Question 1
The $n$th term of an arithmetic series is $a_n$ where
$$ a_{10}+a_{11}+a_{12}=129 \quad \text{and} \quad a_{19}+a_{20}+a_{21}=237 $$Find $a_1$ (4)
Click to view solution of question 1
Solution:
Step 1: Expressing the terms in the arithmetic series
In an arithmetic series, the $n$th term is given by the formula:
$$ a_n = a_1 + (n - 1)d $$where $a_1$ is the first term and $d$ is the common difference.
We are given two equations:
$$ a_{10} + a_{11} + a_{12} = 129 $$and
$$ a_{19} + a_{20} + a_{21} = 237 $$Let's express each of these terms using the general formula for $a_n$.
Step 2: Substituting the expressions for the terms
For the first equation $a_{10} + a_{11} + a_{12} = 129$, we substitute the general formula for each term:
$$ a_{10} = a_1 + 9d, \quad a_{11} = a_1 + 10d, \quad a_{12} = a_1 + 11d $$Thus, the equation becomes:
$$ (a_1 + 9d) + (a_1 + 10d) + (a_1 + 11d) = 129 $$Simplifying:
$$ 3a_1 + 30d = 129 $$Divide through by 3:
$$ a_1 + 10d = 43 \quad \text{(Equation 1)} $$For the second equation $a_{19} + a_{20} + a_{21} = 237$, we substitute the general formula for each term:
$$ a_{19} = a_1 + 18d, \quad a_{20} = a_1 + 19d, \quad a_{21} = a_1 + 20d $$Thus, the equation becomes:
$$ (a_1 + 18d) + (a_1 + 19d) + (a_1 + 20d) = 237 $$Simplifying:
$$ 3a_1 + 57d = 237 $$Divide through by 3:
$$ a_1 + 19d = 79 \quad \text{(Equation 2)} $$Step 3: Solving the system of equations
Now, we solve the system of equations:
$$ a_1 + 10d = 43 \quad \text{(Equation 1)} $$ $$ a_1 + 19d = 79 \quad \text{(Equation 2)} $$Subtract Equation 1 from Equation 2:
$$ (a_1 + 19d) - (a_1 + 10d) = 79 - 43 $$Simplifying:
$$ 9d = 36 $$Solve for $d$:
$$ d = 4 $$Step 4: Finding $a_1$
Substitute $d = 4$ into Equation 1:
$$ a_1 + 10(4) = 43 $$Simplifying:
$$ a_1 + 40 = 43 $$Solving for $a_1$:
$$ a_1 = 43 - 40 = 3 $$Thus, the first term of the arithmetic series is $\boxed{3}$.
Question 2
The point A has coordinates (−5, 3), the point B has coordinates (4, 0) and the point C has coordinates (−1, 5).
The line l passes through C and is perpendicular to AB.
(a) Find an equation of l.
Give your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers.
(4)
The line l intersects AB at the point D.
(b) Show that the coordinates of D are (−2, 2).
(3)
(c) Show that l is not the perpendicular bisector of AB.
(2)
(d) Find the value of $\tan \angle ABC$.
Give your answer in its simplest form.
(4)
Click to view solution of question 2
Solution
Part (a): Find an equation of line $ l $.
The line $ l $ is perpendicular to $ AB $ and passes through point $ C(-1, 5) $. To find the equation of $ l $, we first calculate the slope of line $ AB $.
1. Find the slope of $ AB $:
The slope of a line through two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]For points $ A(-5, 3) $ and $ B(4, 0) $:
\[ m_{AB} = \frac{0 - 3}{4 - (-5)} = \frac{-3}{9} = -\frac{1}{3} \]2. Find the slope of line $ l $:
Since line $ l $ is perpendicular to line $ AB $, the slope of line $ l $, denoted $ m_l $, is the negative reciprocal of $ m_{AB} $:
\[ m_l = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{3}} = 3 \]3. Find the equation of line $ l $:
Using the point-slope form of the equation of a line, $ y - y_1 = m(x - x_1) $, and the point $ C(-1, 5) $, we have:
\[ y - 5 = 3(x + 1) \]Simplifying:
\[ y - 5 = 3x + 3 \quad \Rightarrow \quad y = 3x + 8 \]Rearranging to the form $ ax + by + c = 0 $:
\[ 3x - y + 8 = 0 \]Thus, the equation of line $ l $ is:
\[ 3x - y + 8 = 0 \]Part (b): Show that the coordinates of point $ D $ are $ (-2, 2) $.
The point $ D $ is the intersection of lines $ AB $ and $ l $. We already know the equation of line $ l $ is $ 3x - y + 8 = 0 $. Now, we find the equation of line $ AB $.
1. Find the equation of line $ AB $:
The slope of line $ AB $ is $ m_{AB} = -\frac{1}{3} $. Using the point-slope form of the equation of a line, we get:
\[ y - 3 = -\frac{1}{3}(x + 5) \]Simplifying:
\[ y - 3 = -\frac{1}{3}x - \frac{5}{3} \quad \Rightarrow \quad y = -\frac{1}{3}x + \frac{4}{3} \]Thus, the equation of line $ AB $ is:
\[ y = -\frac{1}{3}x + \frac{4}{3} \]2. Solve the system of equations:
We now solve the system of equations:
\[ y = 3x + 8 \quad \text{(line $ l $)} \] \[ y = -\frac{1}{3}x + \frac{4}{3} \quad \text{(line $ AB $)} \]Set the two expressions for $ y $ equal to each other:
\[ 3x + 8 = -\frac{1}{3}x + \frac{4}{3} \]Multiply through by 3 to eliminate fractions:
\[ 9x + 24 = -x + 4 \]Solve for $ x $:
\[ 9x + x = 4 - 24 \quad \Rightarrow \quad 10x = -20 \quad \Rightarrow \quad x = -2 \]Substitute $ x = -2 $ into the equation of line $ AB $ to find $ y $:
\[ y = -\frac{1}{3}(-2) + \frac{4}{3} = \frac{2}{3} + \frac{4}{3} = 2 \]Thus, the coordinates of $ D $ are $ (-2, 2) $.
Part (c): Show that line $ l $ is not the perpendicular bisector of $ AB $.
To be the perpendicular bisector, line $ l $ must pass through the midpoint of $ AB $ and be perpendicular to $ AB $. We already know that line $ l $ is perpendicular to $ AB $, but we need to check if it passes through the midpoint of $ AB $.
1. Find the midpoint of $ AB $:
The midpoint $ M $ of a line segment with endpoints $ A(x_1, y_1) $ and $ B(x_2, y_2) $ is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]For $ A(-5, 3) $ and $ B(4, 0) $, the midpoint is:
\[ M = \left( \frac{-5 + 4}{2}, \frac{3 + 0}{2} \right) = \left( \frac{-1}{2}, \frac{3}{2} \right) \]2. Check if the midpoint lies on line $ l $:
The equation of line $ l $ is $ y = 3x + 8 $. Substituting $ x = -\frac{1}{2} $ into the equation:
\[ y = 3\left(-\frac{1}{2}\right) + 8 = -\frac{3}{2} + 8 = \frac{13}{2} \]Since the $ y $-coordinate of the midpoint is $ \frac{3}{2} $, which is not equal to $ \frac{13}{2} $, point $ M $ does not lie on line $ l $.
Therefore, line $ l $ is not the perpendicular bisector of $ AB $.
Part (d): Find the value of $ \tan \angle ABC $.
To find $ \tan \angle ABC $, we use the lengths of sides $ AB $, $ BC $, and $ AC $.
1. Find the length of $ AB $:
\[ AB = \sqrt{(4 - (-5))^2 + (0 - 3)^2} = \sqrt{(4 + 5)^2 + (-3)^2} = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \]2. Find the length of $ BC $:
\[ BC = \sqrt{(-1 - 4)^2 + (5 - 0)^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]3. Find the length of $ AC $:
\[ AC = \sqrt{(-1 - (-5))^2 + (5 - 3)^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]4. Use the law of cosines to find $ \cos \angle ABC $:
\[ \cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} \] \[ \cos \angle ABC = \frac{(3\sqrt{10})^2 + (5\sqrt{2})^2 - (2\sqrt{5})^2} {2 \times 3\sqrt{10} \times 5\sqrt{2}} = \frac{120}{30\sqrt{20}} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \]5. Find $ \tan \angle ABC $:
\[ \tan \angle ABC = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]Question 3
Curve C has equation $y=\frac{ax+3}{1-2x}$ where $x\neq\frac{1}{2}$ and $a$ is a constant.
The asymptote to C that is parallel to the $x$-axis has equation $y=4$.
(a) Find the value of $a$.
(2)(b) Write down the equation of the asymptote to C that is parallel to the $y$-axis. (1)
(c) Find the coordinates of the point where C crosses
(i) the $x$-axis,
(ii) the $y$-axis. (2)
(d) Using the axes below, sketch $C$, showing clearly the asymptotes and the coordinates of the points where $C$ crosses the coordinate axes. (4)
Click to view solution of question 3
Solution:
Step (a) Find the value of $a$
We are given the curve equation $y=\frac{ax+3}{1-2x}$, where $x\neq\frac{1}{2}$. The curve has an asymptote that is parallel to the $x$-axis with the equation $y=4$.
As $x\to\infty$, the value of $y$ approaches the horizontal asymptote. To find the asymptote, we consider the behavior of $y$ for large values of $x$. For large $x$, the numerator $ax+3$ behaves like $ax$ and the denominator $1-2x$ behaves like $-2x$. Therefore, we have:
$ y\approx\frac{ax}{-2x}=-\frac{a}{2}. $
For the horizontal asymptote to be $y=4$, we set:
$ -\frac{a}{2}=4. $
Solving for $a$, we get:
$ a=-8. $
Thus, the value of $a$ is $\boxed{-8}$.
Step (b) Write down the equation of the asymptote to $C$ that is parallel to the $y$-axis
Next, we need to find the asymptote that is parallel to the $y$-axis. Vertical asymptotes occur when the denominator of the function becomes zero. The denominator of the given curve is $1-2x$, so we set it equal to zero to find the $x$-value of the vertical asymptote:
$ 1-2x=0 \quad\Rightarrow\quad x=\frac{1}{2}. $
Therefore, the equation of the vertical asymptote is:
$ x=\frac{1}{2}. $
Step (c) Find the coordinates of the point where $C$ crosses
(i) The $x$-axis
The curve crosses the $x$-axis where $y=0$. Setting $y=0$ in the equation of the curve:
$ 0=\frac{ax+3}{1-2x}. $
This implies that the numerator must be zero:
$ ax+3=0 \quad\Rightarrow\quad x=-\frac{3}{a}. $
Substituting $a=-8$, we get:
$ x=-\frac{3}{-8}=\frac{3}{8}. $
Thus, the point where the curve crosses the $x$-axis is $\left(\frac{3}{8},0\right)$.
(ii) The $y$-axis
The curve crosses the $y$-axis where $x=0$. Substituting $x=0$ into the equation of the curve:
$ y=\frac{a(0)+3}{1-2(0)} =\frac{3}{1} =3. $
Thus, the point where the curve crosses the $y$-axis is $(0,3)$.
Step (d) Sketch of the Curve
- The curve has a horizontal asymptote at $y=4$.
- The curve has a vertical asymptote at $x=\frac{1}{2}$.
- The curve crosses the $x$-axis at $\left(\frac{3}{8},0\right)$ and the $y$-axis at $(0,3)$.
Question 4
Problem:
$f(x) = x^3 + px^2 + qx + 6$ where $p$ and $q$ are constants.
Given that $(x - 1)$ is a factor of $f(x)$ and that when $f(x)$ is divided by $(x + 1)$, the remainder is 8.
(a)
(i) Show that $p = -2$
(ii) Find the value of $q$
(6)
(b)
Hence, solve the equation $f(x) = 0$
(3 marks)
Click to view solution of question 4
Problem Solution
We are given the polynomial:
$$ f(x) = x^3 + px^2 + qx + 6 $$
where $p$ and $q$ are constants. It is given that $(x - 1)$ is a factor of $f(x)$, and when $f(x)$ is divided by $(x + 1)$, the remainder is 8.
Part (a) (i): Show that $p = -2$
Since $(x - 1)$ is a factor of $f(x)$, by the Factor Theorem, we know:
$$ f(1) = 0 $$
Substitute $x = 1$ into $f(x)$:
$$ f(1) = 1^3 + p(1)^2 + q(1) + 6 = 1 + p + q + 6 = 0 $$
This simplifies to:
$$ p + q + 7 = 0 \quad \Rightarrow \quad p + q = -7 \quad \text{(Equation 1)} $$
Part (a) (ii): Find the value of $q$
Next, we are told that when $f(x)$ is divided by $(x + 1)$, the remainder is 8. By the Remainder Theorem, we know:
$$ f(-1) = 8 $$
Substitute $x = -1$ into $f(x)$:
$$ f(-1) = (-1)^3 + p(-1)^2 + q(-1) + 6 = -1 + p - q + 6 = 8 $$
Simplifying:
$$ p - q + 5 = 8 \quad \Rightarrow \quad p - q = 3 \quad \text{(Equation 2)} $$
Solving for $p$ and $q$
We now solve the system of equations:
$$ p + q = -7 \quad \text{(Equation 1)} $$
$$ p - q = 3 \quad \text{(Equation 2)} $$
Add these two equations:
$$ (p + q) + (p - q) = -7 + 3 $$
$$ 2p = -4 \quad \Rightarrow \quad p = -2 $$
Substitute $p = -2$ into Equation 1:
$$ -2 + q = -7 \quad \Rightarrow \quad q = -5 $$
Part (b): Solve the equation $f(x) = 0$
Now that we know $p = -2$ and $q = -5$, the polynomial becomes:
$$ f(x) = x^3 - 2x^2 - 5x + 6 $$
Since $(x - 1)$ is a factor, divide $f(x)$ by $(x - 1)$ using synthetic division:
$$ \begin{array}{r|rrrr} 1 & 1 & -2 & -5 & 6 \\ & & 1 & -1 & -6 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array} $$
The quotient is $x^2 - x - 6$, and the remainder is 0, confirming that $x - 1$ is a factor. Thus:
$$ f(x) = (x - 1)(x^2 - x - 6) $$
Factor $x^2 - x - 6$:
$$ x^2 - x - 6 = (x - 3)(x + 2) $$
Thus:
$$ f(x) = (x - 1)(x - 3)(x + 2) $$
The solutions to $f(x) = 0$ are:
$$ x = 1,\; x = 3,\; x = -2 $$
Question 5
Given that $k$ is a non‑zero constant curve $C$ has equation $kx^2 – xy + (k + 1)x = 1$ straight line $l$ has equation $y = (k/2)x + 1$ The point $A$ is the only point that lies on both $C$ and $l.$
(a) Find the value of $k$ (6)
(b) Hence, find the coordinates of $A.$ (2)
Click to view solution of question 5
Problem Solution
We are given:
- The curve $C$ with the equation: $$ kx^2 - xy + (k + 1)x = 1 $$
- The straight line $l$ with the equation: $$ y = \frac{k}{2}x + 1 $$
The point $A$ is the only point that lies on both $C$ and $l$.
Part (a): Find the value of $k$
Substitute $y = \frac{k}{2}x + 1$ from the equation of $l$ into $C$:
$$ kx^2 - x\left(\frac{k}{2}x + 1\right) + (k + 1)x = 1 $$Simplify:
$$ kx^2 - \frac{k}{2}x^2 - x + (k + 1)x = 1 $$Combine like terms:
$$ \frac{k}{2}x^2 + kx = 1 $$Factorize:
$$ x\left(\frac{k}{2}x + k\right) = 1 $$Divide through by $x$ (since $x \neq 0$):
$$ \frac{k}{2}x + k = \frac{1}{x} $$Multiply through by $x$:
$$ kx^2 + 2kx - 2 = 0 $$For $A$ to be the only point of intersection, the discriminant must be zero:
$$ \Delta = b^2 - 4ac $$Here, $a = k$, $b = 2k$, and $c = -2$:
$$ \Delta = (2k)^2 - 4(k)(-2) = 4k^2 + 8k $$Set $\Delta = 0$:
$$ 4k^2 + 8k = 0 $$Factorize:
$$ 4k(k + 2) = 0 $$Since $k \neq 0$, we have:
$$ k + 2 = 0 \quad \Rightarrow \quad k = -2 $$Part (b): Find the coordinates of $A$
Substitute $k = -2$ into the straight line equation:
$$ y = \frac{-2}{2}x + 1 \quad \Rightarrow \quad y = -x + 1 $$Substitute $k = -2$ into the curve equation:
$$ -2x^2 - xy - x = 1 $$Substitute $y = -x + 1$ into the curve equation:
$$ -2x^2 - x(-x + 1) - x = 1 $$Simplify:
$$ -2x^2 + x^2 - x - x = 1 $$ $$ - x^2 - 2x = 1 $$Rearrange:
$$ x^2 + 2x + 1 = 0 $$Factorize:
$$ (x + 1)^2 = 0 $$ $$ x = -1 $$Substitute $x = -1$ into $y = -x + 1$:
$$ y = -(-1) + 1 = 2 $$Thus, the coordinates of $A$ are:
$$ A = (-1, 2) $$Question 6
Given that $$ (8 + 3x)^{\frac{1}{3}} $$ can be expressed in the form $$ p(1 + qx)^{\frac{1}{3}} $$ where $p$ and $q$ are constants,
(a) find the value of $p$ and the value of $q$
(2)
(b) Hence, expand $$ (8 + 3x)^{\frac{1}{3}} $$ in ascending powers of $x$ up to and including the term in $x^2$, expressing each coefficient as an exact fraction in its lowest terms.
(3)
Using the expansion found in part (b) with a suitable value of $x$
(c) show that $$ \sqrt[3]{9} \approx \frac{599}{288} $$
(2)
Click to view solution of question 6
Problem and Solution
We are given:
$$ (8 + 3x)^{\frac{1}{3}} = p(1 + qx)^{\frac{1}{3}} $$where $p$ and $q$ are constants.
Part (a): Finding $p$ and $q$
Factorize $8$ from $(8 + 3x)^{\frac{1}{3}}$:
$$ (8 + 3x)^{\frac{1}{3}} = 8^{\frac{1}{3}} \left(1 + \frac{3x}{8}\right)^{\frac{1}{3}} $$Since $8^{\frac{1}{3}} = 2$, we can write:
$$ (8 + 3x)^{\frac{1}{3}} = 2 \left(1 + \frac{3x}{8}\right)^{\frac{1}{3} } $$Comparing with $p(1 + qx)^{\frac{1}{3}}$, we find:
$$ p = 2, \quad q = \frac{3}{8}. $$Part (b): Expansion of $(8 + 3x)^{\frac{1}{3}$
Using the Binomial Theorem for fractional powers, expand $(1 + qx)^{\frac{1}{3}}$ up to the $x^2$ term:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{\frac{1}{3} \cdot qx}{1!} + \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) (qx)^2}{2!} + \dots $$Substitute $q = \frac{3}{8}$:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{\frac{1}{3} \cdot \frac{3}{8}x}{1} + \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) \left(\frac{3}{8}x\right)^2}{2} $$Simplify:
$$ \text{Linear term:} \quad \frac{\frac{1}{3} \cdot \frac{3}{8}x}{1} = \frac{1}{8}x $$ $$ \text{Quadratic term:} \quad \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) \left(\frac{3}{8}x\right)^2}{2} = \frac{\frac{1}{3} \cdot \left(-\frac{2}{3}\right) \cdot \frac{9}{64}x^2}{2} = -\frac{1}{64}x^2 $$Thus:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{1}{8}x - \frac{1}{64}x^2 $$Now multiply by $p = 2$:
$$ (8 + 3x)^{\frac{1}{3}} = 2 \left(1 + \frac{1}{8}x - \frac{1}{64}x^2\right) = 2 + \frac{1}{4}x - \frac{1}{32}x^2 $$Part (c): Approximation of $\sqrt[3]{9}$
To approximate $\sqrt[3]{9}$, use the expansion with $x = \frac{1}{3}$ (so that $8 + 3x = 9$):
$$ (8 + 3x)^{\frac{1}{3}} \approx 2 + \frac{1}{4}\left(\frac{1}{3}\right) - \frac{1}{32}\left(\frac{1}{3}\right)^2 $$Compute each term:
$$ \text{Constant term:} \quad 2 $$ $$ \text{Linear term:} \quad \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} $$ $$ \text{Quadratic term:} \quad \frac{1}{32} \cdot \frac{1}{9} = \frac{1}{288} $$Add these terms:
$$ \sqrt[3]{9} \approx 2 + \frac{1}{12} - \frac{1}{288} $$Convert all terms to have a denominator of $288$:
$$ 2 = \frac{576}{288}, \quad \frac{1}{12} = \frac{24}{288}, \quad \frac{1}{288} = \frac{1}{288} $$Add them:
$$ \sqrt[3]{9} \approx \frac{576 + 24 - 1}{288} = \frac{599}{288} $$Final Answers
- $p = 2$, $q = \frac{3}{8}$
- Expansion: $$ (8 + 3x)^{\frac{1}{3}} \approx 2 + \frac{1}{4}x - \frac{1}{32}x^2 $$
- Approximation: $$ \sqrt[3]{9} \approx \frac{599}{288} $$
Question 7
(a) Complete the table of values for
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
giving each value to 2 decimal places where appropriate.
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline\\x&-6&-5&-4&-3&-2&-1&0\\ \hline y&4&3.59&3.26&&&&2.5\\ \hline\end{array}$$(2)
(b) On the grid opposite, draw the graph of
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
for $-6 \leqslant x \leqslant 0$
(2)
(c) By drawing a suitable straight line on the grid, obtain an estimate, to one decimal place, of the root of the equation
$\log_2(2x+2)^3 + x + 3 = 0$
in the interval $-6 \leqslant x \leqslant 0$
(6)
Click to view solution of question 7
Solution
Step (a) Complete the table of values for
The given function is
$y = 0.5^{\left(\frac{x}{3}+1\right)} + 2$
We are tasked with completing the table for values of $x$ from $-6$ to $0$. Using the formula for $y$, we calculate the corresponding values of $y$ for each $x$:
For $x = -6$:
$y = 0.5^{\left(\frac{-6}{3}+1\right)} + 2 = 0.5^{(-2+1)} + 2 = 0.5^{-1} + 2 = 2 + 2 = 4$
For $x = -5$:
$y = 0.5^{\left(\frac{-5}{3}+1\right)} + 2 = 0.5^{(-1.6667+1)} + 2 = 0.5^{-0.6667} + 2 \approx 0.792 + 2 = 3.59$
For $x = -4$:
$y = 0.5^{\left(\frac{-4}{3}+1\right)} + 2 = 0.5^{(-1.3333+1)} + 2 = 0.5^{-0.3333} + 2 \approx 0.7937 + 2 = 3.26$
For $x = -3$:
$y = 0.5^{\left(\frac{-3}{3}+1\right)} + 2 = 0.5^{(0)} + 2 = 1 + 2 = 3$
For $x = -2$:
$y = 0.5^{\left(\frac{-2}{3}+1\right)} + 2 = 0.5^{(0.3333)} + 2 \approx 0.7937 + 2 = 2.79$
For $x = -1$:
$y = 0.5^{\left(\frac{-1}{3}+1\right)} + 2 = 0.5^{(0.6667)} + 2 \approx 0.6299 + 2 = 2.6299$
For $x = 0$:
$y = 0.5^{\left(\frac{0}{3}+1\right)} + 2 = 0.5^{(1)} + 2 = 0.5 + 2 = 2.5$
Thus, the completed table of values is:
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline\\x&-6&-5&-4&-3&-2&-1&0\\ \hline y&4&3.59&3.26&3&2.79&2.63&2.5\\ \hline\end{array}$$Step (b) Graph of $y = 0.5^{\left(\frac{x}{3}\right)} + 2$ for $-6 \leq x \leq 0$
Now, we are tasked with drawing the graph of the function $y = 0.5^{\left(\frac{x}{3}\right)} + 2$ for values of $x$ between $-6$ and $0$.
Plot the points from the completed table and join them smoothly to form the curve. The graph is a decreasing exponential curve approaching $y = 2$.
Step (c) Estimating the root of the equation
We are tasked with finding an estimate of the root of the equation
$\log_2\left((2x+2)^3\right) + x + 3 = 0$
This equation is equivalent to
$\log_2\left((2x+2)^3\right) = -x - 3$
$3\log_2(2x+2) = -x - 3$
$\log_2(2x+2) = -\frac{x}{3} - 1$
$2x+2 = 2^{-\frac{x}{3}-1}$
$2x+2 = 2^{-1\left(\frac{x}{3}+1\right)}$
$2x+2 = 0.5^{\frac{x}{3}+1}$
$2x+4 = 0.5^{\frac{x}{3}+1} + 2$
This equation simplifies to the form $y = 2x + 4$ and the corresponding function is $y = 0.5^{\frac{x}{3}+1} + 2$.
To find the root of the equation, we plot both functions $y = 2x + 4$ and $y = 0.5^{\frac{x}{3}+1} + 2$ on the same grid and determine their point of intersection.
From the graph, we can estimate the point of intersection of the two functions. Based on the graph, the approximate value of $x$ where both curves intersect is $x \approx -0.74$.
Thus, the root of the equation is approximately $x = -0.7$, correct to one decimal place.
Question 8
Figure 1 shows a badge, shown shaded, made from two identical rectangles, ABCD and DEFG, and a sector DCG of a circle with centre D.
Each rectangle measures $x\text{ cm}$ by $y\text{ cm}$.
The radius of the sector is $x\text{ cm}$ and the angle $\angle CDG$ is $0.5$ radians.
The area of the badge is $50\text{ cm}^2$
The perimeter of the badge is $P\text{ cm}$.
(a) Show that
$$ P = 2x + \frac{100}{x} $$
(5)
Given that $x$ can vary,
(b) use calculus, to find the exact value of $x$ for which $P$ is a minimum.
Justify that this value of $x$ gives a minimum value for $P$
(6)
(c) Find the minimum value of $P$
Give your answer in the form $k\sqrt{2}$, where $k$ is an integer to be found.
(2)
Click to view solution of question 8
Solution
Derivation of Perimeter $P$
The badge consists of two identical rectangles $ABCD$ and $DEFG$, and a sector $DCG$ of a circle with center $D$.
- Each rectangle has dimensions $x\ \text{cm}$ by $y\ \text{cm}$.
- The radius of the sector is $x\ \text{cm}$.
- The angle of the sector $\angle CDG = 0.5\ \text{radians}$.
- The total area of the badge is given as $50\ \text{cm}^2$.
- The perimeter of the badge is $P\ \text{cm}$.
Step 1: Expression for Total Area
The area of the badge includes:
-
The areas of the two rectangles:
$$ 2 \times \text{Area of one rectangle} = 2(x \cdot y) = 2xy $$
-
The area of the sector:
$$ \text{Area of sector} = \frac{1}{2} \cdot x^2 \cdot 0.5 = 0.25x^2 $$
Thus, the total area is:
$$ \text{Total Area} = 2xy + 0.25x^2 $$
Given that the total area is $50\ \text{cm}^2$, we have:
$$ 2xy + 0.25x^2 = 50 $$
Step 2: Expression for Perimeter $P$
The perimeter of the badge includes:
- The two vertical sides of the rectangles: $2y$
- The two horizontal sides of the rectangles (excluding the part covered by the arc): $x + x = 2x$
-
The arc of the sector:
$$ \text{Arc length} = x \cdot 0.5 = 0.5x $$
Thus, the total perimeter is:
$$ P = 2y + 2x + 0.5x = 2y + 2.5x $$
From the area equation, solve for $y$:
$$ y = \frac{50 - 0.25x^2}{2x} $$
Substitute $y$ into the expression for $P$:
$$ P = 2\left(\frac{50 - 0.25x^2}{2x}\right) + 2.5x $$
Simplify:
$$ P = \frac{50 - 0.25x^2}{x} + 2.5x $$
$$ P = \frac{50}{x} - \frac{0.25x^2}{x} + 2.5x $$
$$ P = \frac{50}{x} + 2x $$
Conclusion
The perimeter of the badge is:
$$ P = 2x + \frac{100}{x} $$
(b) Using Calculus to Minimize $P$
The expression for $P$ is:
$$ P = 2x + \frac{100}{x} $$
To find the value of $x$ that minimizes $P$, we calculate the derivative of $P$ with respect to $x$ and set it equal to zero:
$$ \frac{dP}{dx} = 2 - \frac{100}{x^2} $$
Set $\frac{dP}{dx} = 0$:
$$ 2 - \frac{100}{x^2} = 0 $$
$$ \frac{100}{x^2} = 2 $$
$$ x^2 = \frac{100}{2} = 50 $$
$$ x = \sqrt{50} = 5\sqrt{2} \quad (\text{since } x > 0) $$
Justification for Minimum
To confirm that $P$ has a minimum at $x = 5\sqrt{2}$, we compute the second derivative of $P$:
$$ \frac{d^2P}{dx^2} = \frac{200}{x^3} $$
Since $x > 0$, $\frac{d^2P}{dx^2} > 0$, indicating that $P$ has a local minimum at $x = 5\sqrt{2}$.
(c) Finding the Minimum Value of $P$
Substitute $x = 5\sqrt{2}$ into the expression for $P$:
$$ P = 2x + \frac{100}{x} $$
$$ P = 2(5\sqrt{2}) + \frac{100}{5\sqrt{2}} $$
$$ P = 10\sqrt{2} + \frac{100}{5\sqrt{2}} $$
$$ P = 10\sqrt{2} + \frac{20}{\sqrt{2}} $$
$$ P = 10\sqrt{2} + 10\sqrt{2} $$
$$ P = 20\sqrt{2} $$
Conclusion
The minimum value of $P$ is:
$$ P_{\text{min}} = 20\sqrt{2} $$
where $k = 20$.
Question 9
Giving each value in your solution to 2 decimal places, solve the simultaneous equations
$$ \begin{aligned} e^{2y}-x+2&=0 \\ \ln(x+3)-2y-1&=0 \end{aligned} $$(8)
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Solution
We are given the system of equations:
$$ e^{2y} - x + 2 = 0 \tag{1} $$
$$ \ln(x + 3) - 2y - 1 = 0 \tag{2} $$
Step 1: Solve equation (1) for $x$
From equation (1):
$$ e^{2y} - x + 2 = 0 $$
Rearranging for $x$:
$$ x = e^{2y} + 2 \tag{3} $$
Step 2: Substitute equation (3) into equation (2)
Substitute the expression for $x$ from equation (3) into equation (2):
$$ \ln(e^{2y} + 5) - 2y - 1 = 0 $$
Simplifying:
$$ \ln(e^{2y} + 5) = 2y + 1 $$
$$ e^{2y} + 5 = e^{2y + 1} = e^{2y}e $$
$$ e^{2y}(e - 1) = 5 \tag{4} $$
Step 3: Solve equation (4) numerically
Numerically solving equation (4) gives:
$$ 2y = \ln\left(\frac{5}{e - 1}\right) \rightarrow y \approx 0.53 $$
Step 4: Solve for $x$
Substitute the value of $y$ into equation (3):
$$ x = e^{2(0.53)} + 2 \approx 4.91 $$
Thus, the solutions are:
$$ x \approx 4.91, \quad y \approx 0.53 $$
Question 10
The region $R$, shown shaded in Figure 2, is bounded by the curve with equation
$$ y=x^2+1 $$and the curve with equation
$$ x^2+y^2=11 $$The two curves intersect at the point $A$ and at the point $B$.
(a) Find the $x$ coordinate of the point $A$ and the $x$ coordinate of the point $B$. (4)
The region $R$ is rotated through $360^\circ$ about the $x$-axis.
(b) Use algebraic integration to find the volume, to 2 decimal places, of the solid generated. (5)
Click to view solution of question 10
Solution
Step 1: Finding the $x$ coordinates of the points of intersection
We are given two curves:
$$ y = x^2 + 1 \quad \text{and} \quad x^2 + y^2 = 11. $$
To find the $x$ coordinates of the points of intersection, substitute the expression for $y$ from the first equation into the second equation:
$$ x^2 + (x^2 + 1)^2 = 11. $$
Expanding the square:
$$ x^2 + (x^4 + 2x^2 + 1) = 11. $$
Simplifying:
$$ x^4 + 3x^2 + 1 = 11. $$
Subtracting 11 from both sides:
$$ x^4 + 3x^2 - 10 = 0. $$
Let $z = x^2$. Then the equation becomes:
$$ z^2 + 3z - 10 = 0. $$
Solve this quadratic equation using the quadratic formula:
$$ z = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2}. $$
Thus:
$$ z = \frac{-3 \pm 7}{2}. $$
The two possible solutions for $z$ are:
$$ z = \frac{-3 + 7}{2} = 2 \quad \text{or} \quad z = \frac{-3 - 7}{2} = -5. $$
Since $z = x^2$, and $x^2$ cannot be negative, we take $z = 2$. Therefore, $x^2 = 2$, giving:
$$ x = \pm \sqrt{2}. $$
Thus, the $x$ coordinates of the points of intersection are $x = -\sqrt{2}$ and $x = \sqrt{2}$.
Step 2: Finding the volume of the solid generated
The region $R$ is rotated about the $x$-axis. The volume of the solid can be calculated using the method of cylindrical shells or disks. Here, we use the disk method.
The volume of the solid generated by rotating the region between $y = x^2 + 1$ and the circle $x^2 + y^2 = 11$ about the $x$-axis is given by:
$$ V = \pi \int_{-\sqrt{2}}^{\sqrt{2}} \left[ (11 - x^2) - (x^2 + 1)^2 \right] \, dx. $$
First, simplify the integrand:
$$ (x^2 + 1)^2 = x^4 + 2x^2 + 1, $$
and
$$ 11 - x^2 = 11 - x^2. $$
Thus, the integrand becomes:
$$ -(x^4 + 2x^2 + 1) + (11 - x^2) = -(x^4 + 3x^2 - 10). $$
Therefore, the volume integral is:
$$ V = \pi \int_{-\sqrt{2}}^{\sqrt{2}} -(x^4 + 3x^2 - 10) \, dx. $$
Since the integrand is an even function, we can simplify the integral by doubling the integral from $0$ to $\sqrt{2}$:
$$ V = 2\pi \int_0^{\sqrt{2}} -(x^4 + 3x^2 - 10) \, dx. $$
Now, integrate term by term:
$$ \int x^4 \, dx = \frac{x^5}{5}, \quad \int 3x^2 \, dx = x^3, \quad \int -10 \, dx = -10x. $$
Thus, the volume integral becomes:
$$ V = -2\pi \left[ \frac{x^5}{5} + x^3 - 10x \right]_0^{\sqrt{2}}. $$
Evaluating at the upper limit $x = \sqrt{2}$:
$$ -\frac{(\sqrt{2})^5}{5} - (\sqrt{2})^3 + 10(\sqrt{2}) = -\frac{4\sqrt{2}}{5} - 2\sqrt{2} + 10\sqrt{2}. $$
Simplifying:
$$ -\frac{4\sqrt{2}}{5} - 2\sqrt{2} + 10\sqrt{2} = \left( -\frac{4}{5} - 2 + 10 \right)\sqrt{2} = \left( -\frac{4}{5} + 8 \right)\sqrt{2} = \frac{36}{5}\sqrt{2}. $$
Thus, the volume is:
$$ V = 2\pi \left( \frac{36}{5}\sqrt{2} \right) = \frac{72}{5}\pi\sqrt{2} = 63.977. $$
The final answer for the volume is approximately:
$$ V \approx 63.98 \, \text{cubic units}. $$
Question 11
Figure 3 shows triangle $OAB$ with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$
$M$ is the midpoint of $OA$.
$N$ is the point on $OB$ such that
$$ ON:NB=3:1 $$The lines $AN$ and $BM$ intersect at the point $X$.
(a) Find expressions, in terms of $\mathbf{a}$ and $\mathbf{b}$, for
$$ \text{(i)}\ \overrightarrow{AN} \qquad\qquad \text{(ii)}\ \overrightarrow{BM} $$(3)
(b) Using a vector method, find $AX:XN$ (7)
Click to view solution of question 11
Solution
Given: Triangle $OAB$ with vectors $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Point $M$ is the midpoint of $OA$, and point $N$ divides $OB$ in the ratio $ON : NB = 3 : 1$. The lines $AN$ and $BM$ intersect at point $X$.
(a) Find expressions for:
(i) $\vec{AN}$:
Point $N$ divides $OB$ in the ratio $ON : NB = 3 : 1$. Therefore, the position vector of point $N$ is:
$$ \vec{ON} = \frac{3}{3+1}\vec{OB} = \frac{3}{4}\mathbf{b}. $$
The vector $\vec{AN}$ is the vector from $A$ to $N$, so:
$$ \vec{AN} = \vec{ON} - \vec{OA} = \frac{3}{4}\mathbf{b} - \mathbf{a}. $$
Thus, the expression for $\vec{AN}$ is:
$$ \boxed{\vec{AN} = \frac{3}{4}\mathbf{b} - \mathbf{a}} $$
(ii) $\vec{BM}$:
Since $M$ is the midpoint of $OA$, the position vector of $M$ is:
$$ \vec{OM} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}. $$
The vector $\vec{BM}$ is the vector from $B$ to $M$, so:
$$ \vec{BM} = \vec{OM} - \vec{OB} = \frac{1}{2}\mathbf{a} - \mathbf{b}. $$
Thus, the expression for $\vec{BM}$ is:
$$ \boxed{\vec{BM} = \frac{1}{2}\mathbf{a} - \mathbf{b}} $$
(b) Find $AX : XN$ using vector methods:
To solve this, we will use the fact that $\vec{AB} = \vec{AX} + \vec{XB}$, where $X$ divides $AB$ in some ratio.
Let point $X$ divide $AN$ in the ratio $\lambda : 1$. The position vector of $X$ along $AN$ is:
$$ \vec{AX} = \lambda \vec{AN} = \lambda\left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right). $$
Let point $X$ divide $BM$ in the ratio $\mu : 1$. The position vector of $X$ along $BM$ is:
$$ \vec{BX} = \mu \vec{BM} = \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right). $$
Since the point $X$ is the same in both cases, we use the equation $\vec{AB} = \vec{AX} + \vec{XB}$, where $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$.
Now, we use the relationship:
$$ \vec{AB} = \vec{AX} + \vec{XB}. $$
Substitute for $\vec{AX}$ and $\vec{XB}$:
$$ \mathbf{b} - \mathbf{a} = \lambda\left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right) - \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right). $$
Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$, we solve for $\lambda$ and $\mu$:
For $\mathbf{b}$:
$$ 1 = \lambda\left(\frac{3}{4}\right) + \mu. $$
For $\mathbf{a}$:
$$ -1 = -\lambda - \mu\left(\frac{1}{2}\right). $$
Solving this system of equations gives:
$$ \lambda = \frac{4}{5}, \quad \mu = \frac{2}{5}. $$
Thus, the ratio $AX : XN$ is:
$$ \boxed{AX : XN = 4 : 1} $$
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