Question 1
Given that
$ \frac{a - \sqrt{48}}{\sqrt{3} + 1} $
can be written in the form $ b\sqrt{3} - 9 $ where $ a $ and $ b $ are integers,
find the value of $ a $ and the value of $ b $.
Show your working clearly. (4)
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Solution
We want to rewrite the expression
$$ \frac{a - \sqrt{48}}{\sqrt{3} + 1} $$
in the form $b\sqrt{3} - 9$, where $a$ and $b$ are integers.
Step 1: Simplify the square root and the fraction
First, simplify $\sqrt{48}$:
$$ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} $$
Thus, the expression becomes:
$$ \frac{a - 4\sqrt{3}}{\sqrt{3} + 1} $$
Step 2: Rationalize the denominator
To rationalize the denominator, multiply numerator and denominator by $\sqrt{3} - 1$:
$$ \frac{a - 4\sqrt{3}}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} $$
Simplify the denominator:
$$ (\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 $$
The numerator becomes:
$$ (a - 4\sqrt{3})(\sqrt{3} - 1) $$
Expand this:
$$ (a - 4\sqrt{3})(\sqrt{3} - 1) = a\sqrt{3} - a - 4\sqrt{3}\cdot\sqrt{3} + 4\sqrt{3} $$
Simplify each term:
$$ a\sqrt{3} - a - 4(3) + 4\sqrt{3} = a\sqrt{3} - a - 12 + 4\sqrt{3} $$
Combine like terms:
$$ (a + 4)\sqrt{3} - (a + 12) $$
Thus, the numerator becomes:
$$ (a + 4)\sqrt{3} - (a + 12) $$
Step 3: Simplify the fraction
Divide the numerator by 2 (the denominator):
$$ \frac{(a + 4)\sqrt{3} - (a + 12)}{2} = \frac{a + 4}{2}\sqrt{3} - \frac{a + 12}{2} $$
Step 4: Match the form $b\sqrt{3} - 9$
We are told the expression can be written as:
$$ b\sqrt{3} - 9 $$
Equating coefficients, we get:
$$ \frac{a + 4}{2} = b \quad \text{and} \quad \frac{a + 12}{2} = 9 $$
Step 5: Solve for $a$ and $b$
From
$$ \frac{a + 12}{2} = 9 $$
we get:
$$ a + 12 = 18 \implies a = 6 $$
Substitute $a = 6$ into
$$ \frac{a + 4}{2} = b $$
$$ \frac{6 + 4}{2} = b \implies \frac{10}{2} = b \implies b = 5 $$
Final Answer
$$ a = 6, \quad b = 5 $$
Question 2
In $ \triangle ABC $,
$ \angle BAC = 50^\circ,\quad AB = 10 \, \text{cm},\quad BC = 9 \, \text{cm}. $
Given that $ \angle BCA = x^\circ $,
find the two possible values, to one decimal place, of $ x $. (3)
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Solution
In $\triangle ABC$, we are given the following:
$$ \angle BAC = 50^\circ, \quad AB = 10\ \text{cm}, \quad BC = 9\ \text{cm}. $$
We are tasked with finding the two possible values of $\angle BCA = x^\circ$ to one decimal place.
Solution
We will use the sine rule, which states:
$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}. $$
Here:
$$ A = 50^\circ, \quad a = BC = 9, \quad b = AC\ (\text{unknown}), \quad c = AB = 10. $$
Let $C = x^\circ$ and
$$ B = 180^\circ - (A + C) = 130^\circ - x. $$
Step 1: Applying the Sine Rule
Using the sine rule:
$$ \frac{\sin A}{a} = \frac{\sin C}{c}. $$
Substitute the known values:
$$ \frac{\sin 50^\circ}{9} = \frac{\sin x}{10}. $$
Rearranging for $\sin x$:
$$ \sin x = 10 \cdot \frac{\sin 50^\circ}{9}. $$
Calculate $\sin 50^\circ$:
$$ \sin 50^\circ \approx 0.7660. $$
Thus:
$$ \sin x = 10 \cdot \frac{0.7660}{9} \approx 0.8511. $$
Step 2: Finding $x$
The sine function gives two possible angles between $0^\circ$ and $180^\circ$. Therefore:
$$ x = \arcsin(0.8511). $$
Calculate $\arcsin(0.8511)$:
$$ x \approx 58.3^\circ. $$
The second possible angle is:
$$ x = 180^\circ - 58.3^\circ = 121.7^\circ. $$
Final Answer
The two possible values of $x$ are:
$$ \boxed{ 58.3^\circ \text{ and } 121.7^\circ } $$
Question 3
An arithmetic series has first term 16 and common difference $ -5 $.
The sum to $ n $ terms of this series is $ S_n $.
Given that $ S_n \lt -450 $,
find the least value of $ n $. (4)
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Solution
An arithmetic series has:
$$ \text{First term: } a = 16, \quad \text{Common difference: } d = -5. $$
The sum to $n$ terms of an arithmetic series is given by:
$$ S_n = \frac{n}{2} \left( 2a + (n-1)d \right). $$
We are tasked to find the least value of $n$ such that:
$$ S_n \lt -450. $$
Step 1: Substituting the known values
Substitute $a = 16$ and $d = -5$ into the formula for $S_n$:
$$ S_n = \frac{n}{2} \left( 2(16) + (n-1)(-5) \right). $$
Simplify:
$$ S_n = \frac{n}{2} \left( 32 - 5(n-1) \right). $$
Expand $32 - 5(n-1)$:
$$ 32 - 5(n-1) = 32 - 5n + 5 = 37 - 5n. $$
Thus:
$$ S_n = \frac{n}{2}(37 - 5n). $$
Simplify further:
$$ S_n = \frac{37n - 5n^2}{2}. $$
Step 2: Solve $S_n \lt -450$
We require:
$$ \frac{37n - 5n^2}{2} \lt -450. $$
Multiply through by $2$ to eliminate the fraction:
$$ 37n - 5n^2 \lt -900. $$
Rearrange into standard quadratic form:
$$ 5n^2 - 37n - 900 > 0. $$
Step 3: Solve the quadratic inequality
First, solve the quadratic equation
$$ 5n^2 - 37n - 900 = 0 $$
using the quadratic formula:
$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$
where
$$ a = 5, \quad b = -37, \quad c = -900. $$
Substitute these values:
$$ n = \frac{-(-37) \pm \sqrt{(-37)^2 - 4(5)(-900)}}{2(5)}. $$
Simplify:
$$ n = \frac{37 \pm \sqrt{1369 + 18000}}{10}. $$
$$ n = \frac{37 \pm \sqrt{19369}}{10}. $$
Calculate:
$$ \sqrt{19369} \approx 139.2. $$
Thus:
$$ n = \frac{37 \pm 139.2}{10}. $$
Solve for the two roots:
$$ n = \frac{37 + 139.2}{10} = \frac{176.2}{10} = 17.62 $$
$$ n = \frac{37 - 139.2}{10} = \frac{-102.2}{10} = -10.22 $$
Since $n$ must be a positive integer, we take
$$ n > 17.62. $$
Therefore, the least integer value of $n$ is:
$$ n = 18 $$
Final Answer
The least value of $n$ is:
$$ \boxed{18} $$
Question 4
$ O, A $ and $ B $ are fixed points such that
$ \overrightarrow{OA} = p\mathbf{i} + 2p\mathbf{j}, \quad \overrightarrow{OB} = 5\mathbf{i} + 9p\mathbf{j}. $
Given that $ \overrightarrow{AB} $ is parallel to $ (\mathbf{i} - 2\mathbf{j}) $,
(a) find the value of $ p $. (6)
(b) Hence find $ \overrightarrow{AB} $ as a simplified expression in terms of $ \mathbf{i} $ and $ \mathbf{j} $. (2)
(c) Find a unit vector parallel to $ \overrightarrow{OA} $.
Give your answer in the form
$ \frac{\sqrt{a}}{5}(b\mathbf{i} + c\mathbf{j}) $
where $ a, b $ and $ c $ are integers to be found. (4)
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Solution
Let $O$, $A$, and $B$ be fixed points such that:
$$ \overrightarrow{OA} = p\mathbf{i} + 2p\mathbf{j}, \quad \overrightarrow{OB} = 5\mathbf{i} + 9p\mathbf{j}. $$
We are given that $\overrightarrow{AB}$ is parallel to $(\mathbf{i} - 2\mathbf{j})$.
(a) Find the value of $p$
The vector $\overrightarrow{AB}$ is given by:
$$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}. $$
Substitute the given values:
$$ \overrightarrow{AB} = (5\mathbf{i} + 9p\mathbf{j}) - (p\mathbf{i} + 2p\mathbf{j}). $$
Simplify:
$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + (9p - 2p)\mathbf{j}. $$
$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + 7p\mathbf{j}. $$
Since $\overrightarrow{AB}$ is parallel to $(\mathbf{i} - 2\mathbf{j})$, there exists a scalar $k$ such that:
$$ \overrightarrow{AB} = k(\mathbf{i} - 2\mathbf{j}). $$
Equating components:
$$ 5 - p = k, \quad 7p = -2k. $$
From the first equation:
$$ k = 5 - p. $$
Substitute $k = 5 - p$ into the second equation:
$$ 7p = -2(5 - p). $$
Expand:
$$ 7p = -10 + 2p. $$
Rearrange:
$$ 7p - 2p = -10 \implies 5p = -10 \implies p = -2. $$
(b) Find $\overrightarrow{AB}$ as a simplified expression in terms of $\mathbf{i}$ and $\mathbf{j}$
Substitute $p = -2$ into
$$ \overrightarrow{AB} = (5 - p)\mathbf{i} + 7p\mathbf{j}: $$
$$ \overrightarrow{AB} = (5 - (-2))\mathbf{i} + 7(-2)\mathbf{j}. $$
$$ \overrightarrow{AB} = (5 + 2)\mathbf{i} - 14\mathbf{j}. $$
$$ \overrightarrow{AB} = 7\mathbf{i} - 14\mathbf{j}. $$
(c) Find a unit vector parallel to $\overrightarrow{OA}$
The vector $\overrightarrow{OA}$ is:
$$ \overrightarrow{OA} = p\mathbf{i} + 2p\mathbf{j}. $$
Substitute $p = -2$:
$$ \overrightarrow{OA} = -2\mathbf{i} + 2(-2)\mathbf{j}. $$
$$ \overrightarrow{OA} = -2\mathbf{i} - 4\mathbf{j}. $$
The magnitude of $\overrightarrow{OA}$ is:
$$ |\overrightarrow{OA}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. $$
The unit vector parallel to $\overrightarrow{OA}$ is:
$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \frac{-2\mathbf{i} - 4\mathbf{j}}{2\sqrt{5}}. $$
Simplify:
$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \frac{-1}{\sqrt{5}}\mathbf{i} + \frac{-2}{\sqrt{5}}\mathbf{j}. $$
Rewrite in the form
$$ \left(\frac{\sqrt{a}}{5}\right) (b\mathbf{i} + c\mathbf{j}) $$
$$ \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} = \left( \frac{\sqrt{5}}{5} \right) (-1\mathbf{i} - 2\mathbf{j}). $$
Here:
$$ a = 5, \quad b = -1, \quad c = -2. $$
Final Answers
- (a) $p = -2$
- (b) $\overrightarrow{AB} = 7\mathbf{i} - 14\mathbf{j}$
- (c) The unit vector parallel to $\overrightarrow{OA}$ is:
$$ \left( \frac{\sqrt{5}}{5} \right) (-1\mathbf{i} - 2\mathbf{j}), $$
where
$$ a = 5, \quad b = -1, \quad c = -2. $$
Question 5
$ f(x) = 2ax^3 + x^2 - bx + 3a \quad \text{where } a \text{ and } b \text{ are integers.} $
Given that $(x+2)$ and $(x-1)$ are both factors of $f(x)$,
(a) show that $a=2$ and find the value of $b$. (5)
(b) Hence factorise $f(x)$ completely. (2)
Hence, given that
$ h(y) = 2^{(3y+2)} + 2^{2y} - 11(2^y) + 6, $
(c) solve the equation $h(y) = 0$.
Where appropriate give your answers to 3 decimal places.
(5)
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Solution
The given function is:
$ f(x) = 2a x^3 + x^2 - bx + 3a, $
where $a$ and $b$ are integers, and $(x + 2)$ and $(x - 1)$ are factors of $f(x)$.
(a) Show that $a = 2$ and find the value of $b$
Since $(x + 2)$ and $(x - 1)$ are factors of $f(x)$, substituting $x = -2$ and $x = 1$ into $f(x)$ must result in 0.
Substitute $x = -2$:
$ f(-2) = 2a(-2)^3 + (-2)^2 - b(-2) + 3a = 0. $
Simplify:
$ f(-2) = 2a(-8) + 4 + 2b + 3a = 0. $
$ f(-2) = -16a + 4 + 2b + 3a = 0. $
$ f(-2) = -13a + 4 + 2b = 0. $
Rearrange:
$ 2b = 13a - 4. \tag{1} $
Substitute $x = 1$:
$ f(1) = 2a(1)^3 + (1)^2 - b(1) + 3a = 0. $
Simplify:
$ f(1) = 2a + 1 - b + 3a = 0. $
$ f(1) = 5a + 1 - b = 0. $
Rearrange:
$ b = 5a + 1. \tag{2} $
Solve the equations for $a$ and $b$:
From equation $(2)$, substitute $b = 5a + 1$ into equation $(1)$:
$ 2(5a + 1) = 13a - 4. $
Simplify:
$ 10a + 2 = 13a - 4. $
$ 10a - 13a = -4 - 2. $
$ -3a = -6 \implies a = 2. $
Substitute $a = 2$ into $b = 5a + 1$:
$ b = 5(2) + 1 = 10 + 1 = 11. $
Final Results for (a):
$ a = 2, \quad b = 11. $
(b) Factorise $f(x)$ completely
Substitute $a = 2$ and $b = 11$ into $f(x)$:
$ f(x) = 2(2)x^3 + x^2 - 11x + 3(2). $
$ f(x) = 4x^3 + x^2 - 11x + 6. $
We know that $(x + 2)$ and $(x - 1)$ are factors of $f(x)$. Therefore, divide $f(x)$ by $(x + 2)(x - 1)$.
$ (x + 2)(x - 1) = x^2 + x - 2. $
Perform synthetic or polynomial division:
$ \frac{4x^3 + x^2 - 11x + 6}{x^2 + x - 2}. $
The quotient is:
$ 4x - 3. $
Thus, $f(x)$ factorises as:
$ f(x) = (x + 2)(x - 1)(4x - 3). $
(c) Solve $h(y) = 0$
The function $h(y)$ is:
$ h(y) = 2^{3y + 2} + 2^{2y} - 11(2^y) + 6. $
Let $z = 2^y$, so:
$ 2^{3y + 2} = 4z^3, \quad 2^{2y} = z^2, \quad 2^y = z. $
Substitute into $h(y)$:
$ h(y) = 4z^3 + z^2 - 11z + 6. $
Solve:
$ 4z^3 + z^2 - 11z + 6 = 0. $
Using trial and error or synthetic division, we find that $z = 1$ is a root. Factorise:
$ 4z^3 + z^2 - 11z + 6 = (z - 1)(4z^2 + 5z - 6). $
Factorise $4z^2 + 5z - 6$:
$ 4z^2 + 5z - 6 = (4z - 3)(z + 2). $
Thus:
$ h(y) = (z - 1)(4z - 3)(z + 2). $
Substitute back $z = 2^y$:
$ (2^y - 1)(4(2^y) - 3)(2^y + 2) = 0. $
Solve each factor:
- $2^y - 1 = 0 \implies 2^y = 1 \implies y = 0.$
- $4(2^y) - 3 = 0 \implies 2^y = \frac{3}{4} \implies y = \log_2\left(\frac{3}{4}\right) \approx -0.415.$
- $2^y + 2 = 0 \implies 2^y = -2$ (not possible as $2^y > 0$).
Final Answers:
- (a) $a = 2, \, b = 11.$
- (b) $f(x) = (x + 2)(x - 1)(4x - 3).$
- (c) $y = 0$ or $y \approx -0.415.$
Question 6
The curve $C$ has equation
$ y = \frac{e^{(x^2+1)}}{x^2 + 1}. $
(a) Show that
$ \frac{dy}{dx} = \frac{Kx^3 e^{(x^2+1)}}{(x^2+1)^2} $
where $K$ is a constant whose value is to be found. (5)
(b) Find an equation of the tangent to $C$ at the point on $C$ where $x = -1$. Simplify your answer. (5)
The curve $C$ has the equation:
$ y = \frac{e^{x^2+1}}{x^2 + 1}. $
(a) Show that
$ \frac{dy}{dx} = \frac{Kx^3 e^{x^2+1}}{(x^2 + 1)^2}, $
where $K$ is a constant whose value is to be found.
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6"Solution:
Let $y = \frac{u}{v}$, where $u = e^{x^2+1}$ and $v = x^2 + 1$. Using the quotient rule:
$ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. $
First, find $\frac{du}{dx}$:
$ u = e^{x^2+1}, \quad \frac{du}{dx} = e^{x^2+1} \cdot \frac{d}{dx}(x^2 + 1) = e^{x^2+1} \cdot 2x. $
Next, find $\frac{dv}{dx}$:
$ v = x^2 + 1, \quad \frac{dv}{dx} = 2x. $
Substitute into the quotient rule:
$ \frac{dy}{dx} = \frac{(x^2 + 1)(2x e^{x^2+1}) - e^{x^2+1}(2x)}{(x^2 + 1)^2}. $
Factorize the numerator:
$ \frac{dy}{dx} = \frac{2x e^{x^2+1} \left((x^2 + 1) - 1\right)}{(x^2 + 1)^2}. $
$ \frac{dy}{dx} = \frac{2x e^{x^2+1} \cdot x^2}{(x^2 + 1)^2}. $
Simplify:
$ \frac{dy}{dx} = \frac{2x^3 e^{x^2+1}}{(x^2 + 1)^2}. $
Thus, $K = 2$.
(b) Find an equation of the tangent to $C$ at $x = -1$
Substitute $x = -1$ into $y = \frac{e^{x^2+1}}{x^2 + 1}$ to find the point of tangency:
$ y = \frac{e^{(-1)^2+1}}{(-1)^2 + 1} = \frac{e^{2}}{2}. $
So, the point is:
$ \left(-1, \frac{e^2}{2}\right). $
Next, find the slope of the tangent at $x = -1$ using $\frac{dy}{dx}$:
$ \frac{dy}{dx} = \frac{2x^3 e^{x^2+1}}{(x^2 + 1)^2}. $
Substitute $x = -1$:
$ \frac{dy}{dx} = \frac{2(-1)^3 e^{(-1)^2+1}}{((-1)^2 + 1)^2}. $
$ \frac{dy}{dx} = \frac{2(-1)^3 e^2}{(1 + 1)^2}. $
$ \frac{dy}{dx} = \frac{-2 e^2}{4} = -\frac{e^2}{2}. $
The equation of the tangent is:
$ y - y_1 = m(x - x_1), $
where $\left(x_1, y_1\right) = \left(-1, \frac{e^2}{2}\right)$ and $m = -\frac{e^2}{2}$. Substitute:
$ y - \frac{e^2}{2} = -\frac{e^2}{2}(x + 1). $
Simplify:
$ y = -\frac{e^2}{2}x - \frac{e^2}{2} + \frac{e^2}{2}. $
$ y = -\frac{e^2}{2}x. $
Final Answers:
- (a) $K = 2$.
- (b) The equation of the tangent is:
$ y = -\frac{e^2}{2}x. $
Question 7
A particle $P$ is moving along the $x$-axis. At time $t$ seconds ($t \geq 0$) the velocity of $P$ is $v \, \text{m/s}$ where
$ v = t^2 - 10t + 28. $
(a) Find the velocity of $P$ when $t = 1$. (1)
Given that the distance of $P$ from the origin is $24 \, \text{m}$ when $t = 3$,
(b) find the distance of $P$ from the origin when $t = 5$. (5)
(c) Find the acceleration of $P$ when $t = 9$. (2)
(d) (i) Show that there are no values of $t$ for which $P$ is instantaneously at rest.
(ii) Find the least magnitude of the velocity of $P$.
(3)
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Solution
A particle $P$ is moving along the $x$-axis. At time $t$ seconds ($t \geq 0$), the velocity of $P$ is
$ v \, \text{m/s}, \quad \text{where } v = t^2 - 10t + 28. $
(a) Find the velocity of $P$ when $t = 1$
Substitute $t = 1$ into $v = t^2 - 10t + 28$:
$ v = (1)^2 - 10(1) + 28 = 1 - 10 + 28 = 19. $
The velocity of $P$ when $t = 1$ is $19 \, \text{m/s}$.
(b) Find the distance of $P$ from the origin when $t = 5$
The position of $P$, $x(t)$, is obtained by integrating $v = t^2 - 10t + 28$:
$ x(t) = \int (t^2 - 10t + 28) \, dt = \frac{t^3}{3} - 5t^2 + 28t + C. $
When $t = 3$, the distance of $P$ from the origin is $x(3) = 24$:
$ x(3) = \frac{(3)^3}{3} - 5(3)^2 + 28(3) + C = 24. $
Simplify:
$ x(3) = \frac{27}{3} - 45 + 84 + C = 24. $
$ 9 - 45 + 84 + C = 24. $
$ C = 24 - 48 = -24. $
Thus, the position function is:
$ x(t) = \frac{t^3}{3} - 5t^2 + 28t - 24. $
Substitute $t = 5$ into $x(t)$ to find the distance from the origin:
$ x(5) = \frac{(5)^3}{3} - 5(5)^2 + 28(5) - 24. $
$ x(5) = \frac{125}{3} - 125 + 140 - 24. $
$ x(5) = \frac{125}{3} - 9. $
$ x(5) = \frac{125 - 27}{3} = \frac{98}{3} \approx 32.67. $
The distance of $P$ from the origin when $t = 5$ is approximately $32.67 \, \text{m}$.
(c) Find the acceleration of $P$ when $t = 9$
The acceleration of $P$ is the derivative of $v$:
$ a(t) = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 10t + 28) = 2t - 10. $
Substitute $t = 9$:
$ a(9) = 2(9) - 10 = 18 - 10 = 8. $
The acceleration of $P$ when $t = 9$ is $8 \, \text{m/s}^2$.
(d) (i) Show that there are no values of $t$ for which $P$ is instantaneously at rest
A particle is at rest when $v = 0$. Solve:
$ t^2 - 10t + 28 = 0. $
The discriminant of this quadratic equation is:
$ \Delta = (-10)^2 - 4(1)(28) = 100 - 112 = -12. $
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, there are no values of $t$ for which $P$ is instantaneously at rest.
(d) (ii) Find the least magnitude of the velocity of $P$
The velocity function is $v = t^2 - 10t + 28$. To find the least magnitude, minimize $|v|$. The critical points occur where $\frac{dv}{dt} = 0$:
$ \frac{dv}{dt} = 2t - 10 = 0 \implies t = 5. $
Evaluate $v(t)$ at $t = 5$:
$ v(5) = (5)^2 - 10(5) + 28 = 25 - 50 + 28 = 3. $
The velocity is always positive (since $v > 0$ for all $t \geq 0$), so the least magnitude of the velocity is:
$ \boxed{3 \, \text{m/s}}. $
Final Answers:
- (a) $v = 19 \, \text{m/s}$.
- (b) Distance from origin when $t = 5$: $\approx 32.67 \, \text{m}$.
- (c) Acceleration when $t = 9$: $8 \, \text{m/s}^2$.
- (d) (i) No values of $t$ for which $P$ is instantaneously at rest.
- (d) (ii) Least magnitude of velocity: $3 \, \text{m/s}$.
Question 8
Figure 1 shows part of the curve $ C $ with equation $ y = f(x) $.
The curve $ C $ passes through the points with coordinates
$ (a, 0), \quad (-1, 0), \quad (b, 0) \quad \text{and} \quad (0, c). $
Given that
$ f'(x) = 17 + 2x - 3x^2, $
(a) show that the equation of $ C $ is
$ y = 15 + 17x + x^2 - x^3. $ (4)
(b) Find the value of $ a $, the value of $ b $ and the value of $ c $. (6)
The straight line $ l $ intersects $ C $ at the points with coordinates $ (b, 0) $ and $ (0, c) $.
The region $ R $, shown shaded in Figure 1, is bounded by $ l $ and $ C $.
(c) Use algebraic integration to find the exact area of region $ R $. (5)
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Solutions
The curve $C$ passes through the points with coordinates $(a, 0)$, $(-1, 0)$, $(b, 0)$, and $(0, c)$, where $a < -1$, $b > 0$, $c > 0$.
Given that $f'(x) = 17 + 2x - 3x^2$:
(a) Show that the equation of $C$ is $y = 15 + 17x + x^2 - x^3$
To find $f(x)$, integrate $f'(x)$:
$ f(x) = \int \left(17 + 2x - 3x^2 \right) \, dx. $
$ f(x) = 17x + x^2 - x^3 + C, $
where $C$ is the constant of integration.
Since $C$ passes through the point $(-1, 0)$, substitute $(-1, 0)$ into $f(x)$:
$ 0 = 17(-1) + (-1)^2 - (-1)^3 + C. $
$ 0 = -17 + 1 + 1 + C. $
$ C = 15. $
Thus, the equation of $C$ is:
$ y = f(x) = 15 + 17x + x^2 - x^3. $
(b) Find the values of $a$, $b$, and $c$
The roots of $f(x) = 0$ give the $x$-intercepts $(a, 0)$, $(-1, 0)$, and $(b, 0)$. Solve:
$ 15 + 17x + x^2 - x^3 = 0. $
Factorize:
$ -x^3 + x^2 + 17x + 15 = 0 \implies -(x+1)(x^2 - 2x - 15) = 0. $
The factor $x+1 = 0$ gives $x = -1$. Solve $x^2 - 2x - 15 = 0$ using the quadratic formula:
$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{4 + 60}}{2}. $
$ x = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}. $
Thus:
$ x = \frac{2 + 8}{2} = 5, \quad x = \frac{2 - 8}{2} = -3. $
So, $a = -3$, $b = 5$.
To find $c$, substitute $x = 0$ into $f(x)$:
$ c = f(0) = 15 + 17(0) + (0)^2 - (0)^3 = 15. $
(c) Use algebraic integration to find the exact area of region $R$
The region $R$ is bounded by the curve $C$ and the straight line $l$, which passes through the points $(b, 0)$ and $(0, c)$. The equation of $l$ is:
$ y = mx + c, $
where $m$ is the slope. The slope is:
$ m = \frac{c - 0}{0 - b} = \frac{15}{-5} = -3. $
Thus, the equation of $l$ is:
$ y = -3x + 15. $
The area of $R$ is given by:
$ \text{Area} = \int_0^5 \left(f(x) - l(x)\right) \, dx. $
Substitute $f(x)$ and $l(x)$:
$ \text{Area} = \int_0^5 \left(15 + 17x + x^2 - x^3 - (-3x + 15)\right) dx. $
$ \text{Area} = \int_0^5 \left(-x^3 + x^2 + 20x \right) dx. $
Simplify the integration term by term:
$ \int_0^5 -x^3 \, dx = \left[-\frac{x^4}{4}\right]_0^5 = -\frac{5^4}{4} = -\frac{625}{4}, $
$ \int_0^5 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^5 = \frac{5^3}{3} = \frac{125}{3}, $
$ \int_0^5 20x \, dx = \left[10x^2\right]_0^5 = 10(5^2) = 250. $
Combine the results:
$ \text{Area} = -\frac{625}{4} + \frac{125}{3} + 250. $
Find the common denominator and simplify:
$ \text{Area} = \frac{-1875 + 500 + 3000}{12} = \frac{1625}{12}=135\frac{5}{12}. $
Thus, the exact area of $R$ is:
$ \boxed{135\frac{5}{12}}. $
Final Answers:
- (a) The equation of $C$ is:
$ y = 15 + 17x + x^2 - x^3. $
- (b) $a = -3$, $b = 5$, $c = 15$.
- (c) The exact area of $R$ is:
$ 135\frac{5}{12}. $
Question 9
Figure 2 shows the right pyramid $ OABCD $ with a square base $ ABCD $ of side 12 cm.
$ OA = OB = OC = OD = x \, \text{cm} \quad \text{and} \quad \angle OAC = \angle ODB = \angle OCA = \angle OBD = 30^\circ. $
(a) Find the exact length of $ AC $. (2)
(b) Show that $ x = 4\sqrt{6} $. (2)
(c) Find the total surface area, to the nearest $ \text{cm}^2 $, of the pyramid. (5)
(d) Find the size of the obtuse angle, to the nearest degree, between the plane $ OAB $ and the plane $ OBC $. (4)
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Solution
The right pyramid $OABCD$ has a square base $ABCD$ with side length 12 cm. The edges $OA = OB = OC = OD = x$ cm, and the angles $\angle OAC = \angle ODB = \angle OCA = \angle OBD = 30^\circ$.
(a) Find the exact length of $AC$
The diagonal $AC$ of the square base can be found using the Pythagorean theorem, as $AB = 12$ cm. Since $AC$ is the diagonal of the square:
$ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \text{ cm}. $
(b) Show that $x = 4\sqrt{6}$
From the information given, $\angle OAC = 30^\circ$. We can use trigonometry to relate $x$ and $AC$. In the triangle $OAC$, using the cosine rule:
$ \cos(30^\circ) = \frac{AC}{OA}. $
Substitute the known values:
$ \cos(30^\circ) = \frac{12\sqrt{2}}{x}. $
Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, we have:
$ \frac{\sqrt{3}}{2} = \frac{12\sqrt{2}}{x}. $
Now, solve for $x$:
$ x = \frac{12\sqrt{2} \times 2}{\sqrt{3}} = \frac{24\sqrt{2}}{\sqrt{3}} = 24\sqrt{\frac{2}{3}} = 4\sqrt{6}. $
Thus, $x = 4\sqrt{6}$ cm.
(c) Find the total surface area, to the nearest cm2, of the pyramid
The total surface area of the pyramid consists of the area of the square base and the areas of the four triangular faces.
- The area of the square base:
$ \text{Area of base} = AB^2 = 12^2 = 144 \text{ cm}^2. $
- To find the area of one triangular face, we need to find the slant height $l$ of the triangle. The slant height is the perpendicular distance from the apex $O$ to the midpoint of the side of the square. In the right triangle $OAC$, the slant height $h$ can be found using:
$ h=\sqrt{(4\sqrt{6})^2- (12/2)^2}= 2\sqrt{15}= 7.7459 $
Now, use this to find the area of one triangle. The area of one triangular face is:
$ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 12 \times 2\sqrt{15}= 46.475. $
$ \text{Total surface Area of the pyramid } = 4\times 46.475+144=329.9=330. $
(d) Find the size of the obtuse angle, to the nearest degree, between the plane $OAB$ and the plane $OBC$
In triangle $\triangle OAB$, we are given:
$ OA = OB = 4\sqrt{6}, \quad AB = 12. $
Let $Y$ be the foot of the perpendicular from $A$ to $OB$. We aim to find $AY$ using the area method.
Step 1: Area of $\triangle OAB$ using $AB$ as the base
The area of $\triangle OAB$ can be expressed as:
$ \text{Area} = \frac{1}{2} \times AB \times h, $
where $h = 2\sqrt{15}$ is the perpendicular distance from $O$ to $AB$. Substituting:
$ \text{Area} = \frac{1}{2} \times 12 \times 2\sqrt{15} = 12\sqrt{15}. $
Step 2: Area of $\triangle OAB$ using $OB$ as the base
The area can also be expressed as:
$ \text{Area} = \frac{1}{2} \times OB \times AY, $
where $AY$ is the perpendicular from $A$ to $OB$. Substituting $OB = 4\sqrt{6}$:
$ \text{Area} = \frac{1}{2} \times 4\sqrt{6} \times AY = 2\sqrt{6} \cdot AY. $
Step 3: Equating the two expressions for the area
Equating the two expressions:
$ 12\sqrt{15} = 2\sqrt{6} \cdot AY. $
Solve for $AY$:
$ AY = \frac{12\sqrt{15}}{2\sqrt{6}}. $
Step 4: Simplify $AY$
Simplify the fraction:
$ AY = 6 \cdot \frac{\sqrt{15}}{\sqrt{6}} = 6 \cdot \sqrt{\frac{15}{6}} = 6 \cdot \sqrt{\frac{5}{2}} = 3\sqrt{10}. $
In triangle $\triangle AYC$, we are given:
$ AC = 12\sqrt{2}, \quad AY = YC = 3\sqrt{10}. $
We aim to find $\angle AYC$.
Step 5: Apply the Law of Cosines
Using the Law of Cosines in $\triangle AYC$:
$ AC^2 = AY^2 + YC^2 - 2 \cdot AY \cdot YC \cdot \cos\angle AYC. $
Substitute the given values:
$ (12\sqrt{2})^2 = (3\sqrt{10})^2 + (3\sqrt{10})^2 - 2 \cdot (3\sqrt{10}) \cdot (3\sqrt{10}) \cdot \cos\angle AYC. $
Step 6: Simplify the equation
First, compute the squares:
$ (12\sqrt{2})^2 = 288, \quad (3\sqrt{10})^2 = 90. $
Substitute these into the equation:
$ 288 = 90 + 90 - 2 \cdot 3\sqrt{10} \cdot 3\sqrt{10} \cdot \cos\angle AYC. $
Simplify further:
$ 288 = 180 - 2 \cdot 90 \cdot \cos\angle AYC. $
$ 288 = 180 - 180 \cdot \cos\angle AYC. $
Step 7: Solve for $\cos\angle AYC$
Rearrange to isolate $\cos\angle AYC$:
$ 288 - 180 = -180 \cdot \cos\angle AYC. $
$ 108 = -180 \cdot \cos\angle AYC. $
$ \cos\angle AYC = -\frac{108}{180} = -\frac{3}{5}. $
Step 8: Find $\angle AYC$
Since $\cos\angle AYC = -\frac{3}{5}$, the angle is in the second quadrant:
$ \angle AYC = \cos^{-1}\left(-\frac{3}{5}\right). $
Final Answer
The size of the obtuse angle, to the nearest degree, between the plane $OAB$ and the plane $OBC$ is
$ \boxed{\angle AYC = \cos^{-1}\left(-\frac{3}{5}\right)=126.87=127} $
Question 10
Using formulae from page 2
(a) show that
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B. $ (2)
(b) Hence show that
$ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta. $ (1)
(c) Solve the equation
$ \cos 5\theta - \cos 9\theta = \sqrt{3} \sin 7\theta \quad \text{for } 0 \lt \theta \leq \frac{1}{3}\pi. $
Give your solutions in terms of $\pi$. (7)
(d) Using calculus and showing your working, evaluate
$ \int_{0}^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \tan 2x \, dx. $
Give your answer to 3 decimal places. (6)
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Solution
We use the trigonometric formulae for $ \cos(A - B) $ and $ \cos(A + B) $ to solve the given problems.
Part (a): Show that $ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B $
Using the sum-to-product identities:
$ \cos(A - B) = \cos A \cos B + \sin A \sin B, $
$ \cos(A + B) = \cos A \cos B - \sin A \sin B. $
Subtracting these two:
$ \cos(A - B) - \cos(A + B) = (\cos A \cos B + \sin A \sin B) - (\cos A \cos B - \sin A \sin B). $
Simplify:
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B. $
Part (b): Show that $ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta $
Using the result from Part (a) with $ A = 7\theta $ and $ B = 2\theta $:
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B. $
Substitute $ A = 7\theta $ and $ B = 2\theta $:
$ \cos(7\theta - 2\theta) - \cos(7\theta + 2\theta) = 2 \sin(7\theta) \sin(2\theta). $
Simplify:
$ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta. $
Part (c): Solve $ \cos 5\theta - \cos 9\theta = \sqrt{3} \sin 7\theta $ for $ 0 \lt \theta \leq \frac{\pi}{3} $
Using the result from Part (b):
$ \cos 5\theta - \cos 9\theta = 2 \sin 7\theta \sin 2\theta. $
Substitute into the equation:
$ 2 \sin 7\theta \sin 2\theta = \sqrt{3} \sin 7\theta. $
Solve $ \sin 7\theta = 0 $
We know that $ \sin x = 0 $ when:
$ x = n\pi, \quad n \in \mathbb{Z}. $
For $ \sin 7\theta = 0 $, we have:
$ 7\theta = n\pi, \quad n \in \mathbb{Z}. $
Solve for $ \theta $:
$ \theta = \frac{n\pi}{7}. $
Restrict the solution to $ 0 \lt \theta \leq \frac{\pi}{3} $
From the given condition, $ 0 \lt \theta \leq \frac{\pi}{3} $. Substituting $ \theta = \frac{n\pi}{7} $, we require:
$ 0 \lt \frac{n\pi}{7} \leq \frac{\pi}{3}. $
Divide through by $ \pi $:
$ 0 \lt \frac{n}{7} \leq \frac{1}{3}. $
Multiply through by $ 7 $:
$ 0 \lt n \leq \frac{7}{3}. $
Since $ n $ must be an integer, the only possible values are $ n = 1, \text{ or } n=2 $.
Substitute $ n = 1 $ back into $ \theta = \frac{n\pi}{7} $:
$ \theta = \frac{\pi}{7}. $
Substitute $ n = 2 $ back into $ \theta = \frac{n\pi}{7} $:
$ \theta = \frac{2\pi}{7}. $
Divide through by $ \sin 7\theta $ (assuming $ \sin 7\theta \neq 0 $):
$ 2 \sin 2\theta = \sqrt{3}. $
Solve for $ \sin 2\theta $:
$ \sin 2\theta = \frac{\sqrt{3}}{2}. $
The angle $ 2\theta $ corresponding to $ \sin 2\theta = \frac{\sqrt{3}}{2} $ is:
$ 2\theta = \frac{\pi}{3} \text{ (or) } \frac{2\pi}{3}. $
Solve for $ \theta $:
$ \theta = \frac{\pi}{6} \text{ (or) } \frac{\pi}{3}. $
Final Answer
$ \boxed{\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{7}, \frac{2\pi}{7}.} $
Part (d): Evaluate $ \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \tan 2x \, dx $
We are tasked with evaluating:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \tan 2x \, dx. $
Step 1: Simplify the integrand
Recall that $ \tan 2x = \frac{\sin 2x}{\cos 2x} $. Substituting this into the integral:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \cos 2x \cdot \frac{\sin 2x}{\cos 2x} \, dx. $
Simplify:
$ I = \int_0^{\frac{\pi}{7}} 8 \sin 7x \sin 2x \, dx. $
Step 2: Use the product-to-sum identities
Using the product-to-sum identity:
$ \sin A \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right], $
we rewrite the integrand:
$ \sin 7x \sin 2x = \frac{1}{2} \left[ \cos(7x - 2x) - \cos(7x + 2x) \right]. $
Substitute this into the integral:
$ I = \int_0^{\frac{\pi}{7}} 8 \cdot \frac{1}{2} \left[ \cos(5x) - \cos(9x) \right] \, dx. $
Simplify:
$ I = 4 \int_0^{\frac{\pi}{7}} \cos(5x) \, dx - 4 \int_0^{\frac{\pi}{7}} \cos(9x) \, dx. $
Step 3: Integrate each term
The integral of $ \cos(kx) $ is:
$ \int \cos(kx) \, dx = \frac{\sin(kx)}{k}. $
Using this, we compute each term:
1. For $ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx $:
$ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx = \left[ \frac{\sin(5x)}{5} \right]_0^{\frac{\pi}{7}} = \frac{\sin\left(\frac{5\pi}{7}\right)}{5} - \frac{\sin(0)}{5}. $
Simplify:
$ \int_0^{\frac{\pi}{7}} \cos(5x) \, dx = \frac{\sin\left(\frac{5\pi}{7}\right)}{5}. $
2. For $ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx $:
$ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx = \left[ \frac{\sin(9x)}{9} \right]_0^{\frac{\pi}{7}} = \frac{\sin\left(\frac{9\pi}{7}\right)}{9} - \frac{\sin(0)}{9}. $
Simplify:
$ \int_0^{\frac{\pi}{7}} \cos(9x) \, dx = \frac{\sin\left(\frac{9\pi}{7}\right)}{9}. $
Step 4: Combine the results
Substitute back into $ I $:
$ I = 4 \left( \frac{\sin\left(\frac{5\pi}{7}\right)}{5} \right) - 4 \left( \frac{\sin\left(\frac{9\pi}{7}\right)}{9} \right). $
Step 5: Evaluate numerically
Using numerical approximation (to 3 decimal places):
$ \sin\left(\frac{5\pi}{7}\right) \approx 0.7818, \quad \sin\left(\frac{9\pi}{7}\right) \approx -0.7818. $
Substitute these values:
$ I = 4 \left( \frac{0.7818}{5} \right) - 4 \left( \frac{-0.7818}{9} \right). $
Simplify:
$ I = 4 \cdot 0.1563 + 4 \cdot 0.0868. $
$ I = 0.6254 + 0.3474 = 0.9728. $
Final Answer
$ \boxed{I \approx 0.973} $
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