CIE AS/A level Pure Mathematics Paper 1 Prediction - 2025
CIE AS/A level Pure Mathematics Paper 1 Prediction- 2025
To excel in A level Math Exam, consistent practice with CIE AS & A Level Revision resources is key. CIE AS/A level Pure Mathematics Paper 1 Prediction will guide you for exam pattern.
IITian Academy offers a vast collection of questions that can aid your understanding of specific topics and solidify your concepts. By practicing regularly and focusing on these key areas, you’ll be well-prepared for the A level Math exam
Question 1: Series
The first term of an arithmetic progression is 1.5 and the sum of the first ten terms is 127.5.
Working space:
▶️Answer/Explanation
Answer: \( 2.5 \)
Working:
First term \( a = 1.5 \), sum of first 10 terms \( S_{10} = 127.5 \), number of terms \( n = 10 \).
Sum formula: \( S_n = \frac{n}{2} (2a + (n-1)d) \)
\[ 127.5 = \frac{10}{2} (2 \cdot 1.5 + 9d) \]
\[ 127.5 = 5 (3 + 9d) \]
\[ 127.5 = 15 + 45d \]
\[ 45d = 112.5 \]
\[ d = 2.5 \]
Key Concept:
The sum of an arithmetic progression is calculated using the formula \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference.
Working space:
▶️Answer/Explanation
Answer: \( 1882.5 \)
Working:
Terms: \( a_n = 1.5 + (n-1) \cdot 2.5 = 2.5n – 1 \).
Smallest term \( \geq 25 \):
\[ 2.5n – 1 \geq 25 \]
\[ 2.5n \geq 26 \]
\[ n \geq 10.4 \quad \text{so} \quad n = 11 \]
\[ a_{11} = 2.5 \cdot 11 – 1 = 26.5 \]
Largest term \( \leq 100 \):
\[ 2.5n – 1 \leq 100 \]
\[ 2.5n \leq 101 \]
\[ n \leq 40.4 \quad \text{so} \quad n = 40 \]
\[ a_{40} = 2.5 \cdot 40 – 1 = 99 \]
Terms from \( n = 11 \) to \( n = 40 \), number of terms = \( 40 – 11 + 1 = 30 \).
Sum: \( S = \frac{\text{number of terms}}{2} (\text{first} + \text{last}) \)
\[ S = \frac{30}{2} (26.5 + 99) = 15 \cdot 125.5 = 1882.5 \]
Key Concept:
The sum of a subset of terms in an arithmetic progression is found by identifying the first and last terms within the given range and using the sum formula \( S = \frac{n}{2} (a_{\text{first}} + a_{\text{last}}) \).
Syllabus Reference
Sequences and Series
- (a) SL 1.6 – Arithmetic sequences and series: sum formula
- (b) SL 1.6 – Arithmetic sequences and series: sum of terms in a range
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 2: Differentiation and Integration
The diagram shows the curve with equation \( y = \sqrt{2x^3 + 10} \).
Working space:
▶️Answer/Explanation
Answer: \( 27x – 8y – 17 = 0 \)
Working:
The curve is given by \( y = \sqrt{2x^3 + 10} = (2x^3 + 10)^{\frac{1}{2}} \).
Differentiate to find the gradient:
\[ \frac{dy}{dx} = \frac{1}{2} (2x^3 + 10)^{-\frac{1}{2}} \cdot \frac{d}{dx}(2x^3 + 10) \]
\[ \frac{d}{dx}(2x^3 + 10) = 6x^2 \]
\[ \frac{dy}{dx} = \frac{1}{2} (2x^3 + 10)^{-\frac{1}{2}} \cdot 6x^2 = \frac{6x^2}{2 \sqrt{2x^3 + 10}} = \frac{3x^2}{\sqrt{2x^3 + 10}} \]
At \( x = 3 \):
\[ 2x^3 + 10 = 2 \cdot 3^3 + 10 = 2 \cdot 27 + 10 = 64 \]
\[ \sqrt{2x^3 + 10} = \sqrt{64} = 8 \]
So the point is \( (3, 8) \).
Gradient at \( x = 3 \):
\[ \frac{dy}{dx} = \frac{3 \cdot 3^2}{\sqrt{2 \cdot 3^3 + 10}} = \frac{3 \cdot 9}{\sqrt{64}} = \frac{27}{8} \]
Equation of the tangent at \( (3, 8) \) with slope \( \frac{27}{8} \):
\[ y – 8 = \frac{27}{8} (x – 3) \]
Multiply through by 8 to eliminate the fraction:
\[ 8(y – 8) = 27(x – 3) \]
\[ 8y – 64 = 27x – 81 \]
\[ 27x – 8y – 81 + 64 = 0 \]
\[ 27x – 8y – 17 = 0 \]
Key Concept:
The equation of a tangent line at a point on a curve is found using the point-slope form \( y – y_1 = m(x – x_1) \), where \( m \) is the derivative at \( x = x_1 \).
Working space:
▶️Answer/Explanation
Answer: \( 60\pi \)
Working:
The volume of a solid formed by rotating the region under \( y = \sqrt{2x^3 + 10} \) from \( x = 1 \) to \( x = 3 \) about the x-axis is given by:
\[ V = \pi \int_{1}^{3} y^2 \, dx \]
Since \( y = \sqrt{2x^3 + 10} \), we have:
\[ y^2 = (\sqrt{2x^3 + 10})^2 = 2x^3 + 10 \]
Thus:
\[ V = \pi \int_{1}^{3} (2x^3 + 10) \, dx \]
Integrate:
\[ \int (2x^3 + 10) \, dx = \int 2x^3 \, dx + \int 10 \, dx = \frac{2x^4}{4} + 10x = \frac{x^4}{2} + 10x \]
Evaluate from \( x = 1 \) to \( x = 3 \):
\[ \left[ \frac{x^4}{2} + 10x \right]_{1}^{3} = \left( \frac{3^4}{2} + 10 \cdot 3 \right) – \left( \frac{1^4}{2} + 10 \cdot 1 \right) \]
\[ = \left( \frac{81}{2} + 30 \right) – \left( \frac{1}{2} + 10 \right) \]
\[ = \left( 40.5 + 30 \right) – \left( 0.5 + 10 \right) = 70.5 – 10.5 = 60 \]
\[ V = \pi \cdot 60 = 60\pi \]
Key Concept:
The volume of a solid of revolution about the x-axis is calculated using \( V = \pi \int_{a}^{b} y^2 \, dx \), where \( y \) is the function defining the curve.
Syllabus Reference
Calculus
- (a) SL 1.7 – Differentiation: Tangent lines and derivatives
- (b) SL 1.8 – Integration: Volumes of solids of revolution
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 3
The diagram shows a sector OAB of a circle with centre O and radius r cm. Angle AOB = \theta radians.
It is given that the length of the arc AB is 9.6 cm and that the area of the sector OAB is 76.8 cm².
(a) Find the area of the shaded region.
(b) Find the perimeter of the shaded region.
▶️Answer/Explanation
Solution :-
(a) $\frac{\frac{1}{2}r^2 \theta}{r\theta} = \frac{76.8}{9.6}$
$OR$
$\frac{1}{2} \left( \frac{9.6}{\theta} \right)^2 \theta = 76.8$
$r = 16$
$\theta = 0.6$
$\Delta OAB = \frac{1}{2} \times \text{their } 16^2 \times \sin \text{ their } 0.6$
$\text{Area} = 76.8 – 72.27= 4.53$
(b) $AB = 2 \times 16 \times \sin 0.3$ \quad OR \quad $AB^{2} = 16^{2} + 16^{2} – 2 \times 16^{2} \cos 0.6$
$\text{Perimeter} = 9.6 + 9.46 = 19.1$
Question 4: Coordinate Geometry
The equation of a circle is \((x-3)^2 + y^2 = 18\). The line with equation \(y = mx + c\) passes through the point \((0, -9)\) and is a tangent to the circle.
Find the two possible values of \(m\) and, for each value of \(m\), find the coordinates of the point at which the tangent touches the circle.
Working space:
▶️Answer/Explanation
Circle: \((x – 3)^2 + y^2 = 18\), center \((3, 0)\), radius \(3\sqrt{2}\). Line: \(y = mx + c\), passes through \((0, -9)\), so \(c = -9\). Line is \(y = mx – 9\).
Tangent condition: Distance from \((3, 0)\) to \(y = mx – 9\) equals radius:
\[ \frac{|3m – 9|}{\sqrt{m^2 + 1}} = 3\sqrt{2} \]
\[ |m – 3| = \sqrt{2} \sqrt{m^2 + 1} \]
\[ m^2 – 6m + 9 = 2m^2 + 2 \]
\[ m^2 + 6m – 7 = 0 \]
\[ m = 1 \quad \text{or} \quad m = -7 \]
\(m = 1\):
Line: \(y = x – 9\).
\[(x – 3)^2 + (x – 9)^2 = 18\]
\[ 2x^2 – 24x + 72 = 0 \]
\[ x^2 – 12x + 36 = 0 \]
\[ x = 6, \quad y = -3\]
Point: \((6, -3)\).
\(m = -7\):
Line: \(y = -7x – 9\).
\[(x – 3)^2 + (-7x – 9)^2 = 18\]
\[ 50x^2 + 120x + 54 = 0 \]
\[ x = -\frac{3}{5}, \quad y = -7\left(-\frac{3}{5}\right) – 9 = -\frac{24}{5}\]
Point: \(\left(-\frac{3}{5}, -\frac{24}{5}\right)\).
Final Answer:
\(m = 1\), \((6, -3)\)
\(m = -7\), \(\left(-\frac{3}{5}, -\frac{24}{5}\right)\)
Key Concept:
A line is tangent to a circle if the distance from the circle’s center to the line equals the radius. Solving involves substituting the line equation into the circle equation and using the distance formula.
Syllabus Reference
Coordinate Geometry
- SL 1.3 – Coordinate geometry: equations of circles, tangents to circles
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 5: Differentiation
The equation of a curve is \( y = 2x^{2} – \frac{1}{2x} + 3 \).
Working space:
▶️Answer/Explanation
Answer: \( \left(-\frac{1}{2}, 4.5\right) \)
Working:
Curve: \( y = 2x^2 – \frac{1}{2x} + 3 \).
Derivative:
\[ \frac{dy}{dx} = 4x – \frac{d}{dx}\left(\frac{1}{2x}\right) = 4x + \frac{1}{2x^2} \]
Set \( \frac{dy}{dx} = 0 \):
\[ 4x + \frac{1}{2x^2} = 0 \]
\[ 4x = -\frac{1}{2x^2} \]
\[ 8x^3 = -1 \]
\[ x^3 = -\frac{1}{8} \]
\[ x = -\frac{1}{2} \]
Find \( y \):
\[ y = 2\left(-\frac{1}{2}\right)^2 – \frac{1}{2\left(-\frac{1}{2}\right)} + 3 = 2 \cdot \frac{1}{4} + 1 + 3 = 0.5 + 1 + 3 = 4.5 \]
Working space:
▶️Answer/Explanation
Answer: Minimum
Working:
Second derivative:
\[ \frac{d^2y}{dx^2} = 4 + \frac{d}{dx}\left(\frac{1}{2x^2}\right) = 4 – \frac{1}{x^3} \]
At \( x = -\frac{1}{2} \):
\[ \frac{d^2y}{dx^2} = 4 – \frac{1}{\left(-\frac{1}{2}\right)^3} = 4 – \frac{1}{-\frac{1}{8}} = 4 + 8 = 12 \]
Since \( 12 > 0 \), it’s a minimum.
Working space:
▶️Answer/Explanation
Answer: Increasing, because \( \frac{dy}{dx} > 0 \) for \( x > 0 \).
Working:
For \( x > 0 \), check \( \frac{dy}{dx} = 4x + \frac{1}{2x^2} \):
\( 4x > 0 \)
\( \frac{1}{2x^2} > 0 \)
So, \( \frac{dy}{dx} > 0 \) always.
Since the derivative is positive, the function is increasing.
Syllabus Reference
Differentiation
- SL 1.7 – Differentiation: stationary points, nature of stationary points, increasing and decreasing functions
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 6
The equation of a circle is $(x-3)^2+y^2=18$. The line with equation $y=mx+c$ passes through the point $(0,-9)$ and is a tangent to the circle.
Find the two possible values of $m$ and, for each value of $m$, find the coordinates of the point at which the tangent touches the circle.
▶️Answer/Explanation
Circle: \((x – 3)^2 + y^2 = 18\), center \((3, 0)\), radius \(3\sqrt{2}\). Line: \(y = mx + c\), passes through \((0, -9)\), so \(c = -9\). Line is \(y = mx – 9\).
Tangent condition: Distance from \((3, 0)\) to \(y = mx – 9\) equals radius:
\[ \frac{|3m – 9|}{\sqrt{m^2 + 1}} = 3\sqrt{2} \]
\[ |m – 3| = \sqrt{2} \sqrt{m^2 + 1} \]
\[ m^2 – 6m + 9 = 2m^2 + 2 \]
\[ m^2 + 6m – 7 = 0 \]
\[ m = 1 \quad \text{or} \quad m = -7 \]
\(m = 1\):
Line: \(y = x – 9\).
\[(x – 3)^2 + (x – 9)^2 = 18\]
\[ 2x^2 – 24x + 72 = 0 \]
\[ x^2 – 12x + 36 = 0 \]
\[ x = 6, \quad y = -3\]
Point: \((6, -3)\).
\(m = -7\):
Line: \(y = -7x – 9\).
\[(x – 3)^2 + (-7x – 9)^2 = 18\]
\[ 50x^2 + 120x + 54 = 0 \]
\[ x = -\frac{3}{5}, \quad y = -7\left(-\frac{3}{5}\right) – 9 = -\frac{24}{5}\]
Point: \(\left(-\frac{3}{5}, -\frac{24}{5}\right)\).
Final Answer:
\(m = 1\), \((6, -3)\)
\(m = -7\), \(\left(-\frac{3}{5}, -\frac{24}{5}\right)\)
Question 7: Calculus – Derivative Applications
A function \( f \) is such that \( f'(x) = 6(2x – 3)^2 – 6x \) for \( x \in \mathbb{R} \).
Working space:
▶️Answer/Explanation
Answer: \( 1 < x < \frac{9}{4} \)
Working:
\( f(x) \) is decreasing where \( f'(x) < 0 \):
\[ f'(x) = 6(2x – 3)^2 – 6x \]
Factor out 6:
\[ 6[(2x – 3)^2 – x] \]
\[ (2x – 3)^2 – x = 4x^2 – 12x + 9 – x = 4x^2 – 13x + 9 \]
Solve \( 4x^2 – 13x + 9 < 0 \):
\[ x = \frac{13 \pm \sqrt{(-13)^2 – 4 \cdot 4 \cdot 9}}{8} = \frac{13 \pm \sqrt{169 – 144}}{8} = \frac{13 \pm \sqrt{25}}{8} = \frac{13 \pm 5}{8} \]
\[ x = \frac{18}{8} = 2.25 \quad \text{or} \quad x = \frac{8}{8} = 1 \]
The quadratic \( 4x^2 – 13x + 9 \) opens upwards (coefficient of \( x^2 \) is positive), so it is negative between the roots:
\[ 1 < x < 2.25 \]
\( f(x) \) is decreasing for \( 1 < x < \frac{9}{4} \).
Key Concept:
A function is decreasing where its derivative is negative. Solve the inequality \( f'(x) < 0 \) using quadratic formula and analyze the sign of the quadratic.
Working space:
▶️Answer/Explanation
Answer: \( f(x) = 8x^3 – 39x^2 + 54x – 24 \)
Working:
Integrate \( f'(x) \):
\[ f'(x) = 6(2x – 3)^2 – 6x \]
\[ (2x – 3)^2 = 4x^2 – 12x + 9 \]
\[ 6(4x^2 – 12x + 9) – 6x = 24x^2 – 72x + 54 – 6x = 24x^2 – 78x + 54 \]
\[ f(x) = \int (24x^2 – 78x + 54) \, dx = 24 \cdot \frac{x^3}{3} – 78 \cdot \frac{x^2}{2} + 54x + C \]
\[ = 8x^3 – 39x^2 + 54x + C \]
Use \( f(1) = -1 \):
\[ 8(1)^3 – 39(1)^2 + 54(1) + C = -1 \]
\[ 8 – 39 + 54 + C = -1 \]
\[ 23 + C = -1 \]
\[ C = -24 \]
\[ f(x) = 8x^3 – 39x^2 + 54x – 24 \]
Key Concept:
To find the function, integrate the derivative and use the given point to solve for the constant of integration.
Syllabus Reference
Calculus
- ALV 1.7 – Applications of differentiation: determining intervals where a function is increasing or decreasing
- ALV 1.7 – Integration: finding the function from its derivative
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 8
(a) (i) By first expanding (cos θ + sin θ)², find the three solutions of the equation \[(cos~\theta+sin~\theta)^{2}=1\]
for 0 ≤ θ ≤ π.
(a) (ii) Hence verify that the only solutions of the equation cos θ + sin θ = 1 for 0 ≤ θ ≤ π are 0 and \(\frac{1}{2}\pi\).
(b) Prove the identity
\[\frac{sin~\theta}{cos~\theta+sin~\theta}+\frac{1-cos~\theta}{cos~\theta-sin~\theta}\equiv\frac{cos~\theta+sin~\theta-1}{1-2~sin^{2}\theta}.\]
(c) Using the results of (a) (ii) and (b), solve the equation
\[\frac{sin~\theta}{cos~\theta+sin~\theta}+\frac{1-cos~\theta}{cos~\theta-sin~\theta}=2(cos~\theta+sin~\theta-1)\]
for 0 ≤ θ ≤ π.
▶️Answer/Explanation
(a) (i) Solve \((\cos \theta + \sin \theta)^2 = 1\) for \(0 \leq \theta \leq \pi\)
Expand:
\[ (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta \]
Use \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ 1 + 2 \cos \theta \sin \theta = 1 \]
\[ 2 \cos \theta \sin \theta = 0 \]
\[ \cos \theta \sin \theta = 0 \]
So, either \(\cos \theta = 0\) or \(\sin \theta = 0\). In \(0 \leq \theta \leq \pi\):
\(\sin \theta = 0\): \(\theta = 0, \pi\)
\(\cos \theta = 0\): \(\theta = \frac{\pi}{2}\)
Solutions: \(\theta = 0, \frac{\pi}{2}, \pi\).
(a) (ii) Verify solutions of \(\cos \theta + \sin \theta = 1\)
From (a)(i), \((\cos \theta + \sin \theta)^2 = 1\) means \(\cos \theta + \sin \theta = \pm 1\). Check \(\cos \theta + \sin \theta = 1\):
\(\theta = 0\): \(\cos 0 + \sin 0 = 1 + 0 = 1\) (yes)
\(\theta = \frac{\pi}{2}\): \(\cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1\) (yes)
\(\theta = \pi\): \(\cos \pi + \sin \pi = -1 + 0 = -1\) (no)
Only \(\theta = 0, \frac{\pi}{2}\) work.
(b) Prove the identity
Left side: \(\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 – \cos \theta}{\cos \theta – \sin \theta}\).
Common denominator: \((\cos \theta + \sin \theta)(\cos \theta – \sin \theta) = \cos^2 \theta – \sin^2 \theta = \cos 2\theta\).
Numerator: \(\sin \theta (\cos \theta – \sin \theta) + (1 – \cos \theta)(\cos \theta + \sin \theta)\)
\(= \sin \theta \cos \theta – \sin^2 \theta + \cos \theta + \sin \theta – \cos^2 \theta – \cos \theta \sin \theta\)
\(= -\sin^2 \theta – \cos^2 \theta + \cos \theta + \sin \theta = -1 + \cos \theta + \sin \theta\)
So:
\[ \frac{\cos \theta + \sin \theta – 1}{\cos 2\theta} \]
Right side: \(\frac{\cos \theta + \sin \theta – 1}{1 – 2 \sin^2 \theta}\). Since \(1 – 2 \sin^2 \theta = \cos 2\theta\), they are equal.
(c) Solve the equation
Given: \(\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 – \cos \theta}{\cos \theta – \sin \theta} = 2 (\cos \theta + \sin \theta – 1)\).
From (b), left side = \(\frac{\cos \theta + \sin \theta – 1}{1 – 2 \sin^2 \theta}\), so:
\[ \frac{\cos \theta + \sin \theta – 1}{1 – 2 \sin^2 \theta} = 2 (\cos \theta + \sin \theta – 1) \]
If \(\cos \theta + \sin \theta – 1 = 0\):
\[ \cos \theta + \sin \theta = 1 \]
From (a)(ii), \(\theta = 0, \frac{\pi}{2}\). Check: both sides become \(0 = 0\) (valid).
If \(\cos \theta + \sin \theta – 1 \neq 0\), divide:
\[ \frac{1}{1 – 2 \sin^2 \theta} = 2 \]
\[ 1 = 2 – 4 \sin^2 \theta \]
\[ 4 \sin^2 \theta = 1 \]
\[ \sin \theta = \pm \frac{1}{2} \]
In \(0 \leq \theta \leq \pi\): \(\sin \theta = \frac{1}{2}\) at \(\theta = \frac{\pi}{6}\). Check:
Left: \(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2}} + \frac{1 – \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} – \frac{1}{2}} \approx 0.732\)
Right: \(2 \left( \frac{\sqrt{3}}{2} + \frac{1}{2} – 1 \right) \approx 0.732\) (matches)
Solutions: \(\theta = 0, \frac{\pi}{6}, \frac{\pi}{2}\).
Final Answer:
(a)(i) \(\theta = 0, \frac{\pi}{2}, \pi\)
(a)(ii) Verified: \(\theta = 0, \frac{\pi}{2}\)
(b) Identity proven
(c) \(\theta = 0, \frac{\pi}{6}, \frac{\pi}{2}\)
Question 9: Differentiation and Coordinate Geometry
A function is defined by \( f(x) = \frac{4}{x^3} – \frac{3}{x} + 2 \) for \( x \neq 0 \). The graph of \( y = f(x) \) is shown in the diagram.
Working space:
▶️Answer/Explanation
Solution:
\[ \frac{dy}{dx} = -\frac{12}{x^4} + \frac{3}{x^2} \]
Set the derivative equal to zero to find critical points:
\[ -\frac{12}{x^4} + \frac{3}{x^2} = 0 \]
Simplify:
\[ 3x^4 – 12x^2 = 0 \text{ or } -12 + 3x^2 = 0 \]
\[ 3x^2 (x^2 – 4) = 0 \]
Solving gives:
\[ x = \pm 2 \]
The function is decreasing where \( \frac{dy}{dx} < 0 \). Testing intervals around critical points and considering \( x \neq 0 \):
\[ -2 < x < 0 \text{ and } 0 < x < 2 \]
Thus, the set of values is:
\[ (-2, 0) \cup (0, 2) \text{ or } -2 < x < 2, x \neq 0 \]
Key Concept:
A function is decreasing where its derivative is negative, determined by solving \( \frac{dy}{dx} < 0 \).
Working space:
▶️Answer/Explanation
Solution:
At \( x = 1 \):
\[ f(1) = \frac{4}{1^3} – \frac{3}{1} + 2 = 4 – 3 + 2 = 3 \]
So, the point is \( (1, 3) \).
The derivative at \( x = 1 \):
\[ \frac{dy}{dx} = -\frac{12}{x^4} + \frac{3}{x^2} \]
\[ \frac{dy}{dx} \bigg|_{x=1} = -\frac{12}{1^4} + \frac{3}{1^2} = -12 + 3 = -9 \]
Slope of the tangent at \( x = 1 \), \( m_{\text{tan}} = -9 \).
Slope of the normal:
\[ m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = \frac{1}{9} \]
Equation of the normal at \( (1, 3) \):
\[ y – 3 = \frac{1}{9} (x – 1) \]
\[ y = \frac{1}{9} x + \frac{26}{9} \]
At \( x = -1 \):
\[ f(-1) = \frac{4}{(-1)^3} – \frac{3}{-1} + 2 = -\frac{4}{1} + 3 + 2 = -4 + 3 + 2 = 1 \]
So, the point is \( (-1, 1) \).
The derivative at \( x = -1 \):
\[ \frac{dy}{dx} \bigg|_{x=-1} = -\frac{12}{(-1)^4} + \frac{3}{(-1)^2} = -\frac{12}{1} + \frac{3}{1} = -12 + 3 = -9 \]
Slope of the tangent at \( x = -1 \), \( m = -9 \).
Equation of the tangent at \( (-1, 1) \):
\[ y – 1 = -9 (x + 1) \]
\[ y = -9x – 8 \]
Find the intersection of the normal and tangent:
\[ \frac{1}{9} x + \frac{26}{9} = -9x – 8 \]
Multiply through by 9 to clear denominators:
\[ x + 26 = -81x – 72 \]
\[ 82x = -98 \]
\[ x = -\frac{98}{82} = -\frac{49}{41} \approx -1.19512 \]
Substitute \( x = -\frac{49}{41} \) into the tangent equation:
\[ y = -9 \left(-\frac{49}{41}\right) – 8 = \frac{441}{41} – \frac{328}{41} = \frac{113}{41} \]
Intersection point: \( \left(-\frac{49}{41}, \frac{113}{41}\right) \approx (-1.19512, 2.7561) \).
The triangle is bounded by:
- The y-axis (\( x = 0 \)).
- The normal: \( y = \frac{1}{9} x + \frac{26}{9} \).
- The tangent: \( y = -9x – 8 \).
Vertices of the triangle:
- Y-axis and normal intersection: Set \( x = 0 \):
\[ y = \frac{1}{9} (0) + \frac{26}{9} = \frac{26}{9} \approx 2.8889 \]
Point: \( \left(0, \frac{26}{9}\right) \).
- Y-axis and tangent intersection: Set \( x = 0 \):
\[ y = -9 (0) – 8 = -8 \]
Point: \( (0, -8) \).
- Normal and tangent intersection: \( \left(-\frac{49}{41}, \frac{113}{41}\right) \).
Area of the triangle with vertices \( (0, \frac{26}{9}), (0, -8), \left(-\frac{49}{41}, \frac{113}{41}\right) \):
Using the shoelace formula for vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) \right| \]
Substitute:
\[ \left(0, \frac{26}{9}\right), \left(0, -8\right), \left(-\frac{49}{41}, \frac{113}{41}\right) \]
\[ \text{Area} = \frac{1}{2} \left| 0 \left(-8 – \frac{113}{41}\right) + 0 \left(\frac{113}{41} – \frac{26}{9}\right) + \left(-\frac{49}{41}\right) \left(\frac{26}{9} – (-8)\right) \right| \]
\[ = \frac{1}{2} \left| \left(-\frac{49}{41}\right) \left(\frac{26}{9} + \frac{72}{9}\right) \right| = \frac{1}{2} \left| \left(-\frac{49}{41}\right) \cdot \frac{98}{9} \right| \]
\[ = \frac{1}{2} \cdot \frac{49 \cdot 98}{41 \cdot 9} = \frac{49 \cdot 98}{2 \cdot 41 \cdot 9} = \frac{4802}{738} = \frac{2401}{369} \approx 6.50678 \]
To 3 significant figures:
\[ \text{Area} \approx 6.51 \]
Key Concept:
The area of a triangle formed by lines (tangent, normal, and axis) is calculated by finding the vertices via line intersections and applying the shoelace formula.
Syllabus Reference
Differentiation and Coordinate Geometry
- (a) SL 1.7 – Differentiation
- (b) SL 1.3 – Coordinate geometry
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 10: Series
(a) The first three terms of an arithmetic progression are \(25\), \(4p – 1\), and \(13 – p\), where \(p\) is a constant.
Find the value of the tenth term of the progression.
(b) The first three terms of a geometric progression are \(25\), \(4q – 1\), and \(13 – q\), where \(q\) is a positive constant.
Find the sum to infinity of the progression.
Working space:
▶️Answer/Explanation
Arithmetic progression: Find the tenth term
Given terms: \(25\), \(4p – 1\), \(13 – p\).
In an arithmetic progression, the difference between consecutive terms is constant:
\(d = (4p – 1) – 25 = 4p – 26\)
\(d = (13 – p) – (4p – 1) = 13 – p – 4p + 1 = 14 – 5p\)
Set equal:
\[ 4p – 26 = 14 – 5p \]
\[ 9p = 40 \]
\[ p = \frac{40}{9} \]
First term \(a = 25\), common difference:
\[ d = 4 \cdot \frac{40}{9} – 26 = \frac{160}{9} – \frac{234}{9} = -\frac{74}{9} \]
Tenth term (\(n = 10\)):
\[ a_n = a + (n-1)d = 25 + 9 \left(-\frac{74}{9}\right) = 25 – 74 = -49 \]
Answer: \(-49\)
Key Concept:
The \(n\)-th term of an arithmetic progression is calculated using \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Working space:
▶️Answer/Explanation
Geometric progression: Find the sum to infinity
Given terms: \(25\), \(4q – 1\), \(13 – q\), with \(q > 0\).
In a geometric progression, the ratio is constant:
\[ \frac{4q – 1}{25} = \frac{13 – q}{4q – 1} \]
Cross-multiply:
\[ (4q – 1)^2 = 25 (13 – q) \]
\[ 16q^2 – 8q + 1 = 325 – 25q \]
\[ 16q^2 – 8q + 1 – 325 + 25q = 0 \]
\[ 16q^2 + 17q – 324 = 0 \]
Solve:
\[ q = \frac{-17 \pm \sqrt{289 + 20736}}{32} = \frac{-17 \pm \sqrt{21025}}{32} = \frac{-17 \pm 145}{32} \]
\[ q = \frac{128}{32} = 4 \quad \text{or} \quad q = \frac{-162}{32} = -\frac{81}{16} \]
Since \(q > 0\), \(q = 4\).
First term \(a = 25\), ratio:
\[ r = \frac{4 \cdot 4 – 1}{25} = \frac{15}{25} = \frac{3}{5} \]
Sum to infinity (\(|r| < 1\)):
\[ S = \frac{a}{1 – r} = \frac{25}{1 – \frac{3}{5}} = \frac{25}{\frac{2}{5}} = 25 \cdot \frac{5}{2} = \frac{125}{2} \]
Answer: \(\frac{125}{2}\)
Key Concept:
The sum to infinity of a geometric progression with \(|r| < 1\) is calculated using \(S = \frac{a}{1 – r}\), where \(a\) is the first term and \(r\) is the common ratio.
Syllabus Reference
Series
- (a) SL 1.6 – Arithmetic sequences and series: finding the \(n\)-th term
- (b) SL 1.6 – Geometric sequences and series: sum to infinity
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question 11: Functions and Inverses
The function \( f \) is defined as follows:
\( f(x) = \sqrt{x} – 1 \text{ for } x > 1 \)
The diagram shows the graph of \( y = g(x) \) where \( g(x) = \frac{1}{x^2 + 2} \) for \( x \in \mathbb{R} \).
The function \( h \) is defined by \( h(x) = \frac{1}{x^2 + 2} \) for \( x \geq 0 \).
Working space:
▶️Answer/Explanation
Answer: \( f^{-1}(x) = (x + 1)^2 \)
Working:
To find the inverse, set \( y = f(x) \):
\( y = \sqrt{x} – 1 \)
Solve for \( x \):
\( y + 1 = \sqrt{x} \)
\( x = (y + 1)^2 \)
Thus, the inverse function is:
\( f^{-1}(x) = (x + 1)^2 \)
Key Concept:
The inverse function \( f^{-1}(x) \) is found by swapping \( x \) and \( y \) in the original function and solving for \( y \).
Working space:
▶️Answer/Explanation
Answer:
Range of \( g \): \( 0 < g(x) \leq \frac{1}{2} \) or \( \left(0, \frac{1}{2}\right] \)
\( g^{-1} \) does not exist because \( g \) is not one-to-one.
Working:
For \( g(x) = \frac{1}{x^2 + 2} \):
Since \( x^2 \geq 0 \), \( x^2 + 2 \geq 2 \), so:
\( g(x) = \frac{1}{x^2 + 2} \leq \frac{1}{2} \)
The maximum value occurs when \( x = 0 \):
\( g(0) = \frac{1}{0^2 + 2} = \frac{1}{2} \)
As \( |x| \to \infty \), \( x^2 + 2 \to \infty \), so \( g(x) \to 0^+ \).
Thus, the range is \( \left(0, \frac{1}{2}\right] \).
For \( g^{-1} \) to exist, \( g \) must be one-to-one. However, \( g(x) \) is symmetric about the y-axis (since \( g(-x) = g(x) \)), so it is many-to-one (e.g., \( g(1) = g(-1) = \frac{1}{3} \)). Alternatively, the graph fails the horizontal line test, confirming \( g^{-1} \) does not exist.
Key Concept:
The range of a function is the set of all possible output values. An inverse function exists only if the original function is one-to-one.
Working space:
▶️Answer/Explanation
Answer: \( 3 + 2\sqrt{2} \)
Working:
First, compute \( f\left(\frac{25}{16}\right) \):
\( f(x) = \sqrt{x} – 1 \)
\( f\left(\frac{25}{16}\right) = \sqrt{\frac{25}{16}} – 1 = \frac{5}{4} – 1 = \frac{1}{4} \)
Now, compute \( h(f(x)) \):
\( f(x) = \sqrt{x} – 1 \)
\( h(x) = \frac{1}{x^2 + 2} \)
\( h(f(x)) = h(\sqrt{x} – 1) = \frac{1}{(\sqrt{x} – 1)^2 + 2} \)
Solve the equation \( h(f(x)) = f\left(\frac{25}{16}\right) \):
\( \frac{1}{(\sqrt{x} – 1)^2 + 2} = \frac{1}{4} \)
Take the reciprocal:
\( (\sqrt{x} – 1)^2 + 2 = 4 \)
\( (\sqrt{x} – 1)^2 = 2 \)
\( \sqrt{x} – 1 = \pm \sqrt{2} \)
Since \( x > 1 \), \( \sqrt{x} > 1 \), so consider:
\( \sqrt{x} – 1 = \sqrt{2} \)
\( \sqrt{x} = 1 + \sqrt{2} \)
\( x = (1 + \sqrt{2})^2 \)
Expand:
\( (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2} \)
Verify the domain: \( x = 3 + 2\sqrt{2} > 1 \), which is satisfied.
Alternative approach:
From \( (\sqrt{x} – 1)^2 = 2 \):
\( x – 2\sqrt{x} + 1 = 2 \)
\( x – 2\sqrt{x} – 1 = 0 \)
Let \( u = \sqrt{x} \), so \( x = u^2 \):
\( u^2 – 2u – 1 = 0 \)
\( u = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \)
Since \( u = \sqrt{x} > 1 \), take \( u = 1 + \sqrt{2} \):
\( x = (1 + \sqrt{2})^2 = 3 + 2\sqrt{2} \)
Another approach:
From \( x – 2\sqrt{x} – 1 = 0 \):
\( \sqrt{x} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2} \)
Take \( \sqrt{x} = 1 + \sqrt{2} \):
\( x = (1 + \sqrt{2})^2 = 3 + 2\sqrt{2} \)
Thus, the solution is:
\( x = 3 + 2\sqrt{2} \)
Key Concept:
Solving composite function equations involves substituting and simplifying, ensuring solutions satisfy domain constraints.
Syllabus Reference
Functions and Inverses
- (a) ALV 1.2 – Inverse functions
- (b) ALV 1.2 – Range of functions and conditions for inverse existence
- (c) ALV 1.2 – Composite functions and solving equations
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)