \quad \alpha^2+\beta^2=30. $$
(a) Show that
(i) $\alpha\beta=3$
(ii) $\alpha+\beta=6$
From the above, we already deduced that $$ s=\alpha+\beta=6. $$ Thus: $$ \alpha+\beta=6. $$(b) Without solving $E$
(i) Find $\alpha^4+\beta^4$
Using the identity: $$ \alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2, $$ substitute the known values: $$ \alpha^2+\beta^2=30, \quad \alpha\beta=3. $$ $$ \alpha^4+\beta^4=30^2-2(3^2). $$ $$ \alpha^4+\beta^4=900-18=882. $$ Thus: $$ \alpha^4+\beta^4=882. $$(ii) Find $\alpha^4-\beta^4$
Using the identity: $$ \alpha^4-\beta^4=(\alpha^2+\beta^2)(\alpha^2-\beta^2), $$ we already know $$ \alpha^2+\beta^2=30. $$ To find $\alpha^2-\beta^2$, use the identity: $$ \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta). $$ Substitute $\alpha+\beta=6$ and $\alpha-\beta=2\sqrt{6}$: $$ \alpha^2-\beta^2=6(2\sqrt{6})=12\sqrt{6}. $$ Now substitute: $$ \alpha^4-\beta^4=30\cdot12\sqrt{6}. $$ $$ \alpha^4-\beta^4=360\sqrt{6}. $$(c) Express $\alpha^4=P+Q\sqrt{6}$
We are given $$ \alpha^4=P+Q\sqrt{6}, $$ where $P$ and $Q$ are positive integers. From part (b): $$ \alpha^4+\beta^4=882, \quad \alpha^4-\beta^4=360\sqrt{6}. $$ Using the identities: $$ \alpha^4= \frac{(\alpha^4+\beta^4)+(\alpha^4-\beta^4)}{2}, $$ $$ \beta^4= \frac{(\alpha^4+\beta^4)-(\alpha^4-\beta^4)}{2}. $$ Substitute the values: $$ \alpha^4= \frac{882+360\sqrt{6}}{2}, \quad \beta^4= \frac{882-360\sqrt{6}}{2}. $$ $$ \alpha^4=441+180\sqrt{6}. $$ Thus: $$ P=441, \quad Q=180. $$Final Answers
- (a) (i) $\alpha\beta=3$, (ii) $\alpha+\beta=6$.
- (b) (i) $\alpha^4+\beta^4=882$, (ii) $\alpha^4-\beta^4=360\sqrt{6}$.
- (c) $P=441$, $Q=180$.
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