Question 1
(a) Expand $$ \left(1+\frac{x}{4}\right)^8 $$ in ascending powers of $x$ up to and including the term in $x^3$.
Give each coefficient in its simplest terms.
[3]
(b) Use your expansion with a suitable value of $x$ to obtain an approximation, to 4 decimal places, of $$ (1.035)^8 $$
[3]
Solution
Solution:
Part (a): Expand $ \left( 1 + \frac{x}{4} \right)^8 $ in ascending powers of $ x $
Using the binomial expansion formula:
$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, $$
we expand $ \left( 1 + \frac{x}{4} \right)^8 $ with $ a = 1 $, $ b = \frac{x}{4} $, and $ n = 8 $.
Step 1: Write the terms of the expansion up to $ x^3 $:
$$ \left( 1 + \frac{x}{4} \right)^8 = \binom{8}{0} \cdot 1^8 + \binom{8}{1} \cdot 1^7 \cdot \frac{x}{4} + \binom{8}{2} \cdot 1^6 \cdot \left( \frac{x}{4} \right)^2 + \binom{8}{3} \cdot 1^5 \cdot \left( \frac{x}{4} \right)^3 + \dots $$
Step 2: Compute each term:
- Constant term: $$ \binom{8}{0} \cdot 1^8 = 1 $$
- Term in $ x $: $$ \binom{8}{1} \cdot 1^7 \cdot \frac{x}{4} = 8 \cdot \frac{x}{4} = 2x $$
- Term in $ x^2 $: $$ \binom{8}{2} \cdot 1^6 \cdot \left( \frac{x}{4} \right)^2 = \frac{8 \cdot 7}{2} \cdot \frac{x^2}{16} = 28 \cdot \frac{x^2}{16} = \frac{7x^2}{4} $$
- Term in $ x^3 $: $$ \binom{8}{3} \cdot 1^5 \cdot \left( \frac{x}{4} \right)^3 = \frac{8 \cdot 7 \cdot 6}{6} \cdot \frac{x^3}{64} = 56 \cdot \frac{x^3}{64} = \frac{7x^3}{8} $$
Step 3: Combine the terms:
$$ \left( 1 + \frac{x}{4} \right)^8 = 1 + 2x + \frac{7x^2}{4} + \frac{7x^3}{8} + \dots $$
Part (b): Approximate $ (1.035)^8 $
We know that:
$$ 1.035 = 1 + \frac{x}{4}, \quad \text{so } x = 4 \cdot (1.035 - 1) = 0.14 $$
Step 1: Substitute $ x = 0.14 $ into the expansion:
$$ (1.035)^8 \approx 1 + 2x + \frac{7x^2}{4} + \frac{7x^3}{8} $$
Step 2: Compute each term:
- Constant term: $ 1 $
- Term in $ x $: $$ 2x = 2(0.14) = 0.28 $$
- Term in $ x^2 $: $$ \frac{7x^2}{4} = \frac{7(0.14)^2}{4} = \frac{7(0.0196)}{4} = \frac{0.1372}{4} = 0.0343 $$
- Term in $ x^3 $: $$ \frac{7x^3}{8} = \frac{7(0.14)^3}{8} = \frac{7(0.002744)}{8} = \frac{0.019208}{8} = 0.002401 $$
Step 3: Add the terms:
$$ (1.035)^8 \approx 1 + 0.28 + 0.0343 + 0.002401 = 1.3167 $$
Final Answer
$$ (1.035)^8 \approx \boxed{1.3167} $$
Question 2
Find the set of values of $x$ for which
(a) $3x - 8 < 5x + 3$
[1]
(b) $4x^2 - 7x + 1 > 6 - 2x^2$
[4]
(c) both $3x - 8 < 5x + 3$ and $4x^2 - 7x + 1 > 6 - 2x^2$
[1]
Solution
Solution:
Find the set of values of $x$ for which:
- $3x - 8 \lt 5x + 3$
- $4x^2 - 7x + 1 > 6 - 2x^2$
- Both $3x - 8 < 5x + 3$ and $4x^2 - 7x + 1 > 6 - 2x^2$
Part (a): Solve $3x - 8 \lt 5x + 3$
Step 1: Rearrange the inequality:
$$ 3x - 8 \lt 5x + 3 \implies -8 - 3 \lt 5x - 3x \implies -11 \lt 2x $$
Step 2: Solve for $x$:
$$ x > -\frac{11}{2} $$
The solution to part (a) is:
$$ x > -\frac{11}{2} $$
Part (b): Solve $4x^2 - 7x + 1 > 6 - 2x^2$
Step 1: Rearrange the inequality:
$$ 4x^2 - 7x + 1 > 6 - 2x^2 $$ $$ \implies 4x^2 + 2x^2 - 7x + 1 - 6 > 0 $$ $$ \implies 6x^2 - 7x - 5 > 0 $$
Step 2: Solve the quadratic inequality:
Let $$ f(x) = 6x^2 - 7x - 5 $$ Solve $$ 6x^2 - 7x - 5 = 0 $$ using the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 6,\; b = -7,\; c = -5 $$
$$ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-5)}}{2(6)} = \frac{7 \pm \sqrt{49 + 120}}{12} = \frac{7 \pm \sqrt{169}}{12} = \frac{7 \pm 13}{12} $$
Step 3: Find the roots:
$$ x = \frac{7 + 13}{12} = \frac{20}{12} = \frac{5}{3} $$ $$ x = \frac{7 - 13}{12} = \frac{-6}{12} = -\frac{1}{2} $$
Step 4: Test intervals:
The roots divide the $x$-axis into three intervals:
$$ x \lt -\frac{1}{2}, \quad -\frac{1}{2} \lt x \lt \frac{5}{3}, \quad x > \frac{5}{3} $$
Test a point from each interval in $$ f(x) = 6x^2 - 7x - 5 $$
- For $x = -1$: $$ f(-1) = 6(-1)^2 - 7(-1) - 5 = 6 + 7 - 5 = 8 > 0 $$
- For $x = 0$: $$ f(0) = 6(0)^2 - 7(0) - 5 = -5 \lt 0 $$
- For $x = 2$: $$ f(2) = 6(2)^2 - 7(2) - 5 = 24 - 14 - 5 = 5 > 0 $$
Step 5: Determine the solution:
The solution to $$ 6x^2 - 7x - 5 > 0 $$ is:
$$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
Part (c): Solve both inequalities simultaneously
Combine the solutions:
- From part (a): $$ x > -\frac{11}{2} $$
- From part (b): $$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
The intersection of these solutions is:
$$ -\frac{11}{2} \lt x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
Final Answer
- $$ x > -\frac{11}{2} $$
- $$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
- $$ -\frac{11}{2} \lt x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
Question 3
Problem and Solution
Given that $$ y = e^{3x} \sin 2x $$ show that $$ 13y + \frac{d^2 y}{dx^2} = 6 \frac{dy}{dx} $$
[8]
Solution
Solution:
Step 1: Find $ \frac{dy}{dx} $
We will differentiate $$ y = e^{3x} \sin 2x $$ using the product rule. Recall that $$ \frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v' $$ where $$ u = e^{3x} \quad \text{and} \quad v = \sin 2x $$
First, compute the derivatives:
$$ u' = \frac{d}{dx} e^{3x} = 3e^{3x}, \quad v' = \frac{d}{dx} \sin 2x = 2\cos 2x $$
Now, apply the product rule:
$$ \frac{dy}{dx} = 3e^{3x} \sin 2x + e^{3x} \cdot 2\cos 2x $$
Thus, we have:
$$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$
Step 2: Find $ \frac{d^2 y}{dx^2} $
Next, differentiate $$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$ using the product rule again.
Let $$ u = e^{3x} \quad \text{and} \quad v = 3\sin 2x + 2\cos 2x $$ so
$$ u' = 3e^{3x}, \quad v' = 3 \cdot 2\cos 2x - 2 \cdot 2\sin 2x = 6\cos 2x - 4\sin 2x $$
Now, apply the product rule:
$$ \frac{d^2 y}{dx^2} = 3e^{3x}(3\sin 2x + 2\cos 2x) + e^{3x}(6\cos 2x - 4\sin 2x) $$
Simplify:
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ 9\sin 2x + 6\cos 2x + 6\cos 2x - 4\sin 2x \right] $$
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ (9 - 4)\sin 2x + (6 + 6)\cos 2x \right] $$
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ 5\sin 2x + 12\cos 2x \right] $$
Step 3: Verify the equation $ 13y + \frac{d^2 y}{dx^2} = 6 \frac{dy}{dx} $
Now substitute $$ y = e^{3x}\sin 2x $$ $$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$ and $$ \frac{d^2 y}{dx^2} = e^{3x}(5\sin 2x + 12\cos 2x) $$ into the equation.
The left-hand side:
$$ 13y + \frac{d^2 y}{dx^2} = 13e^{3x}\sin 2x + e^{3x}(5\sin 2x + 12\cos 2x) $$
Factor out $e^{3x}$:
$$ 13e^{3x}\sin 2x + e^{3x}(5\sin 2x + 12\cos 2x) = e^{3x} \left[ 13\sin 2x + 5\sin 2x + 12\cos 2x \right] $$
$$ = e^{3x} \left[ (13 + 5)\sin 2x + 12\cos 2x \right] = e^{3x} \left[ 18\sin 2x + 12\cos 2x \right] $$
The right-hand side:
$$ 6\frac{dy}{dx} = 6e^{3x}(3\sin 2x + 2\cos 2x) $$
$$ = e^{3x} \left[ 18\sin 2x + 12\cos 2x \right] $$
Thus, both sides are equal:
$$ 13y + \frac{d^2 y}{dx^2} = 6\frac{dy}{dx} $$
Final Answer
We have shown that
$$ 13y + \frac{d^2 y}{dx^2} = 6\frac{dy}{dx} $$
Question 4
Figure 1 shows part of the curve $C$ with equation $$ y=\frac{2x+q}{x+r} \quad x \neq -r $$ where $q$ and $r$ are integers.The asymptote to $C$ that is parallel to the $y$-axis has equation $$ x=-4 $$ The asymptote to $C$ that is parallel to the $x$-axis has equation $$ y=p $$
(a) Write down
(i) the value of $p$
(ii) the value of $r$
[2]
Given that $C$ crosses the $y$-axis at the point with coordinates $$ \left(0,\frac{3}{2}\right) $$ (b) find the value of $q$
[2]
Given that $C$ crosses the $x$-axis at the point with coordinates $(s,0)$
(c) find the value of $s$
[4]
Solution
Solution:
We are given the equation of the curve $C$ as
$$ y = \frac{2x + q}{x + r} $$
where $q$ and $r$ are integers. The asymptote parallel to the $y$-axis is $x = -4$, and the asymptote parallel to the $x$-axis is $y = p$. Additionally, the curve crosses the $y$-axis at the point $$ \left(0,\frac{3}{2}\right) $$ and the curve crosses the $x$-axis at the point $$ (s,0) $$
Part (a)
-
(i) The value of $p$:
As $x \to \infty$, the curve tends to the horizontal asymptote, where the ratio of the leading terms in the numerator and denominator gives the asymptote. Therefore, for large values of $x$:
$$ y \approx \frac{2x}{x} = 2 $$
Thus, the horizontal asymptote is $$ y = 2 $$ so the value of $p$ is
$$ p = 2 $$
-
(ii) The value of $r$:
The vertical asymptote occurs when the denominator is zero, i.e. when $$ x + r = 0 $$ which gives $$ x = -r $$
We are told that the vertical asymptote is at $$ x = -4 $$ so
$$ r = 4 $$
Part (b)
We are given that the curve crosses the $y$-axis at the point $$ \left(0,\frac{3}{2}\right) $$ To find $q$, substitute $x = 0$ and $$ y = \frac{3}{2} $$ into the equation of the curve:
$$ \frac{3}{2} = \frac{2(0) + q}{0 + 4} $$
This simplifies to:
$$ \frac{3}{2} = \frac{q}{4} $$
Multiplying both sides by $4$:
$$ q = 6 $$
Thus, the value of $q$ is $6$.
Part (c)
Given that the curve crosses the $x$-axis at the point $$ (s,0) $$ substitute $y = 0$ into the equation of the curve:
$$ 0 = \frac{2x + 6}{x + 4} $$
For the fraction to be zero, the numerator must be zero. Therefore:
$$ 2x + 6 = 0 $$
Solving for $x$:
$$ 2x = -6 \quad \Rightarrow \quad x = -3 $$
Thus, the value of $s$ is $-3$.
Final Answers
$$ \boxed{ p = 2, \quad r = 4, \quad q = 6, \quad s = -3 } $$
Question 5
The line $l$ with gradient $-\frac{1}{12}$ passes through the points $A$ and $B$ with coordinates $(p, 10)$ and $(123, 0)$ respectively.
(a) Show that $p = 3$
[2]
(b) Find an equation for $l$ in the form $rx + sy + t = 0$ where $r$, $s$ and $t$ are integers.
[2]
The line $k$ is perpendicular to $l$ and passes through the point $A$.
(c) Find an equation for $k$ in the form $y = mx + c$
[3]
Line $k$ intersects the $x$-axis at the point $C$.
(d) Find the exact area of triangle $ABC$.
[4]
Solution
Solution:
We are given a line $l$ with gradient $$ -\frac{1}{12} $$ that passes through the points $$ A(p,10) \quad \text{and} \quad B(123,0) $$
Part (a): Show that $p = 3$
The gradient of the line passing through two points $$ (x_1,y_1) \quad \text{and} \quad (x_2,y_2) $$ is given by
$$ \text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} $$
For the points $$ A(p,10) \quad \text{and} \quad B(123,0) $$ the gradient is
$$ \text{Gradient of line} = \frac{0 - 10}{123 - p} $$
We are given that the gradient is $$ -\frac{1}{12} $$ so
$$ \frac{-10}{123 - p} = -\frac{1}{12} $$
Multiplying both sides by $123 - p$:
$$ -10 = \frac{-1}{12}(123 - p) $$
Now multiply both sides by $12$:
$$ -120 = -(123 - p) $$
This simplifies to
$$ -120 = -123 + p $$
Adding $123$ to both sides:
$$ p = 3 $$
Thus, $$ p = 3 $$
Part (b): Find an equation for $l$ in the form $rx + sy + t = 0$
We know the gradient of line $l$ is $$ -\frac{1}{12} $$ and it passes through the point $$ A(3,10) $$
The point-slope form of a line is
$$ y - y_1 = m(x - x_1) $$
Substitute $$ m = -\frac{1}{12}, \quad x_1 = 3, \quad y_1 = 10 $$
$$ y - 10 = -\frac{1}{12}(x - 3) $$
Simplify:
$$ y - 10 = -\frac{1}{12}x + \frac{3}{12} $$
$$ y - 10 = -\frac{1}{12}x + \frac{1}{4} $$
Multiply through by $12$:
$$ 12(y - 10) = -x + 3 $$
$$ 12y - 120 = -x + 3 $$
Rearranging into the form $$ rx + sy + t = 0 $$
$$ x + 12y - 123 = 0 $$
Thus, the equation of line $l$ is
$$ x + 12y - 123 = 0 $$
Part (c): Find an equation for $k$ in the form $y = mx + c$
The line $k$ is perpendicular to $l$ and passes through $$ A(3,10) $$
The gradient of line $l$ is $$ -\frac{1}{12} $$
Therefore, the gradient of line $k$ is the negative reciprocal:
$$ m_k = 12 $$
Using point-slope form:
$$ y - 10 = 12(x - 3) $$
Simplify:
$$ y - 10 = 12x - 36 $$
$$ y = 12x - 26 $$
Thus, the equation of line $k$ is
$$ y = 12x - 26 $$
Part (d): Find the exact area of triangle $ABC$
We are given the points:
- $A(3,10)$
- $B(123,0)$
- $C$ is the x-intercept of line $k$
To find point $C$, set $$ y = 0 $$ in $$ y = 12x - 26 $$
$$ 0 = 12x - 26 $$
Solving for $x$:
$$ 12x = 26 \quad \Rightarrow \quad x = \frac{26}{12} = \frac{13}{6} $$
Thus, $$ C\left(\frac{13}{6},0\right) $$
Use the area formula for a triangle:
$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$
Substitute $$ A(3,10), \quad B(123,0), \quad C\left(\frac{13}{6},0\right) $$
$$ \text{Area} = \frac{1}{2} \left| 3(0 - 0) + 123(0 - 10) + \frac{13}{6}(10 - 0) \right| $$
$$ = \frac{1}{2} \left| 0 + 123(-10) + \frac{130}{6} \right| $$
$$ = \frac{1}{2} \left| -1230 + \frac{65}{3} \right| $$
$$ = \frac{1}{2} \times 1208\frac{1}{3} $$
$$ = 604\frac{1}{6} $$
Thus, the exact area of triangle $ABC$ is
$$ \boxed{604\frac{1}{6}} $$
Question 6
$$ \angle AOB=\frac{\pi}{6}\text{ radians} \quad OA=OB=r\mathrm{~cm} $$
The area of the sector is increasing in such a way that the size of $\angle AOB$ remains constant, and the lengths $OA$ and $OB$ are both increasing at a constant rate of $$ 0.2\mathrm{~cm}/\mathrm{s} $$
Find the exact rate of change, in $$ \mathrm{cm}^2/\mathrm{s}, $$ of the area of the sector when the length of arc $AB$ is $$ \frac{5\pi}{2} $$
[6]
Solution
Solution:
We are given a sector $OAB$ of a circle with center $O$ and radius $r$ cm. The angle $$ \angle AOB = \frac{\pi}{6} $$ radians, and the lengths $$ OA = OB = r $$ cm.
The area of the sector is increasing such that the angle $$ \angle AOB $$ remains constant, and the lengths $$ OA \text{ and } OB $$ are both increasing at a constant rate of $$ 0.2 \text{ cm/s} $$
We are tasked with finding the exact rate of change, in $$ \text{cm}^2/\text{s} $$ of the area of the sector when the length of arc $$ AB $$ is $$ \frac{5\pi}{2} $$
Solution
Step 1: Write the formula for the area of the sector
The area $A$ of a sector of a circle is given by
$$ A = \frac{1}{2}r^2\theta $$
where $r$ is the radius and $\theta$ is the central angle in radians. Here,
$$ \theta = \frac{\pi}{6} $$
which remains constant.
Therefore,
$$ A = \frac{1}{2}r^2 \cdot \frac{\pi}{6} = \frac{\pi}{12}r^2 $$
Step 2: Differentiate the area with respect to time $t$
To find the rate of change of the area, differentiate with respect to $t$:
$$ \frac{dA}{dt} = \frac{d}{dt} \left( \frac{\pi}{12}r^2 \right) = \frac{\pi}{6}r\frac{dr}{dt} $$
We are given
$$ \frac{dr}{dt} = 0.2 \quad \text{cm/s} $$
Therefore,
$$ \frac{dA}{dt} = \frac{\pi}{6}r \cdot 0.2 = \frac{\pi}{30}r $$
Step 3: Find the radius $r$ when the arc length is $ \frac{5\pi}{2} $
The formula for arc length is
$$ \text{Arc length} = r\theta $$
Substitute
$$ \theta = \frac{\pi}{6} $$
so
$$ \text{Arc length} = r \cdot \frac{\pi}{6} $$
We are given that the arc length is
$$ \frac{5\pi}{2} $$
Thus,
$$ r \cdot \frac{\pi}{6} = \frac{5\pi}{2} $$
Solving for $r$:
$$ r = \frac{5\pi}{2} \cdot \frac{6}{\pi} = 15 \text{ cm} $$
Step 4: Substitute $r = 15$ into the rate of change of the area
Now substitute
$$ r = 15 $$
into
$$ \frac{dA}{dt} = \frac{\pi}{30}r $$
$$ \frac{dA}{dt} = \frac{\pi}{30} \cdot 15 = \frac{\pi}{2} \text{ cm}^2/\text{s} $$
Final Answer
Thus, the exact rate of change of the area is
$$ \boxed{ \frac{\pi}{2} \text{ cm}^2/\text{s} } $$
Question 7
The region $R$ is rotated through $360^\circ$ about the $y$-axis to generate a solid with volume $$ 18\pi $$
Use algebraic integration to find the value of $a$
[8]
Solution
Solution:
We are given a curve $S$ with the equation $$ y = x^2 + 2 $$ and a finite region $R$ bounded by the curve, the $y$-axis, and the line $$ y = a $$ where $$ a > 2 $$
The region $R$ is rotated through $360^\circ$ about the $y$-axis to generate a solid with volume $$ 18\pi $$
We are tasked with finding the value of $a$ using algebraic integration.
Solution
Step 1: Set up the volume formula for a solid of revolution
The volume of a solid generated by rotating a region about the $y$-axis is
$$ V = \pi \int_{y_1}^{y_2} x^2 \, dy $$
where:
- $x(y)$ is the radius at height $y$
- $y_1$ and $y_2$ are the limits of integration
The curve is
$$ y = x^2 + 2 $$
so solving for $x$:
$$ x = \sqrt{y - 2} $$
The limits of integration are from $$ y = 2 \quad \text{to} \quad y = a $$
Thus,
$$ V = \pi \int_{2}^{a} \left( \sqrt{y - 2} \right)^2 \, dy = \pi \int_{2}^{a} (y - 2) \, dy $$
Step 2: Compute the integral
Now evaluate
$$ V = \pi \int_{2}^{a} (y - 2) \, dy $$
The antiderivative of $$ y - 2 $$ is
$$ \frac{(y - 2)^2}{2} $$
So,
$$ V = \pi \left[ \frac{(y - 2)^2}{2} \right]_{2}^{a} $$
Substitute the limits:
$$ V = \pi \left( \frac{(a - 2)^2}{2} - 0 \right) $$
$$ V = \frac{\pi}{2} (a - 2)^2 $$
Step 3: Set the volume equal to $18\pi$ and solve for $a$
We are given
$$ V = 18\pi $$
So,
$$ \frac{\pi}{2} (a - 2)^2 = 18\pi $$
Divide both sides by $\pi$:
$$ \frac{1}{2} (a - 2)^2 = 18 $$
Multiply both sides by $2$:
$$ (a - 2)^2 = 36 $$
Take square roots:
$$ a - 2 = 6 \quad \text{or} \quad a - 2 = -6 $$
Thus,
$$ a = 8 \quad \text{or} \quad a = -4 $$
Since $$ a > 2 $$ we take
$$ a = 8 $$
Final Answer
The value of $a$ is
$$ \boxed{8} $$
Question 8
The quadratic equation $3 x^2-k x-1=0$, where $k$ is a positive integer, has roots $\alpha$ and $\beta$
(a) Show that $\alpha^2+\beta^2=\frac{k^2+6}{9}$
[3]
Given that $a^4+\beta^4=\frac{466}{81}$
(b) find the value of $k$
[5]
(c) Hence form an equation, with integer coefficients, which has roots
$$ \frac{\alpha^3+\beta}{\beta} \text { and } \frac{\beta^3+\alpha}{\alpha} $$
[6]
Solution
Solution:
The quadratic equation is $$ 3x^2 - kx - 1 = 0 $$ with roots $$ \alpha \quad \text{and} \quad \beta $$
Part (a): Showing $ \alpha^2 + \beta^2 = \frac{k^2 + 6}{9} $
Using the identity
$$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$
and substituting
$$ \alpha + \beta = \frac{k}{3}, \quad \alpha\beta = \frac{-1}{3} $$
we get
$$ \alpha^2 + \beta^2 = \left(\frac{k}{3}\right)^2 - 2\left(\frac{-1}{3}\right) $$
Simplify:
$$ \alpha^2 + \beta^2 = \frac{k^2}{9} + \frac{2}{3} $$
Converting $$ \frac{2}{3} $$ to denominator $9$:
$$ \alpha^2 + \beta^2 = \frac{k^2}{9} + \frac{6}{9} = \frac{k^2 + 6}{9} $$
Part (b): Finding $k$ given $ \alpha^4 + \beta^4 = \frac{466}{81} $
Using the identity
$$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 $$
and substituting
$$ \alpha^2 + \beta^2 = \frac{k^2 + 6}{9}, \quad (\alpha\beta)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9} $$
we have
$$ \alpha^4 + \beta^4 = \left(\frac{k^2 + 6}{9}\right)^2 - 2\left(\frac{1}{9}\right) $$
Simplify:
$$ \alpha^4 + \beta^4 = \frac{(k^2 + 6)^2}{81} - \frac{2}{9} $$
Convert $$ \frac{2}{9} = \frac{18}{81} $$
Thus
$$ \alpha^4 + \beta^4 = \frac{(k^2 + 6)^2}{81} - \frac{18}{81} $$
Equating to $$ \frac{466}{81} $$ we get
$$ \frac{(k^2 + 6)^2}{81} - \frac{18}{81} = \frac{466}{81} $$
Multiply through by $81$:
$$ (k^2 + 6)^2 - 18 = 466 $$
Simplify:
$$ (k^2 + 6)^2 = 484 $$
Take square roots:
$$ k^2 + 6 = \pm 22 $$
Since $k$ is positive:
$$ k^2 + 6 = 22 \implies k^2 = 16 \implies k = 4 $$
Part (c): Forming the quadratic equation
The roots are
$$ r_1 = \frac{\alpha^3 + \beta}{\beta}, \quad r_2 = \frac{\alpha + \beta^3}{\alpha} $$
Step 1: Sum of the roots $ (r_1 + r_2) $
Using
$$ r_1 + r_2 = \frac{\alpha^4 + \beta^4 + 2\alpha\beta}{\alpha\beta} $$
and substituting
$$ \alpha^4 + \beta^4 = \frac{466}{81}, \quad \alpha\beta = \frac{-1}{3} $$
we get
$$ r_1 + r_2 = \frac{ \frac{466}{81} + 2\left(\frac{-1}{3}\right) }{ \frac{-1}{3} } $$
Simplify:
$$ \frac{-1}{3} = \frac{-27}{81} $$
$$ \frac{466}{81} - \frac{54}{81} = \frac{412}{81} $$
Divide:
$$ r_1 + r_2 = \frac{\frac{412}{81}}{\frac{-1}{3}} = \frac{412}{81}\cdot(-3) = -\frac{1236}{81} = -\frac{412}{27} $$
Step 2: Product of the roots $ (r_1r_2) $
Using
$$ r_1r_2 = \frac{ \alpha^4 + \beta^4 + \alpha\beta + \alpha^3\beta^3 }{ \alpha\beta } $$
and substituting
$$ \alpha^4 + \beta^4 = \frac{466}{81}, \quad \alpha\beta = \frac{-1}{3}, \quad \alpha^3\beta^3 = \left(\frac{-1}{3}\right)^3 = \frac{-1}{27} $$
we get
$$ r_1r_2 = \frac{ \frac{466}{81} + \frac{-1}{3} + \frac{-1}{27} }{ \frac{-1}{3} } $$
Convert to a common denominator:
$$ \frac{-1}{3} = \frac{-27}{81}, \quad \frac{-1}{27} = \frac{-3}{81} $$
Simplify:
$$ \frac{466}{81} - \frac{27}{81} - \frac{3}{81} = \frac{436}{81} $$
Divide:
$$ r_1r_2 = \frac{\frac{436}{81}}{\frac{-1}{3}} = \frac{436}{81}\cdot(-3) = -\frac{1308}{81} = -\frac{436}{27} $$
Step 3: Form the equation
The quadratic equation is
$$ x^2 - (r_1 + r_2)x + r_1r_2 = 0 $$
Substitute:
$$ x^2 - \left( -\frac{412}{27} \right)x - \frac{436}{27} = 0 $$
Simplify:
$$ x^2 + \frac{412}{27}x - \frac{436}{27} = 0 $$
Multiply through by $27$:
$$ 27x^2 + 412x - 436 = 0 $$
Final Answer
The quadratic equation is
$$ \boxed{ 27x^2 + 412x - 436 = 0 } $$
Question 9
A geometric series $G$ has common ratio $r$ where $r \gt 0$
The third term of $G$ is $\frac{27}{2}$ and the sum of the first three terms of $G$ is $\frac{57}{2}$
Given that the sum to $n$ terms of $G$ is $S_n$
(a) show that $$ S_n = \sum_{j=1}^{n} 4\left(\frac{3}{2}\right)^j $$
[8]
Given that $S_k \gt 50000$
(b) show that the least value of $k$ is given by
$$ k \gt \frac{\lg \left(\frac{12503}{3}\right)}{\lg \left(\frac{3}{2}\right)} $$[3]
(c) Hence find the least value of $k$
[1]
Solution
Solution:
A geometric series $G$ has a common ratio $r$, where $r > 0$. The third term of $G$ is $\frac{27}{2}$, and the sum of the first three terms is $\frac{57}{2}$.
(a) Show that $S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j$
Step 1: General formula for the $n$-th term
The general formula for the $n$-th term of a geometric sequence is:
$$ T_n = ar^{n-1}, $$
where $a$ is the first term and $r$ is the common ratio.
Step 2: Solve for $a$ and $r$
The third term is:
$$ T_3 = ar^2 = \frac{27}{2}. $$
The sum of the first three terms is:
$$ S_3 = a(1 + r + r^2) = \frac{57}{2}. $$
From $T_3$, we have:
$$ ar^2 = \frac{27}{2}. $$
From $S_3$, substitute $a(1 + r + r^2) = \frac{57}{2}$ and divide by $ar^2$:
$$ \frac{1 + r + r^2}{r^2} = \frac{\frac{57}{2}}{\frac{27}{2}} = \frac{57}{27} = \frac{19}{9}. $$
Simplify:
$$ 9 + 9r + 9r^2 = 19r^2. $$
Rearranging terms:
$$ 10r^2 - 9r - 9 = 0 \rightarrow (2r - 3)(5r + 3) = 0. $$
Solve for $r > 0$:
$$ r = \frac{3}{2}. $$
Substitute $r = \frac{3}{2}$ into $ar^2 = \frac{27}{2}$:
$$ a \cdot \frac{9}{4} = \frac{27}{2}. $$
Simplify:
$$ a = 6. $$
Step 3: General sum formula
The sum to $n$ terms is:
$$ S_n = \sum_{j=1}^{n} ar^{j-1}. $$
Substitute $a = 6$ and $r = \frac{3}{2}$:
$$ S_n = \sum_{j=1}^{n} 6 \left(\frac{3}{2}\right)^{j-1}. $$
Thus:
$$ S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j. $$
(b) Show that $k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}$
Step 1: Formula for the sum of a geometric series
For $r > 1$, the sum to $n$ terms is:
$$ S_n = \frac{a(r^n - 1)}{r - 1}. $$
Substitute $a = 6$ and $r = \frac{3}{2}$:
$$ S_n = \frac{6\left(\left(\frac{3}{2}\right)^n - 1\right)}{\frac{3}{2} - 1}. $$
Simplify:
$$ S_n = 12 \left(\left(\frac{3}{2}\right)^n - 1\right). $$
Step 2: Condition $S_k > 50,000$
Given $S_k > 50,000$, substitute $S_k$:
$$ 12 \left(\left(\frac{3}{2}\right)^k - 1\right) > 50,000. $$
Divide through by 12:
$$ \left(\frac{3}{2}\right)^k - 1 > \frac{50,000}{12}. $$
Simplify:
$$ \left(\frac{3}{2}\right)^k > \frac{12,503}{3}. $$
Take the logarithm of both sides:
$$ k \log\left(\frac{3}{2}\right) > \log\left(\frac{12,503}{3}\right). $$
Rearrange for $k$:
$$ k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}. $$
(c) Find the least value of $k$
Step 1: Compute $k$
Simplify $\frac{12,503}{3}$:
$$ \frac{12,503}{3} = 4,167.67. $$
Use approximate logarithm values:
$$ \log(4,167.67) \approx 3.62, \quad \log\left(\frac{3}{2}\right) \approx 0.176. $$
Substitute:
$$ k > \frac{3.62}{0.176} \approx 20.57. $$
The least integer $k$ is:
$$ k = 21. $$
Final Answer
- (a) $S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j$
- (b) $k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}$
- (c) The least value of $k$ is $21$
Question 10
(a) Show that $\frac{9^{3 y}}{243}=3^{(6 y-5)}$
[4]
(b) Solve the simultaneous equations
$$ \begin{aligned} \frac{9^{3 y}}{243} & =27^{(x-2)} \\ \log _{10} \sqrt{6 x y} & =\log _4 2 \end{aligned} $$
[9]
Solution
Solution:
(a) Show that $\frac{9^{3y}}{243} = 3^{6y-5}$
Step 1: Rewrite the base $9$ and $243$ in terms of $3$.
We know that:
$9 = 3^2 \quad \text{and} \quad 243 = 3^5.$
Substitute these into the given expression:
$\frac{9^{3y}}{243} = \frac{(3^2)^{3y}}{3^5}.$
Step 2: Simplify the numerator.
Using the power rule $(a^m)^n = a^{mn}$, we get:
$(3^2)^{3y} = 3^{6y}.$
Thus:
$\frac{9^{3y}}{243} = \frac{3^{6y}}{3^5}.$
Step 3: Simplify the division.
Using the rule $\frac{a^m}{a^n} = a^{m-n}$, we have:
$\frac{3^{6y}}{3^5} = 3^{6y-5}.$
Hence:
$\frac{9^{3y}}{243} = 3^{6y-5}.$
(b) Solve the simultaneous equations
The given equations are:
$\frac{9^{3y}}{243} = 27^{x-2},$
$\log_{10} \sqrt{6xy} = \log_4 2.$
Step 1: Rewrite the first equation.
From part (a), we know:
$\frac{9^{3y}}{243} = 3^{6y-5}.$
Rewrite $27$ as $3^3$:
$27^{x-2} = (3^3)^{x-2}.$
Simplify using the power rule:
$27^{x-2} = 3^{3(x-2)} = 3^{3x-6}.$
Equating the exponents:
$6y - 5 = 3x - 6.$
Rearrange:
$6y = 3x - 1.$
Simplify:
$2y = x - \frac{1}{3}.$
$x = 2y + \frac{1}{3}.$
Step 2: Rewrite the second equation.
The second equation is:
$\log_{10} \sqrt{6xy} = \log_4 2.$
Simplify $\sqrt{6xy}$ as $(6xy)^{1/2}$:
$\log_{10} \sqrt{6xy} = \frac{1}{2} \log_{10}(6xy).$
Thus:
$\frac{1}{2} \log_{10}(6xy) = \log_4 2.$
Multiply through by 2:
$\log_{10}(6xy) = 2 \log_4 2.$
Using the change of base formula $\log_a b = \frac{\log_c b}{\log_c a}$, rewrite $\log_4 2$:
$\log_4 2 = \frac{\log_{10} 2}{\log_{10} 4}.$
Since $\log_{10} 4 = \log_{10}(2^2) = 2\log_{10} 2$, we have:
$\log_4 2 = \frac{\log_{10} 2}{2 \log_{10} 2} = \frac{1}{2}.$
Substitute this back:
$\log_{10}(6xy) = 2 \cdot \frac{1}{2} = 1.$
Thus:
$\log_{10}(6xy) = 1.$
Step 3: Solve for $6xy$.
The equation $\log_{10}(6xy) = 1$ implies:
$6xy = 10^1 = 10.$
Simplify:
$xy = \frac{10}{6} = \frac{5}{3}.$
Step 4: Substitute $x = 2y + \frac{1}{3}$ into $xy = \frac{5}{3}$.
Substitute $x$ into $xy = \frac{5}{3}$:
$\left(2y + \frac{1}{3}\right)y = \frac{5}{3}.$
Expand:
$2y^2 + \frac{1}{3}y = \frac{5}{3}.$
Multiply through by 3:
$6y^2 + y = 5.$
Rearrange:
$6y^2 + y - 5 = 0.$
Step 5: Solve the quadratic equation.
Using the quadratic formula:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$
where $a = 6$, $b = 1$, and $c = -5$:
$y = \frac{-1 \pm \sqrt{1^2 - 4(6)(-5)}}{2(6)}.$
Simplify:
$y = \frac{-1 \pm \sqrt{1 + 120}}{12}.$
$y = \frac{-1 \pm \sqrt{121}}{12}.$
$y = \frac{-1 \pm 11}{12}.$
Solve for $y$:
$y = \frac{10}{12} = \frac{5}{6}, \quad \text{or} \quad y = \frac{-12}{12} = -1.$
$y = \frac{5}{6} \text{ (or) } y=-1.$
Step 6: Solve for $x$, when $y=\frac{5}{6}$.
Substitute $y = \frac{5}{6}$ into $x = 2y + \frac{1}{3}$:
$x = 2\left(\frac{5}{6}\right) + \frac{1}{3}.$
Simplify:
$x = \frac{10}{6} + \frac{2}{6} = \frac{12}{6} = 2.$
Step 7: Solve for $x$, when $y=-1$.
Substitute $y = -1$ into the equation $x = 2y + \frac{1}{3}$:
$x = 2(-1) + \frac{1}{3}.$
Simplify:
$x = -2 + \frac{1}{3}.$
Combine the terms:
$x = \frac{-6}{3} + \frac{1}{3} = \frac{-5}{3}.$
Final Answer:
$x = 2, \quad y = \frac{5}{6}.$
$x = \frac{-5}{3}, \quad y = -1.$
Question 11
Figure 4 shows an open container in the shape of a cylinder with radius $r \mathrm{~cm}$ and height $h \mathrm{~cm}$.
Given that the total surface area of the container is $625\pi \mathrm{~cm}^2$
(a) show that
$$ h=\frac{625-r^2}{2r} $$[3]
The volume of the container is $V \mathrm{~cm}^3$
Given that $r$ can vary,
(b) use calculus to find the value, to 3 significant figures, of $r$ for which $V$ is a maximum. Justify that this value of $r$ gives a maximum value of $V$
[6]
(c) For the value of $r$ found in part (b), find the corresponding value, to 3 significant figures, of $h$
[1]
Solution
Solution:
(a) Show that $h = \frac{625 - r^2}{2r}$
The total surface area $S$ of the open cylinder is given by:
$S = \pi r^2 + 2\pi rh,$
where $\pi r^2$ is the area of the circular base and $2\pi rh$ is the area of the curved surface.
We are given that $S = 625\pi$. Substituting this:
$\pi r^2 + 2\pi rh = 625\pi.$
Divide through by $\pi$:
$r^2 + 2rh = 625.$
Rearrange to solve for $h$:
$2rh = 625 - r^2.$
Divide through by $2r$ (where $r > 0$):
$h = \frac{625 - r^2}{2r}.$
(b) Use calculus to find the value of $r$ that maximizes $V$
The volume $V$ of the cylinder is given by:
$V = \pi r^2 h.$
Substitute $h = \frac{625 - r^2}{2r}$ into the volume formula:
$V = \pi r^2 \left(\frac{625 - r^2}{2r}\right).$
Simplify:
$V = \pi \frac{r^2 (625 - r^2)}{2r}.$
$V = \frac{\pi r (625 - r^2)}{2}.$
Expand:
$V = \frac{\pi}{2} (625r - r^3).$
Differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{\pi}{2} \left(625 - 3r^2\right).$
Set $\frac{dV}{dr} = 0$ to find the critical points:
$625 - 3r^2 = 0.$
Solve for $r^2$:
$3r^2 = 625.$
$r^2 = \frac{625}{3}.$
$r = \sqrt{\frac{625}{3}} = \frac{25}{\sqrt{3}} \approx 14.434.$
Thus:
$r \approx 14.4 \, \text{cm (to 3 significant figures)}.$
Justification that this value of $r$ gives a maximum
Differentiate $\frac{dV}{dr}$ again to find $\frac{d^2V}{dr^2}$:
$\frac{d^2V}{dr^2} = \frac{\pi}{2} (-6r).$
Substitute $r = 14.43$:
$\frac{d^2V}{dr^2} = \frac{\pi}{2} (-6 \cdot 14.43) = \frac{\pi}{2} (-86.4) \lt 0.$
Since $\frac{d^2V}{dr^2} \lt 0$, $V$ has a maximum at $r = 14.4$.
(c) Find the corresponding value of $h$
Using $h = \frac{625 - r^2}{2r}$, substitute $r = 14.43$:
$h = \frac{625 - (14.43)^2}{2 \cdot 14.43}.$
Simplify $14.43^2$:
$14.43^2 = 208.22.$
Substitute:
$h = \frac{625 - 208.22}{28.86}.$
$h = \frac{417.64}{28.8} \approx 14.4 \, \text{cm (to 3 significant figures)}.$
Final Answers:
- (b) $r \approx 14.4 \, \text{cm}$.
- (c) $h \approx 14.4 \, \text{cm}$.
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