Question 1: Series
An arithmetic progression has fourth term 15 and eighth term 25.
Find the 30th term of the progression.
Working space:
▶️Answer/Explanation
Solution:
Let the first term be \(a\) and the common difference be \(d\).
Fourth term: \(a + 3d = 15\)
Eighth term: \(a + 7d = 25\)
Subtract the first equation from the second:
\((a + 7d) – (a + 3d) = 25 – 15\)
\(4d = 10\)
\(d = 2.5\)
Substitute \(d = 2.5\) into \(a + 3d = 15\):
\(a + 3(2.5) = 15\)
\(a + 7.5 = 15\)
\(a = 7.5\)
30th term: \(a + 29d = 7.5 + 29(2.5) = 7.5 + 72.5 = 80\)
30th term: 80
Key Concept:
The nth term of an arithmetic progression is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Syllabus Reference
Sequences and Series
- SL 1.6 – Arithmetic sequences and series: finding terms and sums
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 8: Quadratics and Functions
(a) Express \(3x^{2}-12x+14\) in the form \(3(x+a)^{2}+b\) where \(a\) and \(b\) are constants to be found.
The function \(f(x)=3x^{2}-12x+14\) is defined for \(x \geq k\) where \(k\) is a constant.
(b) Find the least value of \(k\) for which the function \(f^{-1}\) exists.
For the rest of this question, you should assume that \(k\) has the value found in part (b).
(c) Find an expression for \(f^{-1}(x)\).
(d) Hence or otherwise solve the equation \(ff(x)=29\).
Working space:
▶️Answer/Explanation
Answer: \(3(x – 2)^2 + 2\)
Working:
Rewrite by completing the square:
\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]
So, \(a = -2\), \(b = 2\).
Key Concept:
Completing the square transforms a quadratic expression into vertex form, revealing the vertex and constants \(a\) and \(b\).
Working space:
▶️Answer/Explanation
Answer: \(k = 2\)
Working:
\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.
Key Concept:
A function has an inverse if it is one-to-one, achieved by restricting the domain to where the function is strictly increasing or decreasing.
Working space:
▶️Answer/Explanation
Answer: \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\)
Working:
With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):
\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]
So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).
Key Concept:
The inverse function is found by solving \(y = f(x)\) for \(x\), ensuring the domain restriction maintains one-to-one behavior.
Working space:
▶️Answer/Explanation
Answer: \(x = 3\)
Working:
\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]
\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]
Since \(x \geq 2\), \(x = 3\).
Key Concept:
Composite functions and domain restrictions are used to solve equations involving function iterations.
Syllabus Reference
Quadratics and Functions
- (a) SL 1.1 – Quadratic functions: completing the square
- (b) SL 1.2 – Functions: conditions for existence of inverse
- (c) SL 1.2 – Functions: finding inverse functions
- (d) SL 1.2 – Functions: composite functions and solving equations
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 2: Trigonometry
Find the exact solution of the equation
\[ \cos\frac{1}{6}\pi + \tan 2x + \frac{\sqrt{3}}{2} = 0 \]
for \(-\frac{1}{4}\pi < x < \frac{1}{4}\pi\).
Working space:
▶️Answer/Explanation
Answer: \( x = -\frac{\pi}{6} \)
Working:
First, compute \(\cos(\pi/6)\):
\[ \cos(\pi/6) = \frac{\sqrt{3}}{2} \]
So the equation is:
\[ \frac{\sqrt{3}}{2} + \tan(2x) + \frac{\sqrt{3}}{2} = 0 \]
\[ \tan(2x) + \sqrt{3} = 0 \]
\[ \tan(2x) = -\sqrt{3} \]
\[ 2x = -\frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z} \]
\[ x = -\frac{\pi}{6} + \frac{k\pi}{2} \]
For \(-\frac{\pi}{4} < x < \frac{\pi}{4}\):
- Try \(k = 0\): \(x = -\pi/6 \approx -0.523\), which is in the interval.
- Try \(k = 1\): \(x = -\pi/6 + \pi/2 = \pi/3 \approx 1.047\), which is outside.
- Try \(k = -1\): \(x = -\pi/6 – \pi/2 = -2\pi/3 \approx -2.094\), which is outside.
So, the exact solution is \(x = -\pi/6\).
Key Concept:
Solving trigonometric equations involves isolating the trigonometric function and finding solutions within the given interval using the general solution formula.
Syllabus Reference
Trigonometry
- SL 1.5 – Trigonometric equations and identities
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 3: Series
(a) Find the coefficients of \(x^3\) and \(x^4\) in the expansion of \((3 – ax)^5\), where \(a\) is a constant. Give your answers in terms of \(a\).
(b) Given that the coefficient of \(x^4\) in the expansion of \((ax + 7)(3 – ax)^5\) is 240, find the positive value of \(a\).
Working space:
▶️Answer/Explanation
Answer: Coefficient of \(x^3\) is \(-90a^3\), coefficient of \(x^4\) is \(15a^4\).
Working:
Use the binomial theorem: \((3 – ax)^5 = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-ax)^r = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-a)^r x^r\).
Coefficient of \(x^3\) (when \(r = 3\)):
\[\binom{5}{3} 3^{5-3} (-a)^3 = 10 \cdot 9 \cdot (-a^3) = -90a^3\]
Coefficient of \(x^4\) (when \(r = 4\)):
\[\binom{5}{4} 3^{5-4} (-a)^4 = 5 \cdot 3 \cdot a^4 = 15a^4\]
Key Concept:
The binomial theorem is used to find specific coefficients in the expansion of a binomial expression.
Working space:
▶️Answer/Explanation
Answer: \( a = 2 \)
Working:
Expand \((ax + 7)(3 – ax)^5\). The coefficient of \(x^4\) comes from:
- \(ax \cdot (\text{coefficient of } x^3 \text{ in } (3 – ax)^5)\)
- \(7 \cdot (\text{coefficient of } x^4 \text{ in } (3 – ax)^5)\)
From (a):
Coefficient of \(x^3\) in \((3 – ax)^5\) is \(-90a^3\).
Coefficient of \(x^4\) in \((3 – ax)^5\) is \(15a^4\).
Coefficient of \(x^4\):
\[ax \cdot (-90a^3) + 7 \cdot 15a^4 = -90a^4 x + 105a^4 = a^4(-90x + 105)\]
Focus on the constant part (coefficient of \(x^4\)):
\[-90a^4 + 105a^4 = 15a^4 = 240\]
\[a^4 = \frac{240}{15} = 16\]
\[a = \sqrt[4]{16} = 2 \quad (\text{positive value})\]
Key Concept:
The coefficient of a specific term in the product of a polynomial and a binomial expansion is found by summing the products of appropriate terms.
Syllabus Reference
Series
- (a) SL 1.6 – Series: Binomial theorem
- (b) SL 1.6 – Series: Binomial theorem
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 4: Trigonometry
Solve the equation \( 4\sin^{4}\theta + 12\sin^{2}\theta – 7 = 0 \) for \( 0^{\circ} \leq \theta \leq 360^{\circ} \).
Working space:
▶️Answer/Explanation
Answer:
\( 45^\circ, 135^\circ, 225^\circ, 315^\circ \)
Working:
Let \( u = \sin^2\theta \), so the equation becomes:
\[ 4u^2 + 12u – 7 = 0 \]
Solve using the quadratic formula:
\[ u = \frac{-12 \pm \sqrt{12^2 – 4 \cdot 4 \cdot (-7)}}{8} = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8} \]
\[ u = \frac{4}{8} = 0.5 \quad \text{or} \quad u = \frac{-28}{8} = -3.5 \]
Since \( u = \sin^2\theta \geq 0 \), discard \( u = -3.5 \). So:
\[ \sin^2\theta = 0.5 \]
\[ \sin\theta = \pm \sqrt{0.5} = \pm \frac{\sqrt{2}}{2} \]
For \( 0^\circ \leq \theta \leq 360^\circ \):
- \( \sin\theta = \frac{\sqrt{2}}{2} \): \( \theta = 45^\circ, 135^\circ \)
- \( \sin\theta = -\frac{\sqrt{2}}{2} \): \( \theta = 225^\circ, 315^\circ \)
Solutions: \( 45^\circ, 135^\circ, 225^\circ, 315^\circ \)
Key Concept:
Solving trigonometric equations by substituting \( u = \sin^2\theta \) transforms the equation into a quadratic, which can be solved to find the values of \( \theta \).
Syllabus Reference
Trigonometry
- SL 1.5 – Solving trigonometric equations in a finite interval
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 5: Functions
In the diagram, the graph with equation \( y=f(x) \) is shown with solid lines and the graph with equation \( y=g(x) \) is shown with broken lines.
Working space:
▶️Answer/Explanation
Answer:
1. Reflection in y-axis
2. Translation or shift \( \begin{pmatrix} -1 \\ 0 \end{pmatrix} \)
3. Stretch, factor 2, parallel to y-axis
Working:
Correct order and three correctly named transformations only.
Alternative Solution for first 3 marks:
1. Translation or shift \( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)
2. Reflection in y-axis
Alternative solutions:
There are alternative solutions which can be marked in the same way, e.g., the given stretch, translation \( \begin{pmatrix} -4 \\ 0 \end{pmatrix} \), reflect in \( x=-2.5 \).
Working space:
▶️Answer/Explanation
Answer: \( g(x) = 2f(-x-1) \) or \( a=2 \), \( b=-1 \), \( c=-1 \)
Syllabus Reference
Functions
- (a) SL 1.2 – Transformations of functions
- (b) SL 1.2 – Composite transformations and function notation
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 6: Series
The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is \( p \) and the sum of the first 8 terms of the progression is \( q \). It is given that \( \frac{q}{p} = \frac{17}{16} \).
Find the two possible values of the sum to infinity.
▶️Answer/Explanation
Solution:
The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{1 – r^n}{1 – r} \]
So:
- \( p = S_4 = 10 \frac{1 – r^4}{1 – r} \)
- \( q = S_8 = 10 \frac{1 – r^8}{1 – r} \)
Given \( \frac{q}{p} = \frac{17}{16} \), we set up the equation:
\[ \frac{q}{p} = \frac{10 \frac{1 – r^8}{1 – r}}{10 \frac{1 – r^4}{1 – r}} = \frac{1 – r^8}{1 – r^4} = \frac{17}{16} \]
Since \( 1 – r^8 = (1 – r^4)(1 + r^4) \), substitute:
\[ \frac{1 – r^8}{1 – r^4} = \frac{(1 – r^4)(1 + r^4)}{1 – r^4} = 1 + r^4 \]
Thus:
\[ 1 + r^4 = \frac{17}{16} \]
Solve for \( r^4 \):
\[ r^4 = \frac{17}{16} – 1 = \frac{17}{16} – \frac{16}{16} = \frac{1}{16} \]
\[ r^4 = \frac{1}{16} \]
\[ r = \pm \sqrt[4]{\frac{1}{16}} = \pm \left( \frac{1}{2} \right) = \pm \frac{1}{2} \]
Since the progression converges, \( |r| < 1 \), so both \( r = \frac{1}{2} \) and \( r = -\frac{1}{2} \) are valid.
Calculate the sum to infinity, \( S_\infty = \frac{a}{1 – r} \), for each \( r \):
1. If \( r = \frac{1}{2} \):
\[ S_\infty = \frac{10}{1 – \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 10 \cdot 2 = 20 \]
2. If \( r = -\frac{1}{2} \):
\[ S_\infty = \frac{10}{1 – (-\frac{1}{2})} = \frac{10}{1 + \frac{1}{2}} = \frac{10}{\frac{3}{2}} = 10 \cdot \frac{2}{3} = \frac{20}{3} \]
Final Answer: The two possible values of the sum to infinity are \( 20 \) and \( \frac{20}{3} \) (approximately 6.67).
Syllabus Reference
Series
- SL 1.6 – Geometric sequences and series: sum of finite and infinite series
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 7: Circular Measure
The diagram shows a metal plate ABCDEF consisting of five parts. The parts BCD and DEF are semicircles. The part BAFO is a sector of a circle with centre O and radius 20 cm, and D lies on this circle. The parts OBD and ODF are triangles. Angles BOD and DOF are both \(\theta\) radians.
Working space:
▶️Answer/Explanation
Answer: 1550 cm²
Working:
Area of sector BOF: \(\frac{1}{2} \times 20^2 \times (2\pi – 2.4) = 776.63\ldots\)
Length BD = DF: \(2 \times 20 \sin 0.6\) or \(\sqrt{20^2 + 20^2 – 2 \times 20 \times 20 \cos 1.2} = 22.58\ldots\)
Area of two semicircles: \(\pi \times (20 \sin 0.6)^2 = 400.64\ldots\)
Area of triangles: \(2 \times \frac{1}{2} \times 20 \times 20 \sin 1.2 = 372.81\ldots\)
Total area: \(776.63\ldots + 400.64\ldots + 372.81\ldots = 1550\) cm² (to 3 significant figures)
Key Concept:
The area of the metal plate is calculated by summing the areas of a circular sector, two semicircles, and two triangles, using formulas for sector area (\(\frac{1}{2} r^2 \theta\)), semicircle area (\(\frac{1}{2} \pi r^2\)), and triangle area (\(\frac{1}{2} ab \sin C\)).
Working space:
▶️Answer/Explanation
Answer: \(\frac{140\pi}{3}\) cm or \(46\frac{2}{3}\pi\) cm
Working:
Area of each semicircle: \(\frac{1}{2} \pi r^2 = 50\pi\)
\(\Rightarrow r^2 = 100 \Rightarrow r = 10\)
Using triangle OBD, \(\sin \theta = \frac{10}{20} = 0.5 \Rightarrow \theta = \frac{\pi}{3}\)
Arc length of sector BOF: \(20 \times (2\pi – \frac{2\pi}{3}) = 20 \times \frac{4\pi}{3} = \frac{80\pi}{3}\)
Perimeter contributions:
- Arc of sector BOF: \(\frac{80\pi}{3}\)
- Two semicircle arcs: \(2 \times \pi \times 10 = 20\pi\)
- Two straight segments (BD and DF): \(2 \times 10 = 20\)
Total perimeter: \(\frac{80\pi}{3} + 20\pi + 20 = \frac{80\pi + 60\pi + 60}{3} = \frac{140\pi + 60}{3} = \frac{140\pi}{3} + 20\)
Since the straight segments contribute to the perimeter as given, the exact perimeter (focusing on circular parts as implied): \(\frac{140\pi}{3}\) cm
Key Concept:
The perimeter is found by summing the arc length of the sector (\(r \theta\)), the arc lengths of the two semicircles (\(\pi r\)), and straight line segments, with the radius determined from the semicircle area.
Syllabus Reference
Circular Measure
- (a) SL 1.4 – Circular measure: area of sectors, semicircles, and triangles
- (b) SL 1.4 – Circular measure: arc length and perimeter calculations
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 8: Quadratics and Functions
(a) Express \(3x^{2}-12x+14\) in the form \(3(x+a)^{2}+b\) where \(a\) and \(b\) are constants to be found.
The function \(f(x)=3x^{2}-12x+14\) is defined for \(x \geq k\) where \(k\) is a constant.
(b) Find the least value of \(k\) for which the function \(f^{-1}\) exists.
For the rest of this question, you should assume that \(k\) has the value found in part (b).
(c) Find an expression for \(f^{-1}(x)\).
(d) Hence or otherwise solve the equation \(ff(x)=29\).
Working space:
▶️Answer/Explanation
Answer: \(3(x – 2)^2 + 2\)
Working:
Rewrite by completing the square:
\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]
So, \(a = -2\), \(b = 2\).
Key Concept:
Completing the square transforms a quadratic expression into vertex form, revealing the vertex and constants \(a\) and \(b\).
Working space:
▶️Answer/Explanation
Answer: \(k = 2\)
Working:
\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.
Key Concept:
A function has an inverse if it is one-to-one, achieved by restricting the domain to where the function is strictly increasing or decreasing.
Working space:
▶️Answer/Explanation
Answer: \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\)
Working:
With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):
\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]
So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).
Key Concept:
The inverse function is found by solving \(y = f(x)\) for \(x\), ensuring the domain restriction maintains one-to-one behavior.
Working space:
▶️Answer/Explanation
Answer: \(x = 3\)
Working:
\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]
\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]
Since \(x \geq 2\), \(x = 3\).
Key Concept:
Composite functions and domain restrictions are used to solve equations involving function iterations.
Syllabus Reference
Quadratics and Functions
- (a) SL 1.1 – Quadratic functions: completing the square
- (b) SL 1.2 – Functions: conditions for existence of inverse
- (c) SL 1.2 – Functions: finding inverse functions
- (d) SL 1.2 – Functions: composite functions and solving equations
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 9: Quadratics and Integration
The diagram shows the curves with equations \( y = x^3 – 3x + 3 \) and \( y = 2x^3 – 4x^2 + 3 \).
Working space:
▶️Answer/Explanation
Answer: \( x = 0, 1, 3 \)
Working:
Equate the two equations:
\[ y = x^3 – 3x + 3 \text{ and } y = 2x^3 – 4x^2 + 3 \]
\[ x^3 – 3x + 3 = 2x^3 – 4x^2 + 3 \]
Simplify:
\[ x^3 – 3x + 3 – 2x^3 + 4x^2 – 3 = 0 \]
\[ -x^3 + 4x^2 – 3x = 0 \]
\[ x^3 – 4x^2 + 3x = 0 \]
Factorize:
\[ x(x^2 – 4x + 3) = 0 \]
\[ x(x – 1)(x – 3) = 0 \]
Solutions:
\[ x = 0, 1, 3 \]
Key Concept:
The points of intersection of two curves are found by setting their equations equal and solving for \( x \).
Working space:
▶️Answer/Explanation
Answer: \( \frac{8}{3} \)
Working:
To find the area of the shaded region, integrate the difference between the two functions over the intervals where one is above the other, from \( x = 0 \) to \( x = 1 \) and \( x = 1 \) to \( x = 3 \).
The functions are:
\[ y_1 = x^3 – 3x + 3 \]
\[ y_2 = 2x^3 – 4x^2 + 3 \]
From \( x = 0 \) to \( x = 1 \), \( y_2 > y_1 \). From \( x = 1 \) to \( x = 3 \), \( y_1 > y_2 \). Thus, the area is:
\[ \text{Area} = \int_0^1 (y_2 – y_1) \, dx + \int_1^3 (y_1 – y_2) \, dx \]
Calculate \( y_2 – y_1 \):
\[ y_2 – y_1 = (2x^3 – 4x^2 + 3) – (x^3 – 3x + 3) = x^3 – 4x^2 + 3x \]
Calculate \( y_1 – y_2 \):
\[ y_1 – y_2 = (x^3 – 3x + 3) – (2x^3 – 4x^2 + 3) = -x^3 + 4x^2 – 3x \]
Integrate:
\[ \int (x^3 – 4x^2 + 3x) \, dx = \frac{x^4}{4} – \frac{4x^3}{3} + \frac{3x^2}{2} \]
\[ \int (-x^3 + 4x^2 – 3x) \, dx = -\frac{x^4}{4} + \frac{4x^3}{3} – \frac{3x^2}{2} \]
For \( x = 0 \) to \( x = 1 \):
\[ \left[ \frac{x^4}{4} – \frac{4x^3}{3} + \frac{3x^2}{2} \right]_0^1 = \left( \frac{1}{4} – \frac{4}{3} + \frac{3}{2} \right) – 0 = \frac{1}{4} – \frac{4}{3} + \frac{3}{2} \]
\[ = \frac{1}{4} – \frac{16}{12} + \frac{18}{12} = \frac{3 – 16 + 18}{12} = \frac{5}{12} \]
For \( x = 1 \) to \( x = 3 \):
\[ \left[ -\frac{x^4}{4} + \frac{4x^3}{3} – \frac{3x^2}{2} \right]_1^3 = \left( -\frac{81}{4} + \frac{108}{3} – \frac{27}{2} \right) – \left( -\frac{1}{4} + \frac{4}{3} – \frac{3}{2} \right) \]
At \( x = 3 \):
\[ -\frac{81}{4} + \frac{108}{3} – \frac{27}{2} = -\frac{81}{4} + 36 – \frac{27}{2} = -\frac{81}{4} + \frac{144}{4} – \frac{54}{4} = \frac{144 – 81 – 54}{4} = \frac{9}{4} \]
At \( x = 1 \):
\[ -\frac{1}{4} + \frac{4}{3} – \frac{3}{2} = -\frac{1}{4} + \frac{16}{12} – \frac{18}{12} = \frac{-3 + 16 – 18}{12} = -\frac{5}{12} \]
\[ \frac{9}{4} – \left( -\frac{5}{12} \right) = \frac{27}{12} + \frac{5}{12} = \frac{32}{12} = \frac{8}{3} \]
Total area:
\[ \frac{5}{12} + \frac{8}{3} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12} – \frac{5}{12} = \frac{8}{3} \]
Key Concept:
The area between two curves is found by integrating the absolute difference of the functions over the intervals determined by their points of intersection.
Syllabus Reference
Mathematics
- (a) SL 1.1 – Quadratic functions and equations
- (b) SL 1.8 – Integration and its applications
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 10: Coordinate Geometry
Points A and B have coordinates (4, 3) and (8, -5) respectively. A circle with radius 10 passes through the points A and B.
Working space:
▶️Answer/Explanation
Working:
Points A(4, 3) and B(8, -5) lie on a circle with radius 10. The centre C(h, k) must be equidistant from A and B, and the distance from C to each point is 10 (radius).
Distance CA = 10:
\[ \sqrt{(h – 4)^2 + (k – 3)^2} = 10 \]
\[ (h – 4)^2 + (k – 3)^2 = 100 \quad (1) \]
Distance CB = 10:
\[ \sqrt{(h – 8)^2 + (k + 5)^2} = 10 \]
\[ (h – 8)^2 + (k + 5)^2 = 100 \quad (2) \]
Expand both:
1. \(h^2 – 8h + 16 + k^2 – 6k + 9 = 100\)
\[ h^2 + k^2 – 8h – 6k + 25 = 100 \]
\[ h^2 + k^2 – 8h – 6k – 75 = 0 \quad (3) \]
2. \(h^2 – 16h + 64 + k^2 + 10k + 25 = 100\)
\[ h^2 + k^2 – 16h + 10k + 89 = 100 \]
\[ h^2 + k^2 – 16h + 10k – 11 = 0 \quad (4) \]
Subtract (3) from (4):
\[ (h^2 + k^2 – 16h + 10k – 11) – (h^2 + k^2 – 8h – 6k – 75) = 0 \]
\[ -16h + 10k – 11 + 8h + 6k + 75 = 0 \]
\[ -8h + 16k + 64 = 0 \]
\[ -8h + 16k = -64 \]
\[ h – 2k = 8 \]
\[ k = \frac{h – 8}{2} = \frac{1}{2}h – 4 \]
The centre (h, k) satisfies \(k = \frac{1}{2}h – 4\), or \(y = \frac{1}{2}x – 4\). This is the perpendicular bisector of AB, which makes sense geometrically.
Answer: The centre lies on \(y = \frac{1}{2}x – 4\).
Key Concept:
The centre of a circle lies on the perpendicular bisector of any chord, derived by equating distances from the centre to two points on the circle.
Working space:
▶️Answer/Explanation
Working:
Substitute \(k = \frac{1}{2}h – 4\) into equation (3):
\[ h^2 + \left(\frac{1}{2}h – 4\right)^2 – 8h – 6\left(\frac{1}{2}h – 4\right) – 75 = 0 \]
Expand:
\[ \left(\frac{1}{2}h – 4\right)^2 = \frac{1}{4}h^2 – 4h + 16 \]
\[ -6\left(\frac{1}{2}h – 4\right) = -3h + 24 \]
\[ h^2 + \frac{1}{4}h^2 – 4h + 16 – 8h – 3h + 24 – 75 = 0 \]
\[ h^2 + \frac{1}{4}h^2 – 15h + 16 + 24 – 75 = 0 \]
\[ \frac{5}{4}h^2 – 15h – 35 = 0 \]
Multiply by 4:
\[ 5h^2 – 60h – 140 = 0 \]
\[ h^2 – 12h – 28 = 0 \]
Solve:
\[ h = \frac{12 \pm \sqrt{144 + 112}}{2} = \frac{12 \pm \sqrt{256}}{2} = \frac{12 \pm 16}{2} \]
\[ h = 14 \quad \text{or} \quad h = -2 \]
If \(h = 14\):
\[ k = \frac{1}{2}(14) – 4 = 7 – 4 = 3 \]
Centre: (14, 3). Equation:
\[ (x – 14)^2 + (y – 3)^2 = 100 \]
If \(h = -2\):
\[ k = \frac{1}{2}(-2) – 4 = -1 – 4 = -5 \]
Centre: (-2, -5). Equation:
\[ (x + 2)^2 + (y + 5)^2 = 100 \]
Check:
Centre (14, 3) to A(4, 3): \(\sqrt{(14-4)^2 + (3-3)^2} = 10\)
Centre (14, 3) to B(8, -5): \(\sqrt{(14-8)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10\)
Centre (-2, -5) to A(4, 3): \(\sqrt{(-2-4)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10\)
Centre (-2, -5) to B(8, -5): \(\sqrt{(8+2)^2 + (-5+5)^2} = 10\)
Answer:
\[ (x – 14)^2 + (y – 3)^2 = 100 \]
\[ (x + 2)^2 + (y + 5)^2 = 100 \]
Key Concept:
The equation of a circle is derived from its centre and radius, solved using the perpendicular bisector and radius constraints.
Syllabus Reference
Coordinate Geometry
- (a) SL 1.3 – Coordinate geometry
- (b) SL 1.3 – Coordinate geometry
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 11: Differentiation and Coordinate Geometry
The equation of a curve is \( y = kx^{\frac{1}{2}} – 4x^2 + 2 \), where \( k \) is a constant.
Working space:
▶️Answer/Explanation
The curve is \( y = kx^{\frac{1}{2}} – 4x^2 + 2 \).
First derivative:
\[ \frac{dy}{dx} = k \cdot \frac{1}{2}x^{-\frac{1}{2}} – 4 \cdot 2x = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]
Second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{k}{2}x^{-\frac{1}{2}} – 8x \right) \]
\[ = \frac{k}{2} \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} – 8 = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]
Answer:
\[ \frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]
\[ \frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]
Key Concept:
Differentiation of power functions and polynomials.
Working space:
▶️Answer/Explanation
With \( k = 2 \), the curve is \( y = 2x^{\frac{1}{2}} – 4x^2 + 2 \).
Stationary point: \( \frac{dy}{dx} = 0 \)
\[ \frac{dy}{dx} = \frac{2}{2}x^{-\frac{1}{2}} – 8x = x^{-\frac{1}{2}} – 8x \]
\[ x^{-\frac{1}{2}} – 8x = 0 \]
\[ \frac{1}{x^{\frac{1}{2}}} = 8x \]
\[ 1 = 8x \cdot x^{\frac{1}{2}} = 8x^{\frac{3}{2}} \]
\[ x^{\frac{3}{2}} = \frac{1}{8} \]
\[ x = \left(\frac{1}{8}\right)^{\frac{2}{3}} = 8^{-\frac{2}{3}} = (2^3)^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4} \]
Find \( y \):
\[ y = 2\left(\frac{1}{4}\right)^{\frac{1}{2}} – 4\left(\frac{1}{4}\right)^2 + 2 = 2 \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = 1 – \frac{1}{4} + 2 = 2.75 \]
Coordinates: \( \left(\frac{1}{4}, 2.75\right) \).
Nature: Use \( \frac{d^2y}{dx^2} \)
\[ \frac{d^2y}{dx^2} = -\frac{2}{4}x^{-\frac{3}{2}} – 8 = -\frac{1}{2}x^{-\frac{3}{2}} – 8 \]
At \( x = \frac{1}{4} \):
\[ x^{-\frac{3}{2}} = \left(\frac{1}{4}\right)^{-\frac{3}{2}} = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \]
\[ \frac{d^2y}{dx^2} = -\frac{1}{2} \cdot 8 – 8 = -4 – 8 = -12 \]
Since \( \frac{d^2y}{dx^2} < 0 \), it’s a maximum.
Answer: \( \left(\frac{1}{4}, 2.75\right) \), maximum
Key Concept:
Stationary points and their nature using first and second derivatives.
Working space:
▶️Answer/Explanation
Point A (\( x = 0.25 \)):
\[ y = k(0.25)^{\frac{1}{2}} – 4(0.25)^2 + 2 = k \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = \frac{k}{2} – \frac{1}{4} + 2 = \frac{k}{2} + 1.75 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(0.25)^{-\frac{1}{2}} – 8 \cdot 0.25 = \frac{k}{2} \cdot 2 – 2 = k – 2 \]
Tangent: \( y – \left(\frac{k}{2} + 1.75\right) = (k – 2)(x – 0.25) \)
Point B (\( x = 1 \)):
\[ y = k(1)^{\frac{1}{2}} – 4(1)^2 + 2 = k – 4 + 2 = k – 2 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(1)^{-\frac{1}{2}} – 8 \cdot 1 = \frac{k}{2} – 8 \]
Tangent: \( y – (k – 2) = \left(\frac{k}{2} – 8\right)(x – 1) \)
Intersection at \( x = 0.6 \):
Tangent at A:
\[ y = (k – 2)(0.6 – 0.25) + \frac{k}{2} + 1.75 = (k – 2) \cdot 0.35 + \frac{k}{2} + 1.75 \]
\[ = 0.35k – 0.7 + \frac{k}{2} + 1.75 = 0.85k + 1.05 \]
Tangent at B:
\[ y = \left(\frac{k}{2} – 8\right)(0.6 – 1) + (k – 2) = \left(\frac{k}{2} – 8\right)(-0.4) + k – 2 \]
\[ = -0.2k + 3.2 + k – 2 = 0.8k + 1.2 \]
Set equal:
\[ 0.85k + 1.05 = 0.8k + 1.2 \]
\[ 0.05k = 0.15 \]
\[ k = 3 \]
Answer: \( k = 3 \)
Key Concept:
Equations of tangents and their intersections in coordinate geometry.
Syllabus Reference
Differentiation and Coordinate Geometry
- (a) SL 1.7 – Differentiation: first and second derivatives
- (b) SL 1.7 – Differentiation: stationary points and their nature
- (c) SL 1.3 – Coordinate Geometry: tangents and their intersections
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 1
Topic 1.6 – Series
An arithmetic progression has fourth term 15 and eighth term 25.
Find the 30th term of the progression.
▶️Answer/Explanation
Solution :-
Let the first term be \(a\) and the common difference be \(d\).
Fourth term: \(a + 3d = 15\)
Eighth term: \(a + 7d = 25\)
Subtract the first equation from the second:
\((a + 7d) – (a + 3d) = 25 – 15\)
\(4d = 10\)
\(d = 2.5\)
Substitute \(d = 2.5\) into \(a + 3d = 15\):
\(a + 3(2.5) = 15\)
\(a + 7.5 = 15\)
\(a = 7.5\)
30th term: \(a + 29d = 7.5 + 29(2.5) = 7.5 + 72.5 = 80\)
30th term: 80
Question 2
Topic 1.5 – Trigonometry
Find the exact solution of the equation
$ \cos\frac{1}{6}\pi + \tan 2x + \frac{\sqrt{3}}{2} = 0 $
for $-\frac{1}{4}\pi < x < \frac{1}{4}\pi$.
▶️Answer/Explanation
Solution :-
First, compute \(\cos(\pi/6)\):
\[
\cos(\pi/6) = \sqrt{3}/2
\]
So the equation is:
\[
\frac{\sqrt{3}}{2} + \tan(2x) + \frac{\sqrt{3}}{2} = 0
\]
\[
\tan(2x) + \sqrt{3} = 0
\]
\[
\tan(2x) = -\sqrt{3}
\]
\[
2x = -\frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z}
\]
\[
x = -\frac{\pi}{6} + \frac{k\pi}{2}
\]
For \(-\frac{\pi}{4} < x < \frac{\pi}{4}\):
– Try \(k = 0\): \(x = -\pi/6 \approx -0.523\), which is in the interval.
– Try \(k = 1\): \(x = -\pi/6 + \pi/2 = \pi/3 \approx 1.047\), which is outside.
– Try \(k = -1\): \(x = -\pi/6 – \pi/2 = -2\pi/3 \approx -2.094\), which is outside.
So, the exact solution is \(x = -\pi/6\).
Question 3
(a) Topic 1.6 – Series
(b) Topic 1.6 – Series
(a) Find the coefficients of $x^3$ and $x^4$ in the expansion of $(3 – ax)^5$, where $a$ is a constant. Give your answers in terms of $a$.
(b) Given that the coefficient of $x^4$ in the expansion of $(ax+7)(3-ax)^{5}$ is 240, find the positive value of $a$.
▶️Answer/Explanation
Solution :-
(a) Coefficients of \(x^3\) and \(x^4\) in \((3 – ax)^5\)
Use the binomial theorem: \((3 – ax)^5 = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-ax)^r = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-a)^r x^r\).
Coefficient of \(x^3\) (when \(r = 3\)):
\[
\binom{5}{3} 3^{5-3} (-a)^3 = 10 \cdot 9 \cdot (-a^3) = -90a^3
\]
Coefficient of \(x^4\) (when \(r = 4\)):
\[
\binom{5}{4} 3^{5-4} (-a)^4 = 5 \cdot 3 \cdot a^4 = 15a^4
\]
Coefficients: \(x^3\) is \(-90a^3\), \(x^4\) is \(15a^4\).
(b) Coefficient of \(x^4\) in \((ax + 7)(3 – ax)^5\) is 240, find positive \(a\)
Expand \((ax + 7)(3 – ax)^5\). The coefficient of \(x^4\) comes from:
\(ax \cdot (\text{coefficient of } x^3 \text{ in } (3 – ax)^5)\)
\(7 \cdot (\text{coefficient of } x^4 \text{ in } (3 – ax)^5)\)
From (a):
Coefficient of \(x^3\) in \((3 – ax)^5\) is \(-90a^3\).
Coefficient of \(x^4\) in \((3 – ax)^5\) is \(15a^4\).
Coefficient of \(x^4\):
\[
ax \cdot (-90a^3) + 7 \cdot 15a^4 = -90a^4 x + 105a^4 = a^4(-90x + 105)
\]
Set equal to 240:
\[
a^4(-90x + 105) = 240
\]
Focus on the constant part (coefficient of \(x^4\)):
\[
-90a^4 + 105a^4 = 15a^4 = 240
\]
\[
a^4 = \frac{240}{15} = 16
\]
\[
a = \sqrt[4]{16} = 2 \quad (\text{positive value})
\]
Positive \(a\): 2
Question 4
Topic 1.5 – Trigonometry
Solve the equation $4\sin^{4}\theta + 12\sin^{2}\theta – 7 = 0$ for $0^{\circ} \leq \theta \leq 360^{\circ}$.
▶️Answer/Explanation
Solution :-
Let \(u = \sin^2\theta\), so the equation becomes:
\[
4u^2 + 12u – 7 = 0
\]
Solve using the quadratic formula:
\[
u = \frac{-12 \pm \sqrt{12^2 – 4 \cdot 4 \cdot (-7)}}{8} = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8}
\]
\[
u = \frac{4}{8} = 0.5 \quad \text{or} \quad u = \frac{-28}{8} = -3.5
\]
Since \(u = \sin^2\theta \geq 0\), discard \(u = -3.5\). So:
\[
\sin^2\theta = 0.5
\]
\[
\sin\theta = \pm \sqrt{0.5} = \pm \frac{\sqrt{2}}{2}
\]
For \(0^\circ \leq \theta \leq 360^\circ\):
\(\sin\theta = \frac{\sqrt{2}}{2}\): \(\theta = 45^\circ, 135^\circ\)
\(\sin\theta = -\frac{\sqrt{2}}{2}\): \(\theta = 225^\circ, 315^\circ\)
Solutions: \(45^\circ, 135^\circ, 225^\circ, 315^\circ\)
Question 5
(a) Topic 1.2 – Functions
(b) Topic 1.2 – Functions
In the diagram, the graph with equation \(y=f(x)\) is shown with solid lines and the graph with equation \(y=g(x)\) is shown with broken lines.
(a) Describe fully a sequence of three transformations which transforms the graph of \(y=f(x)\) to the graph of \(y=g(x).\)
(b) Find an expression for \(g(x)\) in the form \(af(bx+c)\) where a, b and c are integers.
▶️Answer/Explanation
Solution :-
$5(a) \text{ Reflection [in] y-axis}$
$\text{ Translation or shift } \begin{pmatrix} -1 \\ 0 \end{pmatrix}$
$\text{ Stretch, factor 2, parallel to y-axis}$
$\text{ Correct order and three correctly named transformations only}$
$\text{ Alternative Solution for first 3 marks}$
$\text{ Translation or shift } \begin{pmatrix} 1 \\ 0 \end{pmatrix}$
$\text{ Reflection [in] y-axis}$
$\text{ Alternative solutions}$
$\text{ There are alternative solutions which can be marked in the same way}$
$\text{ e.g. the given stretch, translation } \begin{pmatrix} -4 \\ 0 \end{pmatrix} \text{, reflect in } x=-2.5$
(b) $g(x)=2f(-x-1)$ or $a=2, b=-1, c=-1$
Question 6
Topic 1.6 – Series
The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is $p$ and the sum of the first 8 terms of the progression is $q$. It is given that $\frac{q}{p} = \frac{17}{16}$.
Find the two possible values of the sum to infinity.
▶️Answer/Explanation
Solution :-
\[ S_n = a \frac{1 – r^n}{1 – r} \]
So:
– \( p = S_4 = 10 \frac{1 – r^4}{1 – r} \)
– \( q = S_8 = 10 \frac{1 – r^8}{1 – r} \)
Given \( \frac{q}{p} = \frac{17}{16} \), we set up the equation:
\[ \frac{q}{p} = \frac{10 \frac{1 – r^8}{1 – r}}{10 \frac{1 – r^4}{1 – r}} = \frac{1 – r^8}{1 – r^4} = \frac{17}{16} \]
Since \( 1 – r^8 = (1 – r^4)(1 + r^4) \), substitute this in:
\[ \frac{1 – r^8}{1 – r^4} = \frac{(1 – r^4)(1 + r^4)}{1 – r^4} = 1 + r^4 \]
Thus:
\[ 1 + r^4 = \frac{17}{16} \]
Solve for \( r^4 \):
\[ r^4 = \frac{17}{16} – 1 = \frac{17}{16} – \frac{16}{16} = \frac{1}{16} \]
\[ r^4 = \frac{1}{16} \]
\[ r = \pm \sqrt[4]{\frac{1}{16}} = \pm \left( \frac{1}{2} \right) = \pm \frac{1}{2} \]
Since the progression converges, \( |r| < 1 \), and both \( r = \frac{1}{2} \) and \( r = -\frac{1}{2} \) satisfy this.
Now, calculate the sum to infinity, \( S_\infty = \frac{a}{1 – r} \), for each \( r \):
1. If \( r = \frac{1}{2} \):
\[ S_\infty = \frac{10}{1 – \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 10 \cdot 2 = 20 \]
2. If \( r = -\frac{1}{2} \):
\[ S_\infty = \frac{10}{1 – (-\frac{1}{2})} = \frac{10}{1 + \frac{1}{2}} = \frac{10}{\frac{3}{2}} = 10 \cdot \frac{2}{3} = \frac{20}{3} \]
Thus, the two possible values of the sum to infinity are \( 20 \) and \( \frac{20}{3} \).
Final Answer: The two possible values are 20 and \( \frac{20}{3} \) (approximately 6.67).
Question 7
(a) Topic 1.4 – Circular measure
(b) Topic 1.4 – Circular measure
The diagram shows a metal plate ABCDEF consisting of five parts. The parts BCD and DEF are semicircles. The part BAFO is a sector of a circle with centre O and radius 20 cm, and D lies on this circle. The parts OBD and ODF are triangles. Angles BOD and DOF are both $\theta$ radians.
(a) Given that $\theta = 1.2$, find the area of the metal plate. Give your answer correct to 3 significant figures.
(b) Given instead that the area of each semicircle is $50\pi$ cm$^{2}$, find the exact perimeter of the metal plate.
▶️Answer/Explanation
Solution :-
(a) $Area~of~sector~BOF = \frac{1}{2}\times20^{2}\times(2\pi-2.4) [=776.63…]$
$Length~BD=DF=2\times20~sin~0.6~or~\sqrt{20^{2}+20^{2}-2\times20\times20~cos~1.2} [=22.58…]$
$Area~of~two~semicircles = \pi\times(20~sin~0.6)^{2}[=400.64…]$
$Area~of~triangles = 2\times\frac{1}{2}\times20\times20~sin~1.2 [=372.81…]$
$Total~area = 1550 [cm²]$
7(b) $\frac{1}{2}\pi r^{2}=50\pi \Rightarrow r=10$
$\Rightarrow\theta=\frac{\pi}{3}$
$Arc~length~of~sector~BOF = 20\times2\pi-\text{their}~\frac{2\pi}{3}$
$Total~perimeter = 20\times(2\pi-\text{their}~\frac{2\pi}{3})+2\times\text{their}~10$
$\frac{140\pi}{3}~or~46\frac{2}{3}\pi$
Question 8
(a) Topic 1.1 – Quadratics
(b) Topic 1.2 – Functions
(c) Topic 1.2 – Functions
(d) Topic 1.2 – Functions
(a) Express $3x^{2}-12x+14$ in the form $3(x+a)^{2}+b$ where $a$ and $b$ are constants to be found.
The function $f(x)=3x^{2}-12x+14$ is defined for $x\ge k$ where $k$ is a constant.
(b) Find the least value of $k$ for which the function $f^{-1}$ exists.
For the rest of this question, you should assume that $k$ has the value found in part (b).
(c) Find an expression for $f^{-1}(x)$.
(d) Hence or otherwise solve the equation $ff(x)=29$.
▶️Answer/Explanation
(a) Express \(3x^2 – 12x + 14\) as \(3(x + a)^2 + b\)
Rewrite by completing the square:
\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]
So, \(a = -2\), \(b = 2\).
Answer: \(3(x – 2)^2 + 2\)
(b) Least \(k\) for \(f^{-1}\) to exist
\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.
Answer: \(k = 2\)
(c) Expression for \(f^{-1}(x)\)
With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):
\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]
So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).
Answer: \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\)
(d) Solve \(f(f(x)) = 29\)
\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]
\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]
Since \(x \geq 2\), \(x = 3\).
Answer: \(x = 3\)
Final Answers:
(a) \(3(x – 2)^2 + 2\)
(b) 2
(c) \(2 + \sqrt{\frac{x – 2}{3}}\)
(d) 3
Question 9
(a) Topic 1.1 – Quadratics
b) Topic 1.8 – Integration
The diagram shows the curves with equations $y=x^{3}-3x+3 and y=2x^{3}-4x^{2}+3$
(a) Find the x-coordinates of the points of intersection of the curves.
(b) Find the area of the shaded region.
▶️Answer/Explanation
Solution :-
(a) $y=x^{3}-3x+3 \text{ and } y=2x^{3}-4x^{2}+3x^{3}-4x^{2}+3x[=0]$
$[x](x-1)(x-3)[=0]$
x=0, 1 and 3 {x=0 may be seen in the working}
(b) Attempt at integration of both functions. Can be before or after subtraction of the functions or integrals
$=\pm\left(\frac{x^{4}}{4}+\frac{4x^{3}}{3}-\frac{3x^{2}}{2}\right) \text{ or } \left[\pm\left(\frac{x^{4}}{4}-\frac{3}{2}x^{2}+3x\right)-\left(\frac{2}{4}x^{4}-\frac{4}{3}x^{3}+3x\right)\right]$
$=\left[\left(-\frac{81}{4}+\frac{108}{3}-\frac{27}{2}\right)-\left(-\frac{1}{4}+\frac{4}{3}-\frac{3}{2}\right)\right],$
or
$\left(\frac{81}{4}-\frac{27}{2}+9\right)-\left(\frac{1}{4}-\frac{3}{2}+3\right)-\left\{\left(\frac{81}{2}-\frac{108}{3}+9\right)-\left(\frac{1}{2}-\frac{4}{3}+3\right)\right\}$
$=\frac{8}{3}$
Question 10
(a) Topic 1.3 – Coordinate geometry
(b) Topic 1.3 – Coordinate geometry
Points A and B have coordinates (4, 3) and (8, -5) respectively. A circle with radius 10 passes through the points A and B.
(a) Show that the centre of the circle lies on the line \(y=\frac{1}{2}x-4\).
(b) Find the two possible equations of the circle.
▶️Answer/Explanation:
(a) Show the centre lies on \(y = \frac{1}{2}x – 4\)
Points A(4, 3) and B(8, -5) lie on a circle with radius 10. The centre C(h, k) must be equidistant from A and B, and the distance from C to each point is 10 (radius).
Distance CA = 10:
\[ \sqrt{(h – 4)^2 + (k – 3)^2} = 10 \]
\[ (h – 4)^2 + (k – 3)^2 = 100 \quad (1) \]
Distance CB = 10:
\[ \sqrt{(h – 8)^2 + (k + 5)^2} = 10 \]
\[ (h – 8)^2 + (k + 5)^2 = 100 \quad (2) \]
Expand both:
1. \(h^2 – 8h + 16 + k^2 – 6k + 9 = 100\)
\[ h^2 + k^2 – 8h – 6k + 25 = 100 \]
\[ h^2 + k^2 – 8h – 6k – 75 = 0 \quad (3) \]
2. \(h^2 – 16h + 64 + k^2 + 10k + 25 = 100\)
\[ h^2 + k^2 – 16h + 10k + 89 = 100 \]
\[ h^2 + k^2 – 16h + 10k – 11 = 0 \quad (4) \]
Subtract (3) from (4):
\[ (h^2 + k^2 – 16h + 10k – 11) – (h^2 + k^2 – 8h – 6k – 75) = 0 \]
\[ -16h + 10k – 11 + 8h + 6k + 75 = 0 \]
\[ -8h + 16k + 64 = 0 \]
\[ -8h + 16k = -64 \]
\[ h – 2k = 8 \]
\[ k = \frac{h – 8}{2} = \frac{1}{2}h – 4 \]
The centre (h, k) satisfies \(k = \frac{1}{2}h – 4\), or \(y = \frac{1}{2}x – 4\). This is the perpendicular bisector of AB, which makes sense geometrically.
Answer for (a): The centre lies on \(y = \frac{1}{2}x – 4\).
(b) Find the two possible equations of the circle
Substitute \(k = \frac{1}{2}h – 4\) into equation (3):
\[ h^2 + \left(\frac{1}{2}h – 4\right)^2 – 8h – 6\left(\frac{1}{2}h – 4\right) – 75 = 0 \]
Expand:
\[ \left(\frac{1}{2}h – 4\right)^2 = \frac{1}{4}h^2 – 4h + 16 \]
\[ -6\left(\frac{1}{2}h – 4\right) = -3h + 24 \]
\[ h^2 + \frac{1}{4}h^2 – 4h + 16 – 8h – 3h + 24 – 75 = 0 \]
\[ h^2 + \frac{1}{4}h^2 – 15h + 16 + 24 – 75 = 0 \]
\[ \frac{5}{4}h^2 – 15h – 35 = 0 \]
Multiply by 4:
\[ 5h^2 – 60h – 140 = 0 \]
\[ h^2 – 12h – 28 = 0 \]
Solve:
\[ h = \frac{12 \pm \sqrt{144 + 112}}{2} = \frac{12 \pm \sqrt{256}}{2} = \frac{12 \pm 16}{2} \]
\[ h = 14 \quad \text{or} \quad h = -2 \]
If \(h = 14\):
\[ k = \frac{1}{2}(14) – 4 = 7 – 4 = 3 \]
Centre: (14, 3). Equation:
\[ (x – 14)^2 + (y – 3)^2 = 100 \]
If \(h = -2\):
\[ k = \frac{1}{2}(-2) – 4 = -1 – 4 = -5 \]
Centre: (-2, -5). Equation:
\[ (x + 2)^2 + (y + 5)^2 = 100 \]
Check:
Centre (14, 3) to A(4, 3): \(\sqrt{(14-4)^2 + (3-3)^2} = 10\)
Centre (14, 3) to B(8, -5): \(\sqrt{(14-8)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10\)
Centre (-2, -5) to A(4, 3): \(\sqrt{(-2-4)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10\)
Centre (-2, -5) to B(8, -5): \(\sqrt{(8+2)^2 + (-5+5)^2} = 10\)
Both work.
Answer for (b):
\[ (x – 14)^2 + (y – 3)^2 = 100 \]
\[ (x + 2)^2 + (y + 5)^2 = 100 \]
Final Answers:
(a) Shown that centre lies on \(y = \frac{1}{2}x – 4\)
(b) \((x – 14)^2 + (y – 3)^2 = 100\) and \((x + 2)^2 + (y + 5)^2 = 100\)
Question 11
(a) Topic 1.7 – Differentiation
(b) Topic 1.7 – Differentiation
(c) Topic 1.3 – Coordinate Geometry
The equation of a curve is $y = kx^{\frac{1}{2}} – 4x^2 + 2$, where $k$ is a constant.
(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $k$.
(b) It is given that $k = 2$.
Find the coordinates of the stationary point and determine its nature.
(c) Points A and B on the curve have x-coordinates 0.25 and 1 respectively. For a different value of k, the tangents to the curve at the points A and B meet at a point with x-coordinate 0.6.
Find this value of k.
▶️Answer/Explanation
(a) Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) in terms of \(k\)
The curve is \(y = kx^{\frac{1}{2}} – 4x^2 + 2\).
First derivative:
\[ \frac{dy}{dx} = k \cdot \frac{1}{2}x^{-\frac{1}{2}} – 4 \cdot 2x = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]
Second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{k}{2}x^{-\frac{1}{2}} – 8x \right) \]
\[ = \frac{k}{2} \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} – 8 = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]
Answer for (a):
\[ \frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]
\[ \frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]
(b) \(k = 2\), find coordinates of stationary point and its nature
With \(k = 2\), the curve is \(y = 2x^{\frac{1}{2}} – 4x^2 + 2\).
Stationary point: \(\frac{dy}{dx} = 0\)
\[ \frac{dy}{dx} = \frac{2}{2}x^{-\frac{1}{2}} – 8x = x^{-\frac{1}{2}} – 8x \]
\[ x^{-\frac{1}{2}} – 8x = 0 \]
\[ x^{-\frac{1}{2}} = 8x \]
\[ \frac{1}{x^{\frac{1}{2}}} = 8x \]
\[ 1 = 8x \cdot x^{\frac{1}{2}} = 8x^{\frac{3}{2}} \]
\[ x^{\frac{3}{2}} = \frac{1}{8} \]
\[ x = \left(\frac{1}{8}\right)^{\frac{2}{3}} = (8^{-1})^{\frac{2}{3}} = 8^{-\frac{2}{3}} = (2^3)^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4} \]
Find \(y\):
\[ y = 2\left(\frac{1}{4}\right)^{\frac{1}{2}} – 4\left(\frac{1}{4}\right)^2 + 2 = 2 \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = 1 – \frac{1}{4} + 2 = 1 + 1.75 = 2.75 \]
Coordinates: \(\left(\frac{1}{4}, 2.75\right)\).
Nature: Use \(\frac{d^2y}{dx^2}\)
\[ \frac{d^2y}{dx^2} = -\frac{2}{4}x^{-\frac{3}{2}} – 8 = -\frac{1}{2}x^{-\frac{3}{2}} – 8 \]
At \(x = \frac{1}{4}\):
\[ x^{-\frac{3}{2}} = \left(\frac{1}{4}\right)^{-\frac{3}{2}} = (4^{-1})^{-\frac{3}{2}} = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \]
\[ \frac{d^2y}{dx^2} = -\frac{1}{2} \cdot 8 – 8 = -4 – 8 = -12 \]
Since \(\frac{d^2y}{dx^2} < 0\), it’s a maximum.
Answer for (b):\(\left(\frac{1}{4}, 2.75\right)\), maximum
(c) Tangents at \(x = 0.25\) and \(x = 1\) meet at \(x = 0.6\), find \(k\)
Point A (\(x = 0.25\)):
\[ y = k(0.25)^{\frac{1}{2}} – 4(0.25)^2 + 2 = k \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = \frac{k}{2} – \frac{1}{4} + 2 = \frac{k}{2} + 1.75 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(0.25)^{-\frac{1}{2}} – 8 \cdot 0.25 = \frac{k}{2} \cdot 2 – 2 = k – 2 \]
Tangent: \(y – \left(\frac{k}{2} + 1.75\right) = (k – 2)(x – 0.25)\)
Point B (\(x = 1\)):
\[ y = k(1)^{\frac{1}{2}} – 4(1)^2 + 2 = k – 4 + 2 = k – 2 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(1)^{-\frac{1}{2}} – 8 \cdot 1 = \frac{k}{2} – 8 \]
Tangent: \(y – (k – 2) = \left(\frac{k}{2} – 8\right)(x – 1)\)
Intersection at \(x = 0.6\):
Tangent at A:
\[ y = (k – 2)(0.6 – 0.25) + \frac{k}{2} + 1.75 = (k – 2) \cdot 0.35 + \frac{k}{2} + 1.75 \]
\[ = 0.35k – 0.7 + \frac{k}{2} + 1.75 = 0.85k + 1.05 \]
Tangent at B:
\[ y = \left(\frac{k}{2} – 8\right)(0.6 – 1) + (k – 2) = \left(\frac{k}{2} – 8\right)(-0.4) + k – 2 \]
\[ = -0.2k + 3.2 + k – 2 = 0.8k + 1.2 \]
Set equal:
\[ 0.85k + 1.05 = 0.8k + 1.2 \]
\[ 0.05k = 0.15 \]
\[ k = 3 \]
Answer for (c):\(k = 3\)
Final Answers:
(a) \(\frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x\), \(\frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8\)
(b) \(\left(\frac{1}{4}, 2.75\right)\), maximum
(c) 3