Question 1
On the axes below, sketch the graph of $ y = |4 \cos 2x| $ for $ 0 \leqslant x \leqslant \pi $, giving the coordinates of any points where the graph meets the axes. [3]
Solution
Solution:
Step 1: Understand the function
The function to be graphed is
\[ y = |4 \cos 2x|, \]
where $ 0 \leq x \leq \pi $. This function consists of the absolute value of a scaled cosine function with frequency doubled.
Step 2: Analyze the function
The general cosine function, $ \cos 2x $, oscillates between $-1$ and $1$. Scaling it by $4$ gives $4 \cos 2x$, which oscillates between $-4$ and $4$. Applying the absolute value ensures that the negative values are reflected upwards, so $ y = |4 \cos 2x| $ oscillates between $0$ and $4$.
The period of $ \cos 2x $ is given by:
\[ \text{Period} = \frac{2\pi}{\text{frequency}} = \frac{2\pi}{2} = \pi. \]
Thus, within $0 \leq x \leq \pi$, there will be two humps of the cosine graph due to the doubled frequency.
Step 3: Find key points
To sketch the graph, we determine where the function intersects the axes and other key points:
-
When $ y = 0 $: This occurs when $ \cos 2x = 0 $, i.e.
\[ 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \implies x = \frac{\pi}{4}, \frac{3\pi}{4}, \dots \]
Within $ 0 \leq x \leq \pi $, the zeros are at $ x = \frac{\pi}{4} $ and $ x = \frac{3\pi}{4} $.
-
Maximum values: The maximum value of $ y $ is $4$, which occurs when $ \cos 2x = \pm 1 $. These happen at:
\[ 2x = 0, \pi, 2\pi, \dots \implies x = 0, \frac{\pi}{2}, \pi, \dots \]
Within $ 0 \leq x \leq \pi $, the maxima occur at $ (0, 4) $, $ \left(\frac{\pi}{2}, 4\right) $, and $ (\pi, 4) $.
Step 4: Sketch the graph
Below is the sketch of the graph $ y = |4 \cos 2x| $ for $ 0 \leq x \leq \pi $. Key points are labeled on the axes.
Question 2
DO NOT USE A CALCULATOR IN THIS QUESTION.
Expand and simplify $$\left(\frac{x \sqrt{11}}{2 \sqrt{3}-1}\right)^2,$$ giving your answer with a rational denominator. [4]
Solution
Solution:
Step 1: Write the expression
We are tasked to expand and simplify the following expression:
Our goal is to express the final result with a rational denominator.
Step 2: Rationalize the denominator
First, we rationalize the denominator of the fraction:
To do this, we multiply both the numerator and denominator by the conjugate of the denominator, $2 \sqrt{3} + 1$:
Step 3: Simplify the denominator
The denominator becomes:
Thus, the expression simplifies to:
Step 4: Expand the numerator
Now expand the numerator:
The expression becomes:
Step 5: Square the entire expression
Now square the entire fraction:
First, square the numerator:
Simplify each term:
Add these terms together:
Thus, the squared numerator is:
The denominator squared is:
Step 6: Final simplified result
Combine the squared numerator and denominator:
Question 3
Solve the inequality $|5 x+4| \leqslant|2 x-3|$. [4]
Solution
Solution:
Step 1: Analyze the inequality
The given inequality is:
To solve, we need to determine where the absolute values are equal, i.e., solve:
Step 2: Solve $ |5x + 4| = |2x - 3| $
This equality breaks into two cases:
Case 1: $ 5x + 4 = 2x - 3 $
Simplify:
Case 2: $ 5x + 4 = -(2x - 3) $
Simplify:
Rearrange:
Thus, the critical values are:
Step 3: Divide the real line into intervals
The critical values divide the real line into three intervals:
We analyze $ |5x + 4| \leqslant |2x - 3| $ on each interval.
Step 4: Solve on each interval
Interval 1: $ (-\infty, -\frac{7}{3}) $
On this interval:
The inequality becomes:
Simplify:
Rearrange:
Divide by $-3$ (reverse the inequality):
On this interval, $ x \geqslant -\frac{7}{3} $ conflicts with $ x < -\frac{7}{3} $. Hence, no solution exists on this interval.
Interval 2: $ \left(-\frac{7}{3}, -\frac{1}{7}\right) $
On this interval:
The inequality becomes:
Simplify:
Rearrange:
Divide by 7:
Thus, the solution on this interval is:
Interval 3: $ \left(-\frac{1}{7}, \infty\right) $
On this interval:
The inequality becomes:
Simplify:
Divide by 3:
On this interval, $ x > -\frac{1}{7} $ conflicts with $ x \leqslant -\frac{7}{3} $. Hence, no solution exists on this interval.
Step 5: Combine the results
The only valid solution is from Interval 2:
Thus, the solution to the inequality is:
Question 4
\[ y=\frac{\sec ^2 5 x-\tan ^2 5 x}{\operatorname{cosec} 5 x} \]
Show that $ y=a \sin b x $, where $ a $ and $ b $ are integers, and hence find the value of \[ \int_0^{\frac{\pi}{5}} y \, \mathrm{d}x. \] [4]
Solution
Solution:
Step 1: Simplify the expression for $ y $
We are given the expression:
\[ y = \frac{\sec^2(5x) - \tan^2(5x)}{\csc(5x)} \]
Using the identity $ \sec^2 \theta - \tan^2 \theta = 1 $, we can simplify the numerator:
\[ \sec^2(5x) - \tan^2(5x) = 1. \]
Thus, the expression for $ y $ becomes:
\[ y = \frac{1}{\csc(5x)}. \]
Since $ \csc \theta = \frac{1}{\sin \theta} $, we can rewrite the denominator as:
\[ y = \sin(5x). \]
Step 2: Express $ y $ in the form $ a \sin(bx) $
We now have:
\[ y = \sin(5x). \]
This is already in the required form $ a \sin(bx) $, where $ a = 1 $ and $ b = 5 $.
Step 3: Compute the integral
We are tasked with finding the value of the integral:
\[ \int_0^{\frac{\pi}{5}} y \, dx = \int_0^{\frac{\pi}{5}} \sin(5x) \, dx. \]
The antiderivative of $ \sin(5x) $ is $ -\frac{1}{5} \cos(5x) $. Thus, we evaluate the integral:
\[ \int_0^{\frac{\pi}{5}} \sin(5x) \, dx = \left[ -\frac{1}{5}\cos(5x) \right]_0^{\frac{\pi}{5}}. \]
Substituting the limits of integration:
\[ = -\frac{1}{5} \left[ \cos\left(5 \times \frac{\pi}{5}\right) - \cos(0) \right] \]
\[ = -\frac{1}{5} \left[ \cos(\pi) - \cos(0) \right]. \]
Since $ \cos(\pi) = -1 $ and $ \cos(0) = 1 $, we get:
\[ = -\frac{1}{5}(-1 - 1) \]
\[ = -\frac{1}{5}(-2) \]
\[ = \frac{2}{5}. \]
Conclusion
\[ \int_0^{\frac{\pi}{5}} y \, dx = \frac{2}{5}. \]
Question 5
DO NOT USE A CALCULATOR IN THIS QUESTION.
(a) Show that $x-1$ is a factor of the expression $x^3-2x^2-19x+20$. [1]
(b) Hence write $x^3-2x^2-19x+20$ as a product of its linear factors. [3]
(c) Hence find the exact solutions of the equation $\mathrm{e}^{3y}-2\mathrm{e}^{2y}-19\mathrm{e}^y+20=0$. [2]
Solution
Solution:
Step (a): Show that $x-1$ is a factor of the expression $x^3-2x^2-19x+20$
To show that $x-1$ is a factor of the expression $x^3 - 2x^2 - 19x + 20$, we can use the Factor Theorem. According to the Factor Theorem, if $x-1$ is a factor of the polynomial, then substituting $x=1$ into the polynomial should result in zero.
We evaluate the polynomial at $x=1$:
\[ P(1) = (1)^3 - 2(1)^2 - 19(1) + 20 = 1 - 2 - 19 + 20 = 0. \]
Since $P(1)=0$, we can conclude that $x-1$ is indeed a factor of the polynomial $x^3 - 2x^2 - 19x + 20$.
Step (b): Write $x^3 - 2x^2 - 19x + 20$ as a product of its linear factors
Now that we know $x-1$ is a factor, we can perform polynomial division to divide $x^3 - 2x^2 - 19x + 20$ by $x-1$.
We divide $x^3 - 2x^2 - 19x + 20$ by $x-1$ using synthetic division:
\[ \begin{array}{r|rrrr} 1 & 1 & -2 & -19 & 20 \\ & & 1 & -1 & -20 \\ \hline & 1 & -1 & -20 & 0 \\ \end{array} \]
The quotient is $x^2 - x - 20$, and the remainder is $0$.
Now, we factor $x^2 - x - 20$. To factor this quadratic, we need two numbers that multiply to $-20$ and add to $-1$. These numbers are $-5$ and $4$, so:
\[ x^2 - x - 20 = (x - 5)(x + 4). \]
Thus, we can express the original cubic polynomial as:
\[ x^3 - 2x^2 - 19x + 20 = (x - 1)(x - 5)(x + 4). \]
Step (c): Find the exact solutions of the equation $\mathrm{e}^{3y} - 2\mathrm{e}^{2y} - 19\mathrm{e}^y + 20 = 0$
To solve the equation $\mathrm{e}^{3y} - 2\mathrm{e}^{2y} - 19\mathrm{e}^y + 20 = 0$, we make the substitution:
\[ z = \mathrm{e}^y. \]
This transforms the equation into:
\[ z^3 - 2z^2 - 19z + 20 = 0. \]
From part (b), we already know that:
\[ z^3 - 2z^2 - 19z + 20 = (z - 1)(z - 5)(z + 4). \]
Thus:
\[ (z - 1)(z - 5)(z + 4) = 0. \]
This gives:
\[ z = 1, \quad z = 5, \quad z = -4. \]
Recall that $z=\mathrm{e}^y$, so:
For $z=1$:
\[ \mathrm{e}^y = 1 \]
\[ y = 0. \]
For $z=5$:
\[ \mathrm{e}^y = 5 \]
\[ y = \ln 5. \]
For $z=-4$:
\[ \mathrm{e}^y = -4. \]
Since $\mathrm{e}^y$ is always positive, there is no real solution in this case.
Conclusion
\[ y = 0 \quad \text{or} \quad y = \ln 5. \]
Question 6
(a) A geometric progression has first term 64 and common ratio 0.5.
(i) Find the 10th term. [2]
(ii) Find the sum of the first 10 terms. [2]
(iii) Find the sum to infinity. [1]
(b) An arithmetic progression is such that $S_{20} - 400 = 2S_{10}$ and $u_1 : u_6$ is $1 : 5$. Find the sum of the first 3 terms of this progression. [6]
Solution
Solution:
Step (a): Geometric Progression
A geometric progression has first term 64 and common ratio 0.5.
(i) Find the 10th term.
The general formula for the $n$-th term of a geometric progression is:
$u_n = a r^{n-1}$
Given $a = 64$, $r = 0.5$, and $n = 10$:
$u_{10} = 64 \times (0.5)^{10-1} = 64 \times (0.5)^9 = 64 \times \frac{1}{512} = \frac{64}{512} = \frac{1}{8}$
Thus, the 10th term is $u_{10} = \frac{1}{8}$.
(ii) Find the sum of the first 10 terms.
The sum of the first $n$ terms of a geometric progression is:
$S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \lt 1)$
For $n=10$:
$S_{10} = \frac{64(1 - (0.5)^{10})}{1 - 0.5} = \frac{64(1 - \frac{1}{1024})}{0.5} = 128 \left(1 - \frac{1}{1024}\right)$
$S_{10} = 128 \times \frac{1023}{1024} = \frac{1023}{8} = 127.875$
Thus, $S_{10} = 127.875$.
(iii) Find the sum to infinity.
The sum to infinity of a geometric progression is:
$S_{\infty} = \frac{a}{1 - r} \quad (|r| \lt 1)$
$S_{\infty} = \frac{64}{1 - 0.5} = \frac{64}{0.5} = 128$
Thus, $S_{\infty} = 128$.
Step (b): Arithmetic Progression
An arithmetic progression is such that $S_{20} - 400 = 2S_{10}$ and $u_1 : u_6 = 1 : 5$.
Find the sum of the first 3 terms.
The sum of the first $n$ terms of an arithmetic progression is:
$S_n = \frac{n}{2}(2a + (n-1)d)$
From $u_1 : u_6 = 1 : 5$:
$\frac{a + 5d}{a} = 5 \Rightarrow a + 5d = 5a \Rightarrow 5d = 4a \Rightarrow d = \frac{4a}{5}$
Now:
$S_{20} = 10(2a + 19d), \quad S_{10} = 5(2a + 9d)$
Substitute $d = \frac{4a}{5}$:
$S_{20} = 10\left(2a + \frac{76a}{5}\right) = \frac{860a}{5}$
$S_{10} = 5\left(2a + \frac{36a}{5}\right) = 46a$
Given $S_{20} - 400 = 2S_{10}$:
$\frac{860a}{5} - 400 = 92a$
$\frac{860a - 460a}{5} = 400 \Rightarrow \frac{400a}{5} = 400$
$400a = 2000 \Rightarrow a = 5$
Then:
$d = \frac{4a}{5} = 4$
Now the first 3 terms sum:
$S_3 = \frac{3}{2}(2a + 2d) = \frac{3}{2}(10 + 8) = 27$
Thus, $S_3 = 27$.
Question 7
- (a) Variables $x$ and $y$ are such that $y=\frac{1+\cos^2 x}{\tan x}$. Use differentiation to find the approximate change in $y$ as $x$ increases from $\frac{\pi}{4}$ to $\frac{\pi}{4}+h$, where $h$ is small. [5]
- (b) Given that $y=\frac{1}{(x-3)^3}$ show that $y-\frac{dy}{dx}-\frac{1}{3}\left(\frac{d^2y}{dx^2}\right)$ can be written as $\frac{(x+1)(x-4)}{(x-3)^5}$. [4]
Solution
Solution:
Step (a)
We are given:
$y = \frac{1 + \cos^2 x}{\tan x}$
We find the derivative using the quotient rule:
$\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$
where $f(x)=1+\cos^2 x$ and $g(x)=\tan x$.
$f'(x) = \frac{d}{dx}(1+\cos^2 x) = -2\cos x \sin x$
$g'(x) = \sec^2 x$
So,
$\frac{dy}{dx} = \frac{(-2\cos x \sin x)(\tan x) - (1+\cos^2 x)(\sec^2 x)}{\tan^2 x}$
For small change $h$:
$\Delta y \approx \left.\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} h$
At $x=\frac{\pi}{4}$:
$\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \tan \frac{\pi}{4}=1, \quad \sec^2 \frac{\pi}{4}=2$
Substituting:
$\frac{dy}{dx} = \frac{(-2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2})(1) - (1+\frac{1}{2})(2)}{1} = -4$
Thus,
$\Delta y \approx -4h$
Step (b)
Given:
$y = \frac{1}{(x-3)^3}$
First derivatives:
$\frac{dy}{dx} = -3(x-3)^{-4}$
$\frac{d^2y}{dx^2} = 12(x-3)^{-5}$
Now consider:
$y - \frac{dy}{dx} - \frac{1}{3}\frac{d^2y}{dx^2}$
$= \frac{1}{(x-3)^3} + 3(x-3)^{-4} - 4(x-3)^{-5}$
Factor $(x-3)^{-5}$:
$= (x-3)^{-5}\left((x-3)^2 + 3(x-3) - 4\right)$
$= (x-3)^{-5}\left(x^2 - 6x + 9 + 3x - 9 - 4\right)$
$= (x-3)^{-5}(x^2 - 3x - 4)$
$= (x-3)^{-5}(x+1)(x-4)$
Thus,
$y - \frac{dy}{dx} - \frac{1}{3}\frac{d^2y}{dx^2} = \frac{(x+1)(x-4)}{(x-3)^5}$
Question 8
The function f is defined for $x \geqslant 0$ by $f(x)=5-2e^{-x}$.
(a) (i) Find the domain of $f^{-1}$. [2]
(ii) Solve $ff^{-1}(x)=\sqrt{5x-4}$. [3]
(iii) On the axes, sketch the graph of $y=f(x)$ and hence sketch the graph of $y=f^{-1}(x)$.
Show clearly the positions of any points where the graphs meet the coordinate axes and any asymptotes.
[4]
(b) The function g is defined for $0 \leqslant x \leqslant 0.2$ by $g(x)=\frac{3}{1-x}$. Find and simplify an expression for $f^{-1}g(x)$. [4]
Solution
Solution:
Step (a)(i): Domain of $f^{-1}$
$f(x)=5-2e^{-x}$ is defined for $x \geqslant 0$.
As $x \to 0$, $f(x)\to 5-2=3$.
As $x \to \infty$, $f(x)\to 5$.
So the range of $f(x)$ is $[3,5)$.
Hence the domain of $f^{-1}$ is $[3,5)$.
Step (a)(ii)
By definition, $f(f^{-1}(x)) = x$.
So the equation becomes:
$x = \sqrt{5x-4}$
Square both sides:
$x^2 = 5x - 4$
$x^2 - 5x + 4 = 0$
$(x-4)(x-1)=0$
$x=4 \text{ or } x=1$
Since domain of $f^{-1}$ is $3 \le x \lt 5$, the valid solution is:
$\boxed{x=4}$
Step (a)(iii): Sketch
The graph of $f(x)=5-2e^{-x}$:
- At $x=0$, $f(0)=3$ so point $(0,3)$
- As $x \to \infty$, $f(x)\to 5$ so horizontal asymptote $y=5$
The graph is an increasing exponential curve starting at $(0,3)$ and approaching $y=5$.
The graph of $y=f^{-1}(x)$ is the reflection in the line $y=x$.
- Point $(0,3)$ becomes $(3,0)$
- Point $(3,0)$ becomes $(0,3)$
- Asymptote $y=5$ becomes $x=5$
Step (b): Find $f^{-1}g(x)$
Let $y=f^{-1}(g(x))$, so $g(x)=f(y)$.
$\frac{3}{1-x} = 5 - 2e^{-y}$
$2e^{-y} = 5 - \frac{3}{1-x}$
$2e^{-y} = \frac{5(1-x)-3}{1-x} = \frac{2-5x}{1-x}$
$e^{-y} = \frac{2-5x}{2(1-x)}$
$-y = \ln\left(\frac{2-5x}{2(1-x)}\right)$
$y = -\ln\left(\frac{2-5x}{2(1-x)}\right)$
Therefore:
$f^{-1}(g(x)) = -\ln\left(\frac{2-5x}{2(1-x)}\right)$
Question 9
In this question, all lengths are in centimetres and all angles are in radians.
-
The diagram shows sectors $AOB$ and $COD$ of two circles with the same centre, $O$. Angle $AOB$ is $\frac{3\pi}{8}$ and the length of $OC$ is $6.5$. It is given that $OAC$ and $OBD$ are straight lines and $OA : OC = 4 : 5$. Find the perimeter of the shaded region. [3]
-
The diagram shows a circle with centre $O$ and radius $a$. Sector $PQR$ is a sector of a different circle with centre $R$ and radius $y$. Angle $OPR$ is $\phi$. Find, in terms of $a$ and $\phi$ only, the total area of the three shaded regions. Simplify your answer. [4]
Solution
Solution: (a)
Step 1: Understanding the problem (Part a)
The problem involves two sectors $AOB$ and $COD$ of circles with the same center $O$. The following details are given:
- Angle $\angle AOB = \frac{3\pi}{8}$ radians.
- Length $OC = 6.5$ cm.
- The ratio $OA : OC = 4 : 5$.
- $OAC$ and $OBD$ are straight lines.
The goal is to find the perimeter of the shaded region.
Step 2: Calculating $OA$
Using the ratio $OA : OC = 4 : 5$, we calculate $OA$:
$$ OA = \frac{4}{5} \times OC = \frac{4}{5} \times 6.5 = 5.2 \text{ cm} $$
Step 3: Finding arc lengths
The lengths of the arcs are calculated using the formula $\text{Arc Length} = r\theta$:
- For sector $AOB$ (radius $OA = 5.2$ cm): $$ \text{Arc Length}_{AOB} = 5.2 \times \frac{3\pi}{8} = \frac{15.6\pi}{8} \text{ cm} $$
- For sector $COD$ (radius $OC = 6.5$ cm): $$ \text{Arc Length}_{COD} = 6.5 \times \frac{3\pi}{8} = \frac{19.5\pi}{8} \text{ cm} $$
Step 4: Total perimeter of the shaded region
The perimeter of the shaded region includes:
- $AC$: Since $OA + AC = OC$, $$ AC = OC - OA = 6.5 - 5.2 = 1.3 \text{ cm} $$
- $BD$: Since $AC = BD$, $$ BD = 1.3 \text{ cm} $$
- The arc lengths $\text{Arc Length}_{AOB}$ and $\text{Arc Length}_{COD}$.
Adding these together:
$$ \text{Perimeter} = 2 \times 1.3 + \frac{15.6\pi}{8} + \frac{19.5\pi}{8} $$
$$ = 2.6 + \frac{35.1\pi}{8} \approx 16.38 \approx 16.4 \text{ cm} $$
Solution: (b)
Step 1: Analyze the geometry of the problem
The given diagram consists of:
- A circle with center $O$ and radius $a$.
- A sector $PQR$ of a different circle with center $R$ and radius $y$.
- Angle $OPR$ is $\phi$.
- Triangles $\triangle OPR$ and $\triangle ORQ$ are congruent, and angle $\angle PRQ = 2\phi$.
We aim to find the total area of the three shaded regions in terms of $a$ and $\phi$.
Step 2: Determine $PM$ and $y$
Let $M$ be the midpoint of $PR$. Since $\triangle OPR$ is isosceles ($OP = OR = a$), $PM$ is the perpendicular bisector of $PR$.
Using trigonometry in $\triangle OPM$:
$$ PM = a \cos \phi $$
$$ PR = 2a \cos \phi $$
Since $y = PM + MR$, and $MR = PM$, we find:
$$ y = 2a \cos \phi $$
Step 3: Calculating the areas of the shaded regions
- Area of the circle with radius $a$: $$ \text{Area}_{\text{circle}} = \pi a^2 $$
- Area of sector $PQR$ (radius $y$, angle $2\phi$): $$ \text{Area}_{\text{sector}} = \frac{1}{2} y^2 (2\phi) = \frac{1}{2}(2a\cos\phi)^2(2\phi) = 4a^2\phi\cos^2\phi $$
- Thus, the area of the three shaded regions is: $$ \text{Area} = \pi a^2 - 4a^2\phi\cos^2\phi $$
$$ \boxed{ \pi a^2 - 4a^2\phi\cos^2\phi } $$
Question 10
A particle $P$ moves in a straight line such that, $t$ seconds after passing a fixed point $O$, its acceleration, $a \; \mathrm{ms}^{-2}$, is given by
\[ a = \begin{cases} 6t, & 0 \leqslant t \leqslant 3 \\[8pt] \displaystyle \frac{18e^3}{e^t}, & t \geqslant 3 \end{cases} \]When $t=1$, the velocity of $P$ is $2 \; \mathrm{ms}^{-1}$ and its displacement from $O$ is $-4 \; \mathrm{m}$.
-
(i) Find the velocity of $P$ when $t=3$. [3]
(ii) Find the displacement of $P$ from $O$ when $t=3$.[3] - Find an expression in terms of $t$ for the displacement of $P$ from $O$ when $t \geqslant 3$. [4]
Solution
Solution:
Step 1: Find the velocity of $P$ when $t = 3$
Acceleration $a$ is the derivative of velocity $v$: $a = \dfrac{dv}{dt}$. For $0 \le t \le 3$, $a = 6t$. Integrate:
\[ v = \int 6t \, dt = 3t^2 + C_1, \]where $C_1$ is constant. Using $v(1) = 2$:
\[ 2 = 3(1)^2 + C_1 \quad\Longrightarrow\quad C_1 = -1. \]Thus for $0 \le t \le 3$: $v(t) = 3t^2 - 1$. At $t = 3$:
\[ v(3) = 3(3)^2 - 1 = 27 - 1 = 26 \; \mathrm{ms}^{-1}. \]Answer (a)(i): $v(3) = 26 \; \mathrm{ms}^{-1}$.
Step 2: Find the displacement of $P$ from $O$ when $t = 3$
Velocity $v = \dfrac{ds}{dt}$. For $0 \le t \le 3$, $v = 3t^2 - 1$. Integrate:
\[ s = \int (3t^2 - 1) \, dt = t^3 - t + C_2, \]where $C_2$ is constant. Using $s(1) = -4$:
\[ -4 = (1)^3 - 1 + C_2 \quad\Longrightarrow\quad C_2 = -4. \]Hence for $0 \le t \le 3$: $s(t) = t^3 - t - 4$. At $t = 3$:
\[ s(3) = (3)^3 - 3 - 4 = 27 - 7 = 20 \; \mathrm{m}. \]Answer (a)(ii): $s(3) = 20 \; \mathrm{m}$.
Step 3: Find an expression for the displacement when $t \geqslant 3$
For $t \ge 3$, acceleration is $a = \dfrac{18e^3}{e^t} = 18e^{3-t}$. Integrate to get velocity:
\[ v = \int 18e^{3-t} \, dt = -18e^{3-t} + C_3. \]Velocity is continuous at $t=3$; from part (a)(i) $v(3)=26$:
\[ 26 = -18e^{0} + C_3 \quad\Longrightarrow\quad C_3 = 44. \]Thus for $t \ge 3$: $v(t) = -18e^{3-t} + 44$.
Integrate velocity to get displacement:
\[ s = \int \bigl(-18e^{3-t} + 44\bigr) \, dt = 18e^{3-t} + 44t + C_4. \]Displacement continuity at $t=3$: $s(3)=20$. Hence:
\[ 20 = 18e^{0} + 44(3) + C_4 = 18 + 132 + C_4, \] \[ C_4 = 20 - 150 = -130. \]Therefore, for $t \ge 3$:
\[ s(t) = 18e^{3-t} + 44t - 130 = \frac{18e^3}{e^t} + 44t - 130. \]Answer (b): $\displaystyle s(t) = \frac{18e^3}{e^t} + 44t - 130 \quad (t \ge 3)$.
Final boxed results
- (a)(i) $v(3) = 26 \; \mathrm{ms}^{-1}$
- (a)(ii) $s(3) = 20 \; \mathrm{m}$
- (b) $\displaystyle s(t) = \frac{18e^3}{e^t} + 44t - 130$ for $t \ge 3$
\[ \boxed{\frac{18e^3}{e^t}+44t-130} \]
Question 11
The normal to the curve $y = \sin(4x - \pi)$ at the point $A(a, 0)$, where $\frac{\pi}{2} \lt a \lt \pi$, meets the $y$-axis at the point $B$. Find the exact area of triangle $OAB$, where $O$ is the origin. [9]
Solution
Solution:
Problem Statement
The normal to the curve $ y = \sin(4x - \pi) $ at the point $ A(a, 0) $, where $ \frac{\pi}{2} \lt a \lt \pi $, meets the $ y $-axis at the point $ B $. Find the exact area of triangle $ OAB $, where $ O $ is the origin.
Step 1: Finding $ a $ where $ y = 0 $
At point $ A(a, 0) $, we have:
\[ y = \sin(4a - \pi) = 0. \]The sine function is zero when:
\[ 4a - \pi = n\pi, \quad n \in \mathbb{Z}. \]Rearranging for $ a $:
\[ a = \frac{n\pi + \pi}{4} = \frac{(n+1)\pi}{4}. \]
Given $ \frac{\pi}{2} < a < \pi $, we test values of $ n $:
- For $ n = 1 $: $ a = \frac{2\pi}{4} = \frac{\pi}{2} $ (not valid as $ a > \frac{\pi}{2} $).
- For $ n = 2 $: $ a = \frac{3\pi}{4} $ (valid).
Thus, $ a = \frac{3\pi}{4} $.
Step 2: Derivative of the curve
The slope of the tangent to the curve $ y = \sin(4x - \pi) $ is:
\[ \frac{dy}{dx} = \cos(4x - \pi) \cdot 4 = 4\cos(4x - \pi). \]At $ x = a = \frac{3\pi}{4} $, the slope of the tangent is:
\[ m_{\text{tangent}} = 4\cos\left(4 \cdot \frac{3\pi}{4} - \pi\right) = 4\cos\left(3\pi - \pi\right) = 4\cos(2\pi) = 4 \cdot 1 = 4. \]Step 3: Slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ m_{\text{normal}} = -\frac{1}{4}. \]Step 4: Equation of the normal
The equation of the normal at point $ A\left(\frac{3\pi}{4}, 0\right) $ is:
\[ y - 0 = m_{\text{normal}} \left(x - \frac{3\pi}{4}\right), \]or equivalently:
\[ y = -\frac{1}{4}\left(x - \frac{3\pi}{4}\right). \]Step 5: Finding point $ B $
The normal intersects the $ y $-axis at point $ B $, where $ x = 0 $. Substituting $ x = 0 $ into the equation of the normal:
\[ y = -\frac{1}{4}\left(0 - \frac{3\pi}{4}\right) = \frac{3\pi}{16}. \]Thus, point $ B $ is:
\[ B\left(0, \frac{3\pi}{16}\right). \]Step 6: Area of triangle $ OAB $
The vertices of triangle $ OAB $ are $ O(0, 0) $, $ A\left(\frac{3\pi}{4}, 0\right) $, and $ B\left(0, \frac{3\pi}{16}\right) $. The area of the triangle is:
\[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1) \right|, \]where $ (x_1, y_1) = (0, 0) $, $ (x_2, y_2) = \left(\frac{3\pi}{4}, 0\right) $, and $ (x_3, y_3) = \left(0, \frac{3\pi}{16}\right) $.
Substituting these values:
\[ \text{Area} = \frac{1}{2} \left| 0 \cdot 0 + \frac{3\pi}{4} \cdot \frac{3\pi}{16} + 0 \cdot 0 - (0 \cdot \frac{3\pi}{4} + 0 \cdot 0 + \frac{3\pi}{16} \cdot 0) \right|. \]Simplify:
\[ \text{Area} = \frac{1}{2} \left| \frac{9\pi^2}{64} \right| = \frac{9\pi^2}{128}. \]Final Answer
The exact area of triangle $ OAB $ is:
\[ \boxed{\frac{9\pi^2}{128}}. \]
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