Question 1
Question 1
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
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
Problem Statement
$\text{A circle, centre } O, \text{ with radius } r \,\mathrm{cm}. \\ \text{The points } A, P, \text{ and } B \text{ lie on the circle.} \\ \text{The obtuse angle } \angle AOB = \theta \text{ radians.}$
$\text{The area of the major sector } APBO \text{ is } 372.4 \,\mathrm{cm}^2, \\ \text{and the length of the arc } APB \text{ is } 53.2 \,\mathrm{cm}.$
$\text{Find, to 3 significant figures where appropriate, the value of:}$
- $(i)\; r,$
- $(ii)\; \theta.$
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
- $(i)\; r = 14.0 \,\mathrm{cm},$
- $(ii)\; \theta = 2\pi - 3.80 = 2.48 \,\mathrm{radians}.$
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