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GCSE Nov 2019 Edexcel Higher Paper 1 (Non-Calc)
How to use this paper
- Paper Type: Non-Calculator (Show all arithmetic methods)
- Structure: Try the question first, then check the solution.
- Solutions: Click “Show Solution” to see the “Why”, “How”, and “What” of each step.
- Visuals: All diagrams are mathematically accurate representations.
Table of Contents
- Question 1 (LCM)
- Question 2 (Ratio & Proportion)
- Question 3 (Fraction Arithmetic)
- Question 4 (Construction)
- Question 5 (Geometry & Angles)
- Question 6 (Mean Averages)
- Question 7 (Algebraic Indices)
- Question 8 (Bearings & Speed)
- Question 9 (Speed Conversion)
- Question 10 (Cumulative Frequency)
- Question 11 (Probability/Estimation)
- Question 12 (Algebraic Errors)
- Question 13 (Algebraic Fractions)
- Question 14 (Area & Quadratics)
- Question 15 (Recurring Decimals)
- Question 16 (Surds)
- Question 17 (Similarity – Area/Vol)
- Question 18 (Functions)
- Question 19 (Indices & Exact Values)
- Question 20 (Graph Transformations)
- Question 21 (Quadratic Graphs)
- Question 22 (Congruency Proof)
Question 1 (3 marks)
Find the Lowest Common Multiple (LCM) of 108 and 120.
Worked Solution
Step 1: Write numbers as products of prime factors
Why: The LCM is the smallest number that divides by both. Breaking them into prime building blocks helps us construct this number.
For 108:
\( 108 = 2 \times 54 = 2 \times 2 \times 27 = 2 \times 2 \times 3 \times 9 = 2 \times 2 \times 3 \times 3 \times 3 \)
\( 108 = 2^2 \times 3^3 \)
For 120:
\( 120 = 10 \times 12 = (2 \times 5) \times (3 \times 4) = 2 \times 5 \times 3 \times 2 \times 2 \)
\( 120 = 2^3 \times 3 \times 5 \)
Step 2: Construct the LCM
Method: We take the highest power of each prime factor present in either list.
- Primes involved: 2, 3, 5
- Highest power of 2: \( 2^3 \) (from 120)
- Highest power of 3: \( 3^3 \) (from 108)
- Highest power of 5: \( 5^1 \) (from 120)
Step 3: Calculate the final value
Tip: Multiply 8 and 5 first to make it easier.
\[ 8 \times 5 = 40 \] \[ 40 \times 27 = 4 \times 27 \times 10 \] \[ 4 \times 27 = 108 \] \[ 108 \times 10 = 1080 \]Final Answer:
1080
Question 2 (4 marks)
There are 60 people in a choir.
Half of the people in the choir are women.
The number of women in the choir is 3 times the number of men in the choir.
The rest of the people in the choir are children.
the number of children in the choir : the number of men in the choir = \( n : 1 \)
Work out the value of \( n \).
Worked Solution
Step 1: Find the number of women
Why: We are told half the total people are women.
Step 2: Find the number of men
Why: We know the number of women is 3 times the number of men.
Step 3: Find the number of children
Why: The “rest” are children. We subtract men and women from the total.
Step 4: Find the ratio and value of n
Goal: We need the ratio Children : Men in the form \( n : 1 \).
Divide both sides by 10 to get the form \( n : 1 \):
\[ 2 : 1 \]Therefore, \( n = 2 \).
Final Answer:
\( n = 2 \)
Question 3 (3 marks)
Work out \( 1\frac{3}{4} \times 1\frac{1}{3} \)
Give your answer as a mixed number.
Worked Solution
Step 1: Convert to improper fractions
Why: It is much easier to multiply improper fractions than mixed numbers.
Step 2: Multiply the fractions
Method: Multiply numerators together and denominators together. Look for cancellations first.
Notice the 4 in the denominator and the 4 in the numerator cancel out:
\[ = \frac{7}{\cancel{4}^1} \times \frac{\cancel{4}^1}{3} = \frac{7 \times 1}{1 \times 3} = \frac{7}{3} \]Step 3: Convert back to a mixed number
Why: The question asks for the answer as a mixed number.
How many 3s go into 7?
\( 7 \div 3 = 2 \) remainder \( 1 \)
\[ \frac{7}{3} = 2\frac{1}{3} \]Final Answer:
\( 2\frac{1}{3} \)
Question 4 (2 marks)
Use a ruler and compasses to construct the line from the point \( P \) perpendicular to the line \( CD \).
You must show all construction lines.
Worked Solution
Construction Steps
Method: To drop a perpendicular from a point to a line:
- Place compass point on \( P \).
- Draw an arc that cuts line \( CD \) in two places.
- From these two intersection points, draw two crossing arcs below the line.
- Draw a straight line from \( P \) through the crossing point of these arcs.
Answer: A straight line drawn from P intersecting CD at 90 degrees, with visible construction arcs.
Question 5 (4 marks)
The diagram shows triangle \( ABC \).
\( ADB \) is a straight line.
The size of angle \( DCB \) : the size of angle \( ACD = 2 : 1 \)
Work out the size of angle \( BDC \).
Worked Solution
Step 1: Calculate Angle \( ACB \)
Why: We know the angles in a triangle add up to \( 180^\circ \). We have two angles of the large triangle \( ABC \).
Step 2: Split Angle \( ACB \) using the ratio
Why: We are told \( \angle DCB : \angle ACD = 2 : 1 \). This splits the \( 54^\circ \) angle.
Total parts = \( 2 + 1 = 3 \)
Value of one part = \( 54^\circ \div 3 = 18^\circ \)
Angle \( ACD \) (1 part) = \( 18^\circ \)
Angle \( DCB \) (2 parts) = \( 18^\circ \times 2 = 36^\circ \)
Step 3: Calculate Angle \( BDC \)
Method: We can use triangle \( BDC \). Angles inside sum to \( 180^\circ \).
In triangle \( BDC \):
- \( \angle DBC = 51^\circ \) (given)
- \( \angle DCB = 36^\circ \) (calculated)
Step 4: Check
Method: We can check using triangle \( ADC \). Angles on straight line \( ADB \) sum to 180.
In triangle \( ADC \): \( 180 – (75 + 18) = 87^\circ \)
Angles on line: \( 87^\circ + 93^\circ = 180^\circ \). (Correct)
Final Answer:
93°
Question 6 (3 marks)
4 red bricks have a mean weight of 5 kg.
5 blue bricks have a mean weight of 9 kg.
1 green brick has a weight of 6 kg.
Donna says,
“The mean weight of the 10 bricks is less than 7 kg.”
Is Donna correct? You must show how you get your answer.
Worked Solution
Step 1: Calculate total weight of each colour
Why: To find the combined mean, we first need the total weight of all bricks. Total = Mean × Count.
Red: \( 4 \times 5 = 20 \text{ kg} \)
Blue: \( 5 \times 9 = 45 \text{ kg} \)
Green: \( 1 \times 6 = 6 \text{ kg} \)
Step 2: Calculate overall mean
Method: Sum all weights and divide by the total number of bricks (10).
Step 3: Compare and Conclude
Question: Donna says it is “less than 7 kg”.
Our calculated mean is \( 7.1 \text{ kg} \).
\( 7.1 \) is not less than \( 7 \).
Therefore, Donna is incorrect.
Final Answer:
No, Donna is not correct because the mean is 7.1 kg.
Question 7 (3 marks)
(a) Simplify \( (p^2)^5 \)
(b) Simplify \( 12x^7y^3 \div 6x^3y \)
Worked Solution
Part (a): Power of a Power
Rule: \( (a^m)^n = a^{m \times n} \)
Part (b): Division of Indices
Method: Deal with numbers, then x’s, then y’s separately.
Rule: \( \frac{a^m}{a^n} = a^{m-n} \)
Numbers: \( 12 \div 6 = 2 \)
x terms: \( x^7 \div x^3 = x^{7-3} = x^4 \)
y terms: \( y^3 \div y^1 = y^{3-1} = y^2 \) (Remember \( y \) is \( y^1 \))
Combine them: \( 2x^4y^2 \)
Final Answer:
(a) \( p^{10} \)
(b) \( 2x^4y^2 \)
Question 8 (5 marks)
The accurate scale drawing shows the positions of port \( P \) and a lighthouse \( L \).
Scale: 1 cm represents 4 km.
Aleena sails her boat from port \( P \) on a bearing of \( 070^\circ \).
She sails for \( 1\frac{1}{2} \) hours at an average speed of 12 km/h to a port \( Q \).
Find:
(i) the distance, in km, of port \( Q \) from lighthouse \( L \),
(ii) the bearing of port \( Q \) from lighthouse \( L \).
Worked Solution
Step 1: Calculate real distance traveled to Q
Formula: \( \text{Distance} = \text{Speed} \times \text{Time} \)
Step 2: Convert to Scale Distance
Scale: 1 cm represents 4 km.
Step 3: Plot point Q
Action:
- From P, measure angle \( 070^\circ \) clockwise from North.
- Draw a line 4.5 cm long. Mark this point Q.
Step 4: Measure QL
Action: Measure the distance from Q to L on your diagram in cm, then convert back to km.
Typical measurement for QL is approx 5.5 cm.
(Acceptable range: 20 – 23 km)
Step 5: Measure Bearing of Q from L
Action: At L, measure the angle clockwise from North to the line LQ.
The angle is large (reflex). It is roughly \( 320^\circ \).
(Acceptable range: 317° – 330°)
Final Answer:
(i) Distance: ~22 km
(ii) Bearing: ~322°
Question 9 (4 marks)
A car travels for 18 minutes at an average speed of 72 km/h.
(a) How far will the car travel in these 18 minutes?
David says,
“72 kilometres per hour is faster than 20 metres per second.”
(b) Is David correct? You must show how you get your answer.
Worked Solution
Part (a): Distance Calculation
Note: Time is in minutes, Speed is in km/h. We must convert time to hours.
Simplify fraction (divide by 6):
\[ \frac{18}{60} = \frac{3}{10} = 0.3 \text{ hours} \] \[ \text{Distance} = \text{Speed} \times \text{Time} \] \[ \text{Distance} = 72 \times 0.3 \] \[ 72 \times 3 = 216 \] \[ \text{So, } 72 \times 0.3 = 21.6 \text{ km} \]Part (b): Unit Conversion
Strategy: Convert 20 m/s to km/h OR 72 km/h to m/s to compare them.
Method 1: Convert m/s to km/h
20 metres in 1 second.
\( \times 60 \) for metres in 1 minute: \( 20 \times 60 = 1200 \) m/min.
\( \times 60 \) for metres in 1 hour: \( 1200 \times 60 = 72,000 \) m/hour.
Convert to km: \( 72,000 \div 1000 = 72 \) km/h.
Comparison:
David compares 72 km/h with 72 km/h.
They are exactly the same speed.
Therefore, 72 km/h is NOT faster.
Final Answer:
(a) 21.6 km
(b) No, David is incorrect. They are the same speed.
Question 10 (6 marks)
The cumulative frequency table shows information about the times, in minutes, taken by 40 people to complete a puzzle.
| Time (m minutes) | Cumulative frequency |
|---|---|
| 20 < m ≤ 40 | 5 |
| 20 < m ≤ 60 | 25 |
| 20 < m ≤ 80 | 35 |
| 20 < m ≤ 100 | 38 |
| 20 < m ≤ 120 | 40 |
(a) On the grid below, draw a cumulative frequency graph for this information.
(b) Use your graph to find an estimate for the interquartile range.
(c) Use your graph to find an estimate for the probability that this person took between 50 minutes and 90 minutes.
Worked Solution
Part (a): Plotting the Graph
Rule: Plot cumulative frequency against the UPPER bound of the class interval.
Points to plot:
- (40, 5)
- (60, 25)
- (80, 35)
- (100, 38)
- (120, 40)
Also, the graph starts at (20, 0) because no one took less than 20 minutes.
Part (b): Interquartile Range (IQR)
Formula: \( \text{IQR} = \text{Upper Quartile} – \text{Lower Quartile} \)
Lower Quartile (LQ): Position = \( \frac{1}{4} \times 40 = 10 \). Read from y=10 across to curve, then down.
Upper Quartile (UQ): Position = \( \frac{3}{4} \times 40 = 30 \). Read from y=30 across to curve, then down.
From graph:
LQ (at cf=10) ≈ 44 minutes
UQ (at cf=30) ≈ 71 minutes
\[ \text{IQR} = 71 – 44 = 27 \text{ minutes} \](Allowed range: 21 to 28 minutes)
Part (c): Probability 50 < t < 90
Method: Find the number of people at 90 mins and subtract the number of people at 50 mins.
Read up from x=50 to curve, then across: cf ≈ 13
Read up from x=90 to curve, then across: cf ≈ 37
Number of people = \( 37 – 13 = 24 \)
Probability = \( \frac{24}{40} \)
Simplify: \( \frac{3}{5} \) or 0.6
(Allowed range for answer: 0.475 to 0.6)
Final Answer:
(b) Approx 27 minutes
(c) \( \frac{24}{40} \) (or approx 0.6)
Question 11 (2 marks)
There are \( p \) counters in a bag.
12 of the counters are yellow.
Shafiq takes at random 30 counters from the bag.
5 of these 30 counters are yellow.
Work out an estimate for the value of \( p \).
Worked Solution
Step 1: Set up the proportion
Why: The proportion of yellow counters in the sample should estimate the proportion of yellow counters in the whole bag.
Step 2: Solve for p
Method: We can simplify the fraction or cross-multiply.
Simplify \( \frac{5}{30} \) to \( \frac{1}{6} \):
\[ \frac{1}{6} = \frac{12}{p} \]Multiply both sides by \( p \):
\[ p \times \frac{1}{6} = 12 \]Multiply both sides by 6:
\[ p = 12 \times 6 \] \[ p = 72 \]Final Answer:
72
Question 12 (1 mark)
\( T = \frac{q}{2} + 5 \)
Here is Spencer’s method to make \( q \) the subject of the formula.
\( 2 \times T = q + 5 \)
\( q = 2T – 5 \)
What mistake did Spencer make in the first line of his method?
Worked Solution
Identify the Error
Analysis: Spencer wanted to remove the fraction by multiplying by 2. However, he must multiply every term in the equation by 2.
Equation: \( T = \frac{q}{2} + 5 \)
Correct step: \( 2 \times T = 2 \times (\frac{q}{2}) + 2 \times 5 \)
Correct result: \( 2T = q + 10 \)
Spencer’s Mistake: He multiplied \( T \) and \( \frac{q}{2} \) by 2, but he failed to multiply the 5 by 2.
Final Answer:
Spencer did not multiply the 5 by 2.
Question 13 (3 marks)
(a) Write \( \frac{5}{x+1} + \frac{2}{3x} \) as a single fraction in its simplest form.
(b) Factorise \( (x + y)^2 + 3(x + y) \)
Worked Solution
Part (a): Algebraic Fractions
Method: Find a common denominator by multiplying the two denominators together.
Common Denominator: \( 3x(x+1) \)
Adjust numerators:
\[ \frac{5(3x)}{3x(x+1)} + \frac{2(x+1)}{3x(x+1)} \]Combine into one fraction:
\[ \frac{15x + 2(x+1)}{3x(x+1)} \]Expand the numerator:
\[ \frac{15x + 2x + 2}{3x(x+1)} \]Simplify:
\[ \frac{17x + 2}{3x(x+1)} \]Part (b): Factorising
Method: Look for a common factor. The term \( (x+y) \) is present in both parts of the expression.
Expression: \( (x + y)^2 + 3(x + y) \)
Let \( A = (x+y) \). Then we have \( A^2 + 3A \).
Factorise out \( A \): \( A(A + 3) \).
Substitute \( x+y \) back in:
\[ (x + y)( (x + y) + 3 ) \] \[ (x + y)(x + y + 3) \]Final Answer:
(a) \( \frac{17x + 2}{3x(x+1)} \)
(b) \( (x + y)(x + y + 3) \)
Question 14 (4 marks)
The diagram shows a right-angled triangle.
All the measurements are in centimetres.
The area of the triangle is \( 27.5 \text{ cm}^2 \).
Work out the length of the shortest side of the triangle.
You must show all your working.
Worked Solution
Step 1: Set up the Area Equation
Formula: Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \).
We are given Area = 27.5:
\[ \frac{1}{2} (x + 4)(x – 2) = 27.5 \]Step 2: Form a Quadratic Equation
Method: Multiply by 2 to remove the fraction, then expand the brackets.
Expand:
\[ x^2 – 2x + 4x – 8 = 55 \] \[ x^2 + 2x – 8 = 55 \]Rearrange to equal zero:
\[ x^2 + 2x – 8 – 55 = 0 \] \[ x^2 + 2x – 63 = 0 \]Step 3: Solve the Quadratic
Method: Factorise the quadratic. We need two numbers that multiply to make -63 and add to make +2.
Factors of 63: 1, 63; 3, 21; 7, 9.
Using +9 and -7 gives +2.
\[ (x + 9)(x – 7) = 0 \]Solutions:
\[ x = -9 \quad \text{or} \quad x = 7 \]Since a length cannot be negative (e.g., \( x-2 \) would be -11), we must use \( x = 7 \).
Step 4: Calculate Shortest Side
Method: Substitute \( x = 7 \) into the side lengths.
Height: \( x – 2 = 7 – 2 = 5 \text{ cm} \)
Base: \( x + 4 = 7 + 4 = 11 \text{ cm} \)
The shortest side is 5 cm.
Final Answer:
5 cm
Question 15 (3 marks)
Express \( 0.4\dot{1}\dot{8} \) as a fraction.
You must show all your working.
Worked Solution
Step 1: Set up the equation
Method: Let \( x \) equal the recurring decimal. Identify the repeating part (18).
Step 2: Multiply to align the decimals
Goal: Get the recurring part immediately after the decimal point in two different equations so they cancel out when subtracted.
Multiply by 10 to move the non-repeating ‘4’ past the decimal:
\[ 10x = 4.181818… \]Multiply by 1000 (since the repeat length is 2 digits, multiply previous line by 100):
\[ 1000x = 418.181818… \]Step 3: Subtract and Solve
Subtract the equations:
1000x = 418.1818... - 10x = 4.1818... --------------------- 990x = 414
Step 4: Simplify the fraction
Method: Divide numerator and denominator by common factors.
Both are even, divide by 2:
\[ \frac{207}{495} \]Check sum of digits for divisibility by 9: \( 2+0+7=9 \) and \( 4+9+5=18 \). Both divisible by 9.
\[ 207 \div 9 = 23 \] \[ 495 \div 9 = 55 \] \[ \text{Fraction} = \frac{23}{55} \]Final Answer:
\( \frac{23}{55} \) (or \( \frac{414}{990} \))
Question 16 (5 marks)
(a) Rationalise the denominator of \( \frac{22}{\sqrt{11}} \)
Give your answer in its simplest form.
(b) Show that \( \frac{\sqrt{3}}{2\sqrt{3}-1} \) can be written in the form \( \frac{a+\sqrt{3}}{b} \) where \( a \) and \( b \) are integers.
Worked Solution
Part (a): Rationalise
Method: Multiply numerator and denominator by \( \sqrt{11} \) to remove the surd from the bottom.
Simplify by dividing 22 by 11:
\[ \frac{22}{11} = 2 \] \[ \text{Answer} = 2\sqrt{11} \]Part (b): Rationalise Binomial Denominator
Method: Multiply by the conjugate of the denominator. The conjugate of \( 2\sqrt{3}-1 \) is \( 2\sqrt{3}+1 \).
Step 1: Expand Numerator
\[ \sqrt{3}(2\sqrt{3} + 1) = ( \sqrt{3} \times 2\sqrt{3} ) + ( \sqrt{3} \times 1 ) \] \[ = (2 \times 3) + \sqrt{3} \] \[ = 6 + \sqrt{3} \]Step 2: Expand Denominator (Difference of Two Squares)
\[ (2\sqrt{3}-1)(2\sqrt{3}+1) = (2\sqrt{3})^2 – 1^2 \] \[ (2\sqrt{3})^2 = 4 \times 3 = 12 \] \[ 12 – 1 = 11 \]Step 3: Combine
\[ \frac{6+\sqrt{3}}{11} \]This matches the form \( \frac{a+\sqrt{3}}{b} \) with \( a=6 \) and \( b=11 \).
Final Answer:
(a) \( 2\sqrt{11} \)
(b) Shown: \( \frac{6+\sqrt{3}}{11} \)
Question 17 (4 marks)
A and B are two similar cylindrical containers.
The surface area of container A : the surface area of container B = \( 4 : 9 \)
Tyler fills container A with water.
She then pours all the water into container B.
Tyler repeats this and stops when container B is full of water.
Work out the number of times that Tyler fills container A with water.
You must show all your working.
Worked Solution
Step 1: Find the Linear Scale Factor (k)
Concept: If shapes are similar, the ratio of their areas is the square of the ratio of their lengths.
So, linear scale factor \( k = \frac{3}{2} = 1.5 \).
Step 2: Find the Volume Ratio
Concept: The ratio of volumes is the cube of the ratio of lengths.
This means Volume B is \( \frac{27}{8} \) times Volume A.
Step 3: Calculate number of fills
Question: How many times does A fit into B?
Tyler fills A and pours. She needs to do this 3 full times (which gives 24 parts). After 3 pours, B is not yet full.
She fills A a 4th time to pour the remaining required amount.
She stops when B is full.
Final Answer:
4 times
Question 18 (6 marks)
The function \( f \) is given by \( f(x) = 2x^3 – 4 \)
(a) Show that \( f^{-1}(50) = 3 \)
The functions \( g \) and \( h \) are given by \( g(x) = x + 2 \) and \( h(x) = x^2 \)
(b) Find the values of \( x \) for which \( hg(x) = 3x^2 + x – 1 \)
Worked Solution
Part (a): Inverse Function
Method 1: Calculate \( f(3) \)
If \( f^{-1}(50) = 3 \), then \( f(3) \) must equal 50.
Since \( f(3) = 50 \), it follows that \( f^{-1}(50) = 3 \).
Method 2: Find \( f^{-1}(x) \)
Let \( y = 2x^3 – 4 \). Rearrange for \( x \).
Substitute \( y = 50 \):
\[ f^{-1}(50) = \sqrt[3]{\frac{50+4}{2}} = \sqrt[3]{27} = 3 \]Part (b): Composite Function
Step 1: Find \( hg(x) \)
\( hg(x) \) means apply \( g \) first, then \( h \). Put \( g(x) \) inside \( h(x) \).
Expand brackets:
\[ (x+2)(x+2) = x^2 + 4x + 4 \]Step 2: Solve the Equation
Set \( hg(x) = 3x^2 + x – 1 \):
\[ x^2 + 4x + 4 = 3x^2 + x – 1 \]Rearrange to form a quadratic = 0 (move terms to the right):
\[ 0 = 3x^2 – x^2 + x – 4x – 1 – 4 \] \[ 0 = 2x^2 – 3x – 5 \]Step 3: Solve Quadratic
Method: Factorise \( 2x^2 – 3x – 5 \). We need numbers that multiply to \( 2 \times -5 = -10 \) and add to -3.
The numbers are -5 and +2.
\[ 2x^2 – 5x + 2x – 5 = 0 \] \[ x(2x – 5) + 1(2x – 5) = 0 \] \[ (x + 1)(2x – 5) = 0 \]Solutions:
\[ x + 1 = 0 \Rightarrow x = -1 \] \[ 2x – 5 = 0 \Rightarrow x = 2.5 \]Final Answer:
\( x = -1 \) and \( x = 2.5 \)
Question 19 (3 marks)
Given that \( 9^{-\frac{1}{2}} = 27^{\frac{1}{4}} \div 3^{x+1} \)
find the exact value of \( x \).
Worked Solution
Step 1: Convert all bases to 3
Why: 9, 27, and 3 are all powers of 3. Making bases the same allows us to equate indices.
Step 2: Rewrite the equation
Left Hand Side:
\[ 9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{2 \times -0.5} = 3^{-1} \]Right Hand Side:
\[ 27^{\frac{1}{4}} \div 3^{x+1} = (3^3)^{\frac{1}{4}} \div 3^{x+1} \] \[ = 3^{\frac{3}{4}} \div 3^{x+1} \]Using division law of indices (subtract powers):
\[ = 3^{\frac{3}{4} – (x+1)} \] \[ = 3^{0.75 – x – 1} = 3^{-0.25 – x} \]Step 3: Equate powers and solve
Powers must be equal:
\[ -1 = -0.25 – x \]Add \( x \) to both sides:
\[ x – 1 = -0.25 \]Add 1 to both sides:
\[ x = 1 – 0.25 \] \[ x = 0.75 \]Final Answer:
\( x = \frac{3}{4} \) (or 0.75)
Question 20 (3 marks)
The graph of \( y = f(x) \) is shown on the grid.
(a) On the grid, draw the graph with equation \( y = f(x+1) – 3 \).
Point \( A(-2, 1) \) lies on the graph of \( y = f(x) \).
When the graph of \( y = f(x) \) is transformed to the graph with equation \( y = f(-x) \), point \( A \) is mapped to point \( B \).
(b) Write down the coordinates of point \( B \).
Worked Solution
Part (a): Drawing the transformed graph
Transformation Logic:
- \( f(x+1) \): Translation by vector \( \begin{pmatrix} -1 \\ 0 \end{pmatrix} \) (Shift Left 1).
- \( -3 \): Translation by vector \( \begin{pmatrix} 0 \\ -3 \end{pmatrix} \) (Shift Down 3).
- Combined: Shift every point Left 1 and Down 3. Vector \( \begin{pmatrix} -1 \\ -3 \end{pmatrix} \).
Key Points:
- (-4, 3) \( \rightarrow \) (-5, 0)
- (-3, 3) \( \rightarrow \) (-4, 0)
- (-2, 1) \( \rightarrow \) (-3, -2)
- (0, 3) \( \rightarrow \) (-1, 0)
- (1, 4) \( \rightarrow \) (0, 1)
- (3, 4) \( \rightarrow \) (2, 1)
Part (b): Coordinate Transformation
Rule: \( y = f(-x) \) is a reflection in the y-axis.
The x-coordinate changes sign: \( (x, y) \rightarrow (-x, y) \).
Point A is \( (-2, 1) \).
Reflect in y-axis:
\[ x: -2 \rightarrow -(-2) = 2 \] \[ y: 1 \rightarrow 1 \]Point B is \( (2, 1) \).
Final Answer:
(b) (2, 1)
Question 21 (5 marks)
Sketch the graph of \( y = 2x^2 – 8x – 5 \)
Showing the coordinates of the turning point and the exact coordinates of any intercepts with the coordinate axes.
Worked Solution
Step 1: Find the Turning Point (Completing the Square)
Complete the square for \( x^2 – 4x \):
\[ x^2 – 4x = (x-2)^2 – 4 \]Substitute back:
\[ y = 2[ (x-2)^2 – 4 ] – 5 \] \[ y = 2(x-2)^2 – 8 – 5 \] \[ y = 2(x-2)^2 – 13 \]Turning Point is at \( (2, -13) \).
Step 2: Find the y-intercept
Set \( x = 0 \).
Intercept: \( (0, -5) \)
Step 3: Find x-intercepts
Set \( y = 0 \) and solve. Use the completed square form or quadratic formula.
Intercepts: \( (2 + \sqrt{\frac{13}{2}}, 0) \) and \( (2 – \sqrt{\frac{13}{2}}, 0) \)
Step 4: Sketch
Summary:
Minimum: \( (2, -13) \)
y-intercept: \( (0, -5) \)
x-intercepts: \( 2 \pm \sqrt{\frac{13}{2}} \)
Question 22 (4 marks)
A, B, C and D are four points on a circle.
AEC and DEB are straight lines.
Triangle \( AED \) is an equilateral triangle.
Prove that triangle \( ABC \) is congruent to triangle \( DCB \).
Worked Solution
Step 1: Identify Key Information
Given: Triangle \( AED \) is equilateral.
Implies: \( AE = ED = AD \) and all angles in \( AED \) are \( 60^\circ \).
Step 2: Find Angles in the Same Segment
Theorem: Angles subtended by the same arc at the circumference are equal.
Arc AB subtends \( \angle ADB \) and \( \angle ACB \).
\[ \angle ACB = \angle ADB = 60^\circ \](Since \( \angle ADE = 60^\circ \) as AED is equilateral).
Arc DC subtends \( \angle DAC \) and \( \angle DBC \).
\[ \angle DBC = \angle DAC = 60^\circ \](Since \( \angle DAE = 60^\circ \) as AED is equilateral).
Step 3: Construct the Proof for Congruency (AAS)
We need 3 facts (Side, Angle, Angle) to prove \( \triangle ABC \) and \( \triangle DCB \) are congruent.
- Angle: \( \angle ACB = \angle DBC = 60^\circ \). (Proved above).
- Angle: Vertically opposite angles at E? No, let’s look at the full triangles ABC and DCB.
Let’s use a different set of properties for triangles ABC and DCB.
1. \( BC \) is common to both.
2. \( \angle BAC = \angle BDC \) (Angles in same segment from arc BC).
Let’s use the \( 60^\circ \) facts again.
- \( \angle ABC = \angle ABD + \angle DBC \).
- \( \angle DCB = \angle DCA + \angle ACB \).
Actually, simpler path:
1. \( \angle ACB = 60^\circ \)
2. \( \angle DBC = 60^\circ \)
Therefore \( \triangle EBC \) is equilateral (since two angles are 60, the third is 60). So \( EB = EC = BC \).
Now consider sides:
\( AC = AE + EC \)
\( DB = DE + EB \)
Since \( AE=DE \) (equilateral) and \( EC=EB \) (equilateral), then \( AC = DB \).
Final Proof (SAS or SSS):
- Side: \( AC = DB \) (proven above)
- Side: \( BC \) is common.
- Angle: \( \angle ACB = \angle DBC = 60^\circ \).
This is SAS (Side-Angle-Side).
Alternatively: \( AB = DC \) (can be proven similarly). Then SSS.
Conclusion:
1. \( \angle ADE = \angle DAE = 60^\circ \) (AED is equilateral).
2. \( \angle ACB = \angle ADE = 60^\circ \) (Angles in same segment).
3. \( \angle DBC = \angle DAE = 60^\circ \) (Angles in same segment).
4. Since base angles \( \angle ECB \) and \( \angle EBC \) are both \( 60^\circ \), \( \triangle CEB \) is equilateral, so \( EC = EB \).
5. \( AC = AE + EC \) and \( DB = DE + EB \). Since \( AE=DE \) and \( EC=EB \), \( AC = DB \).
6. In \( \triangle ABC \) and \( \triangle DCB \):
– \( AC = DB \)
– \( BC \) is common
– \( \angle ACB = \angle DBC = 60^\circ \)
Therefore, congruent by SAS.