Question 9
(7+6=13)
The straight line $ABC$ is the diameter of the semicircle $ADC$.
$AEB$ is an arc of a circle with centre $O$.
All angles are measured in radians.
$BC = 2x$ cm
$OA = OB = x$ cm
length of arc $AEB = 1.8x$ cm
The perimeter of the logo is $P$.
(a) Show that $P = a x (\pi + \pi \sin 0.9 + b)$ where $a$ and $b$ are constants to be found. (7)
Given that $x = 10$ cm,
(b) find, in cm$^2$ to 3 significant figures, the area of the logo. (6)
Problem Statement
A logo, $AEBACD$, is shown shaded in Figure 2. The straight line $ABC$ is the diameter of the semicircle $ADC$. $AEB$ is an arc of a circle with center $O$.
All angles are measured in radians, and the following properties are given:
- $BC = 2x$ cm
- $OA = OB = x$ cm
- The length of arc $AEB = 1.8x$ cm
The perimeter of the logo is $P$.
- Show that $P = ax(\pi + \pi \sin(0.9) + b)$ where $a$ and $b$ are constants to be found.
- Given $x = 10$ cm, find the area of the logo in cm$^2$, rounded to three significant figures.
Solution
Step 1: Calculate angle $\angle AOB$ using $l = r\theta$
The angle subtended by arc $AEB$ at the center $O$ can be calculated as:
$$ \theta = \frac{l}{r} = \frac{1.8x}{x} = 1.8 \text{ radians} $$Thus, $\angle AOB = 1.8$ radians.
Step 2: Find $AB$ in terms of $\sin(0.9)$
Using triangle $\triangle AOB$ with $\angle AOB = 1.8$, and noting that $OA = OB = x$, the chord $AB$ can be calculated using:
$$ AB = 2 \cdot OA \cdot \sin\left(\frac{\angle AOB}{2}\right) = 2x \sin(0.9) $$Step 3: Find the diameter of the larger semicircle $ADC$
The diameter of the semicircle $ADC$ includes $AB$ and $BC$. Thus:
$$ \text{Diameter of semicircle} = AB + BC = 2x \sin(0.9) + 2x $$The radius of the semicircle is:
$$ \text{Radius} = x + x \sin(0.9) $$Step 4: Calculate the perimeter $P$
The perimeter of the logo consists of:
- The semicircular arc $ADC$, with radius $r = x + x \sin(0.9)$
- The straight segment $BC$, with length $2x$
- The arc $AEB$, with given length $1.8x$
The length of the semicircular arc $ADC$ is:
$$ \text{Length of semicircle } ADC = \pi (x + x \sin(0.9)) = \pi x (1 + \sin(0.9)) $$Thus, the total perimeter is:
$$ P = \pi x (1 + \sin(0.9)) + 1.8x + 2x $$ $$ = x\big(\pi + \pi \sin(0.9) + 3.8 \big) $$Comparing with $P = ax(\pi + \pi \sin(0.9) + b)$, we obtain:
$$ a = 1, \quad b = 3.8. $$Step 5: Calculate the area of the logo
The area of the logo consists of:
- The area of semicircle $ADC$
- Subtracting the area of the smaller segment $AEB$
Step 5.1: Area of semicircle $ADC$
The area of the semicircle is:
$$ \text{Area of semicircle } ADC = \frac{1}{2} \pi \big(x + x \sin(0.9)\big)^2 $$Step 5.2: Area of sector $AEB$
The area of the smaller sector $AEB$ is:
$$ \text{Area of sector } AEB = \frac{1}{2} x^2 \theta $$Substituting $\theta = 1.8$:
$$ \text{Area of sector } AEB = \frac{1}{2} x^2 (1.8) = 0.9x^2 $$Step 5.3: Area of the logo
The area of triangle $OAB$ is:
$$ \text{Area of triangle } OAB = \frac{1}{2} x^2 \sin(1.8) $$Hence, the area of the circular segment $AEB$ is:
$$ \text{Area of segment } AEB = 0.9x^2 - \frac{1}{2} x^2 \sin(1.8) $$Therefore, the area of the logo is:
$$ \text{Area of logo} = \frac{1}{2} \pi \big(x + x \sin(0.9)\big)^2 - 0.9x^2 + \frac{1}{2} x^2 \sin(1.8) $$Step 5.4: Substitute $x = 10$ cm
$$ \text{Area of logo} = \frac{1}{2} \pi \big(10 + 10 \sin(0.9)\big)^2 - 0.9(10)^2 + \frac{1}{2}(10)^2 \sin(1.8) $$ $$ = 0.5 \pi (10 + 7.8332)^2 - 41.307 $$ $$ = 0.5 \pi (17.8332)^2 - 41.307 $$ $$ \approx 0.5 \pi (318.02) - 41.307 $$ $$ \approx 499.55 - 41.307 $$ $$ = 458.2 \text{ cm}^2 $$Thus, the area of the logo is approximately:
$$ \boxed{458 \text{ cm}^2} $$Question 10
(3+7=10)The roots of a quadratic equation are $\alpha$ and $\beta$ where \[\alpha + \beta = -\frac{5}{2} \mbox{ and }\alpha^3 + \beta^3 = \frac{115}{8}\]
(a) Show that $\alpha \beta = 4$. (3)
(b) Form a quadratic equation with integer coefficients that has roots \[\frac{\alpha^2 + 1}{\beta} \mbox{ and } \frac{\beta^2 + 1}{\alpha}\] (7)
Solution
Step (a): Show that $\alpha \beta = 4$
We are given the following information:
$$ \alpha + \beta = -\frac{5}{2}, $$ $$ \alpha^3 + \beta^3 = \frac{115}{8}. $$Using the identity for the sum of cubes:
$$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta), $$substitute $\alpha + \beta = -\frac{5}{2}$:
$$ \alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)^3 - 3\alpha \beta \left(-\frac{5}{2}\right). $$Simplify:
$$ \left(-\frac{5}{2}\right)^3 = -\frac{125}{8}. $$Hence:
$$ \alpha^3 + \beta^3 = -\frac{125}{8} + \frac{15}{2}\alpha \beta. $$Given that $\alpha^3 + \beta^3 = \frac{115}{8}$:
$$ \frac{115}{8} = -\frac{125}{8} + \frac{15}{2}\alpha \beta. $$Simplify:
$$ \frac{115}{8} + \frac{125}{8} = \frac{15}{2}\alpha \beta $$ $$ \frac{240}{8} = \frac{15}{2}\alpha \beta $$ $$ 30 = \frac{15}{2}\alpha \beta $$Multiply through by $2$:
$$ 60 = 15\alpha \beta $$Divide by $15$:
$$ \alpha \beta = 4. $$Thus, we have shown that $\alpha \beta = 4$.
Step (b): Form a quadratic equation with integer coefficients
The roots of the quadratic equation are:
$$ a = \frac{\alpha^2 + 1}{\beta}, \quad b = \frac{\beta^2 + 1}{\alpha}. $$First, find $a + b$:
$$ a + b = \frac{\alpha^2 + 1}{\beta} + \frac{\beta^2 + 1}{\alpha}. $$Bring to a common denominator:
$$ a + b = \frac{\alpha(\alpha^2 + 1) + \beta(\beta^2 + 1)}{\alpha \beta}. $$Simplify the numerator:
$$ \alpha(\alpha^2 + 1) + \beta(\beta^2 + 1) = \alpha^3 + \alpha + \beta^3 + \beta $$ $$ = (\alpha^3 + \beta^3) + (\alpha + \beta). $$Substitute the known values:
$$ \alpha^3 + \beta^3 + \alpha + \beta = \frac{115}{8} - \frac{5}{2}. $$Simplify:
$$ \frac{115}{8} - \frac{5}{2} = \frac{115}{8} - \frac{20}{8} = \frac{95}{8}. $$Thus:
$$ a + b = \frac{\frac{95}{8}}{4} = \frac{95}{32}. $$Next, find $ab$:
$$ ab = \frac{(\alpha^2 + 1)(\beta^2 + 1)}{\alpha \beta}. $$ $$ ab = \frac{\alpha^2\beta^2 + \alpha^2 + \beta^2 + 1}{4}. $$Use $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$:
$$ \alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2(4). $$ $$ = \frac{25}{4} - 8 = \frac{25}{4} - \frac{32}{4} = -\frac{7}{4}. $$Substitute into $ab$:
$$ ab = \frac{4^2 + \alpha^2 + \beta^2 + 1}{4} $$ $$ = \frac{16 - \frac{7}{4} + 1}{4} $$ $$ = \frac{17 - \frac{7}{4}}{4} $$ $$ = \frac{\frac{68}{4} - \frac{7}{4}}{4} $$ $$ = \frac{\frac{61}{4}}{4} = \frac{61}{16}. $$The quadratic equation is:
$$ x^2 - (a+b)x + ab = 0 $$ $$ x^2 - \frac{95}{32}x + \frac{61}{16} = 0. $$Multiply through by $32$:
$$ \boxed{32x^2 - 95x + 122 = 0} $$Question 11
(3+6=9)
The point $M$ is the midpoint of $OC$.
(a) Find $\overrightarrow{MA}$ as a simplified expression in terms of $\mathbf{p}$ and $\mathbf{q}$. (3)
The point $N$ lies on $OB$ such that $M, N$ and $A$ are collinear.
(b) Find the ratio $MN : NA$. (6)
Solution
Step 1: Understanding the Problem
We are given quadrilateral $OABC$ with the following vector relations:
$$ \overrightarrow{OA} = 4 \mathbf{p} + 5 \mathbf{q}, \quad \overrightarrow{OB} = 3 \mathbf{p} + \mathbf{q}, \quad \overrightarrow{OC} = 2 \mathbf{p} - 4 \mathbf{q} $$The point $M$ is the midpoint of $OC$, and the point $N$ lies on $OB$ such that the points $M$, $N$, and $A$ are collinear.
Step 2: Find $\overrightarrow{MA}$
Since $M$ is the midpoint of $OC$, the position vector of $M$ is:
$$ \overrightarrow{OM} = \frac{1}{2}\overrightarrow{OC} = \frac{1}{2}(2\mathbf{p} - 4\mathbf{q}) = \mathbf{p} - 2\mathbf{q} $$Thus,
$$ \overrightarrow{MA} = \overrightarrow{OA} - \overrightarrow{OM} $$ $$ = (4\mathbf{p} + 5\mathbf{q}) - (\mathbf{p} - 2\mathbf{q}) $$ $$ = 3\mathbf{p} + 7\mathbf{q} $$Step 3: Find the ratio $MN : NA$
Since $M$, $N$, and $A$ are collinear, suppose $N$ divides $MA$ in the ratio $k : 1$. Then:
$$ MN : NA = k : 1 $$Using the section formula:
$$ \overrightarrow{ON} = \frac{1}{k+1} \left( k\overrightarrow{OA} + \overrightarrow{OM} \right) $$Substitute $\overrightarrow{OA} = 4\mathbf{p} + 5\mathbf{q}$ and $\overrightarrow{OM} = \mathbf{p} - 2\mathbf{q}$:
$$ \overrightarrow{ON} = \frac{1}{k+1} \left( k(4\mathbf{p} + 5\mathbf{q}) + (\mathbf{p} - 2\mathbf{q}) \right) $$Simplify:
$$ \overrightarrow{ON} = \frac{1}{k+1} \left( (4k+1)\mathbf{p} + (5k-2)\mathbf{q} \right) $$Since $N$ lies on $OB$, its position vector is also a scalar multiple of $\overrightarrow{OB} = 3\mathbf{p} + \mathbf{q}$:
$$ \overrightarrow{ON} = \lambda(3\mathbf{p} + \mathbf{q}) $$Equating coefficients of $\mathbf{p}$ and $\mathbf{q}$:
$$ \frac{4k+1}{k+1} = 3\lambda, \quad \frac{5k-2}{k+1} = \lambda $$Step 4: Solve for $k$
From the second equation:
$$ \lambda = \frac{5k-2}{k+1} $$Substitute into the first equation:
$$ \frac{4k+1}{k+1} = 3\left( \frac{5k-2}{k+1} \right) $$Since both sides have denominator $k+1$, cancel it:
$$ 4k + 1 = 3(5k - 2) $$Expand the right-hand side:
$$ 4k + 1 = 15k - 6 $$Rearrange:
$$ 4k - 15k = -6 - 1 $$ $$ -11k = -7 $$ $$ k = \frac{7}{11} $$Step 5: Find $\lambda$
Substitute $k = \frac{7}{11}$ into $\lambda = \frac{5k-2}{k+1}$:
$$ \lambda = \frac{5\left(\frac{7}{11}\right)-2}{\frac{7}{11}+1} $$ $$ = \frac{\frac{35}{11} - 2}{\frac{7}{11} + \frac{11}{11}} $$ $$ = \frac{\frac{13}{11}}{\frac{18}{11}} = \frac{13}{18} $$Step 6: State the required ratio
Since the ratio is $MN : NA = k : 1$:
$$ MN : NA = \frac{7}{11} : 1 = 7 : 11 $$Hence,
$$ \boxed{MN : NA = 7 : 11} $$
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