IGCSE MATH: A&G TESTS

TERM 2 MIRROR TEST 1 (FEBRUARY 2026)

Subject: MATHEMATICS	Class: YEAR 10	Stream: __
Time: 1 HOUR 40 MINUTES	Total Marks: 80

INSTRUCTIONS

Answer all questions.
Use a black or dark blue pen.
Write your name and admission number in the boxes above.
Write your answer to each question in the space provided.
You should use a calculator where appropriate.
You must show all necessary working clearly.
Give non-exact numerical answers correct to 3 significant figures.
For π, use either your calculator value or 3.142.

1. Factorise Completely

(a) 4m + 5p - 12km - 15kp [2]

(b) 5x² - 45y² [2]

2. Solve

(4x - 5)(3x + 8) = 0 [2]

3. Factorise

8x² + 14x - 15 [3]

4. Solve

(a) 5(w + 9) = 75 [1]

(b) (x - 4)/2 = 5 [1]

5. Solve Simultaneous Equations

2x + 3y = 7

5x - 2y = 16

You must show all your working. [4]

[Show working here]

6. Solve

x² + 6x + 5 = 20

x = _______________ [4]

or x = _______________ [2]

7. Coordinate Geometry

The line y = 2x + 3 intersects the graph of y = x² - 1 at points A and B.

Find coordinates of A and B.

You must show all your working. [4]

A ( ______ , ______ )

B ( ______ , ______ )

[Show working here]

8. Inequality from Number Line

[Diagram of number line would be here]

Write the inequality shown by the number line. [1]

9. Solve Inequality

4m + 10 ≤ 9m - 8 [2]

10. Integer Values

Find integer values of n: -5 < 3n ≤ 9 [2]

11. Integer Values

Find integer values of n: 12 ≤ 3n < 27 [2]

12. Solve Equation

2/(x - 2) + 3/(x + 4) = 1 [5]

x = _______________ or x = _______________

13. Solve Equation

(15 - 5x)/4 = 5 - x [3]

14. Inequalities Region

The region R satisfies:

y > 2, y < x + 4, x + 2y ≤ 6

By drawing three suitable lines and shading unwanted regions, find and label region R. [4]

[Grid for drawing would be here]

15. Find Inequalities

[Diagram of shaded region R would be here]

Find the three inequalities that define region R. [3]

[Write inequalities here]

16. Cubic Graph

Table for y = x³ - 3x + 1:

x	-2	-1.5	-1	-0.5	0	0.5	1	1.5	2
y	-1	_____	3	_____	1	_____	-1	_____	3

(a) Complete the table [3]

(b) Draw graph for -2 ≤ x ≤ 2 [4]

[Graph grid would be here]

17. Quadratic Graph

Table for y = x² + 4x - 140:

x	-20	-15	-10	-5	0	5	10	15
y	100	_____	-80	_____	-140	_____	_____	145

(a) Complete the table [3]

(b) Draw graph for -20 ≤ x ≤ 15 [4]

[Graph grid would be here]

18. Reciprocal Graph

Table for y = -8/x:

x	-8	-5	-4	-2	-1.6	-1	1	1.6	2	4	5	8
y	1	1.6	_____	4	_____	8	-8	_____	-4	_____	-1.6	-1

(a) Complete the table [3]

(b) Draw graph for -8 ≤ x ≤ -1 and 1 ≤ x ≤ 8 [4]

(d) Equation(s) of line symmetry: _____ [2]

(e) Draw line y = 3 [1]

(f) Solve -8/x = 3: x = _____ [1]

[Graph grid would be here]

END OF TEST - TOTAL MARKS: 80

📚 MATHEMATICS REVISION NOTES

Term 2 Test Preparation - Year 10

Essential Concepts, Formulas, and Practice Tips

⚠️ Test Information

Duration: 1 hour 40 minutes
Total Marks: 80
Calculator: Allowed (non-programmable)
Topics: Algebra, Equations, Graphs, Geometry

📑 QUICK NAVIGATION

Algebraic Factorization
Solving Equations
Simultaneous Equations
Quadratic Equations
Inequalities

Coordinate Geometry
Graph Plotting
Graphical Solutions
Essential Formulas
Exam Tips

1. ALGEBRAIC FACTORIZATION

🔹 Common Factor Method

Example: 4m + 5p - 12km - 15kp

Group terms: (4m - 12km) + (5p - 15kp)
Factor: 4m(1 - 3k) + 5p(1 - 3k)
Answer: (1 - 3k)(4m + 5p)

🔹 Difference of Two Squares

Formula: a² - b² = (a - b)(a + b)

Example: 5x² - 45y² = 5(x² - 9y²) = 5(x - 3y)(x + 3y)

🔹 Quadratic Trinomials (ac-method)

For ax² + bx + c:

1. Multiply a × c
2. Find numbers that multiply to a×c and add to b
3. Split middle term
4. Factor by grouping

Example: 8x² + 14x - 15

a×c = 8 × (-15) = -120

Numbers: 20 and -6 (20 × -6 = -120, 20 + (-6) = 14)

= 8x² + 20x - 6x - 15

= 4x(2x + 5) - 3(2x + 5)

= (2x + 5)(4x - 3)

2. SOLVING EQUATIONS

🔹 Linear Equations

Example 1: 5(w + 9) = 75

5w + 45 = 75 → 5w = 30 → w = 6

Example 2: (x - 4)/2 = 5

x - 4 = 10 → x = 14

Example 3: 6x - 7 = 23

6x = 30 → x = 5

🔹 Zero Product Property

Rule: If A × B = 0, then A = 0 OR B = 0

Example: (4x - 5)(3x + 8) = 0

4x - 5 = 0 → x = 5/4 = 1.25

OR 3x + 8 = 0 → x = -8/3 ≈ -2.667

3. SIMULTANEOUS EQUATIONS

🔹 Elimination Method

Given: 2x + 3y = 7 ... (1)
5x - 2y = 16 ... (2)

Multiply (1) by 2: 4x + 6y = 14 ... (3)

Multiply (2) by 3: 15x - 6y = 48 ... (4)

Add (3) + (4): 19x = 62

x = 62/19

Substitute: 2(62/19) + 3y = 7

124/19 + 3y = 133/19

3y = 9/19 → y = 3/19

4. QUADRATIC EQUATIONS

🔹 Quadratic Formula

x = [-b ± √(b² - 4ac)] / 2a

For equation: ax² + bx + c = 0

Example: x² + 6x - 15 = 0

a = 1, b = 6, c = -15

Discriminant: b² - 4ac = 36 + 60 = 96

x = [-6 ± √96] / 2 = [-6 ± 4√6] / 2

x = -3 ± 2√6

x ≈ 1.899 or x ≈ -7.899 (3 s.f.)

5. INEQUALITIES

🔹 Solving Linear Inequalities

Example: 4m + 10 ≤ 9m - 8

10 + 8 ≤ 9m - 4m

18 ≤ 5m

5m ≥ 18

m ≥ 18/5 or m ≥ 3.6

🔹 Integer Solutions

Example: -5 < 3n ≤ 9

Divide by 3: -5/3 < n ≤ 3

Integer values: -1, 0, 1, 2, 3

6. COORDINATE GEOMETRY

🔹 Intersection of Line and Curve

Line: y = 2x + 3

Curve: y = x² - 1

Set equal: x² - 1 = 2x + 3

x² - 2x - 4 = 0

Solve: x = 1 ± √5

Points: A(1+√5, 5+2√5) and B(1-√5, 5-2√5)

7. GRAPH PLOTTING TIPS

📈 Cubic Graphs

Calculate y-values carefully
Draw smooth "S" curve
Plot points accurately

📊 Quadratic Graphs

U-shaped if a > 0
∩-shaped if a < 0
Find vertex

📉 Reciprocal Graphs

Two separate curves
Never touches axes
Symmetry: y = x and y = -x

8. GRAPHICAL SOLUTIONS

To solve f(x) = k graphically:

Plot y = f(x)
Draw horizontal line y = k
Find intersection points
Read x-values from graph

9. ESSENTIAL FORMULAS

Algebra

a² - b² = (a - b)(a + b)

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Quadratic

x = [-b ± √(b² - 4ac)] / 2a

Discriminant: Δ = b² - 4ac

Straight Line

y = mx + c

m = (y₂ - y₁)/(x₂ - x₁)

Midpoint formula

10. EXAM TIPS & STRATEGY

📝 Before Test

Review notes 2-3 days before
Practice past papers
Get good sleep
Eat healthy breakfast

⏱️ During Test

Read instructions carefully
Show ALL working
Answer every question
Check answers if time

❌ Avoid Mistakes

Forgetting ± in roots
Wrong inequality signs
Misreading graphs
Rounding too early

✅ PRACTICE CHECKLIST

Can you:

✓ Factorise quadratics?
✓ Solve simultaneous equations?
✓ Use quadratic formula?
✓ Solve inequalities?
✓ Plot different graphs?

Need more practice?

Review sections where you answered "No"

Ask teacher for help

Practice with classmates

Do extra problems

abel.masitsa.com Works!

Preparation + Practice + Precision = Success

Revision Notes • February 2026

SOLUTIONS

TERM 2 MIRROR TEST 1 MATHEMATICS

YEAR 10 - FEBRUARY 2026

1. Factorise Completely

(a) 4m + 5p - 12km - 15kp

Solution:

= 4m - 12km + 5p - 15kp

= 4m(1 - 3k) + 5p(1 - 3k)

= (1 - 3k)(4m + 5p)

(b) 5x² - 45y²

Solution:

= 5(x² - 9y²)

= 5[x² - (3y)²]

= 5(x - 3y)(x + 3y)

2. Solve

(4x - 5)(3x + 8) = 0

Solution:

Using zero product property:

4x - 5 = 0 OR 3x + 8 = 0

4x = 5 OR 3x = -8

x = 5/4 OR x = -8/3

x = 1.25 OR x = -2.667 (to 3 s.f.)

3. Factorise

8x² + 14x - 15

Solution:

Using the ac-method:

a × c = 8 × (-15) = -120

Find two numbers that multiply to -120 and add to 14: 20 and -6

= 8x² + 20x - 6x - 15

= 4x(2x + 5) - 3(2x + 5)

= (2x + 5)(4x - 3)

4. Solve

(a) 5(w + 9) = 75

5w + 45 = 75

5w = 75 - 45

5w = 30

w = 6

(b) (x - 4)/2 = 5

x - 4 = 5 × 2

x - 4 = 10

x = 14

(c) 6x - 7 = 23

6x = 23 + 7

6x = 30

x = 5

5. Solve Simultaneous Equations

2x + 3y = 7 ... (1)

5x - 2y = 16 ... (2)

Solution:

Multiply (1) by 2: 4x + 6y = 14 ... (3)

Multiply (2) by 3: 15x - 6y = 48 ... (4)

Add (3) and (4): 19x = 62

x = 62/19 ≈ 3.263

Substitute into (1): 2(62/19) + 3y = 7

124/19 + 3y = 133/19

3y = 133/19 - 124/19 = 9/19

y = 3/19 ≈ 0.158

Answer: x = 62/19, y = 3/19

6. Solve

x² + 6x + 5 = 20

Solution:

x² + 6x + 5 - 20 = 0

x² + 6x - 15 = 0

Using quadratic formula: a=1, b=6, c=-15

x = [-b ± √(b² - 4ac)] / 2a

x = [-6 ± √(36 + 60)] / 2

x = [-6 ± √96] / 2

x = [-6 ± 4√6] / 2

x = -3 ± 2√6

x = -3 + 2√6 OR x = -3 - 2√6

x ≈ 1.899 OR x ≈ -7.899 (to 3 s.f.)

7. Coordinate Geometry

Line: y = 2x + 3

Curve: y = x² - 1

Solution:

Set equations equal: x² - 1 = 2x + 3

x² - 2x - 4 = 0

Using quadratic formula: a=1, b=-2, c=-4

x = [2 ± √(4 + 16)] / 2

x = [2 ± √20] / 2

x = [2 ± 2√5] / 2

x = 1 ± √5

When x = 1 + √5: y = 2(1 + √5) + 3 = 5 + 2√5

When x = 1 - √5: y = 2(1 - √5) + 3 = 5 - 2√5

A(1 + √5, 5 + 2√5)

B(1 - √5, 5 - 2√5)

Approximately: A(3.236, 9.472), B(-1.236, 0.528)

9. Solve Inequality

4m + 10 ≤ 9m - 8

Solution:

4m + 10 ≤ 9m - 8

10 + 8 ≤ 9m - 4m

18 ≤ 5m

5m ≥ 18

m ≥ 18/5 OR m ≥ 3.6

10. Integer Values

-5 < 3n ≤ 9

Solution:

Divide all parts by 3: -5/3 < n ≤ 3

-1.667 < n ≤ 3

Integer values: -1, 0, 1, 2, 3

11. Integer Values

12 ≤ 3n < 27

Solution:

Divide all parts by 3: 4 ≤ n < 9

Integer values: 4, 5, 6, 7, 8

12. Solve Equation

2/(x - 2) + 3/(x + 4) = 1

Solution:

Multiply through by (x-2)(x+4):

2(x+4) + 3(x-2) = (x-2)(x+4)

2x + 8 + 3x - 6 = x² + 4x - 2x - 8

5x + 2 = x² + 2x - 8

0 = x² + 2x - 8 - 5x - 2

0 = x² - 3x - 10

0 = (x - 5)(x + 2)

x = 5 OR x = -2

13. Solve Equation

(15 - 5x)/4 = 5 - x

Solution:

Multiply both sides by 4: 15 - 5x = 4(5 - x)

15 - 5x = 20 - 4x

15 - 20 = -4x + 5x

-5 = x

x = -5

14. Inequalities Region

Inequalities: y > 2, y < x + 4, x + 2y ≤ 6

Solution:

Step 1: Draw boundary lines:

y = 2 (horizontal line, dashed)
y = x + 4 (line with gradient 1, dashed)
x + 2y = 6 → y = (6 - x)/2 (solid line)

Step 2: Shade regions:

For y > 2: Shade ABOVE y = 2
For y < x + 4: Shade BELOW y = x + 4
For x + 2y ≤ 6: Shade BELOW y = (6 - x)/2

Step 3: The region R is where all three shaded areas overlap.

Key intersection points:

y = 2 and y = x + 4 → 2 = x + 4 → x = -2 → Point (-2, 2)
y = 2 and x + 2y = 6 → x + 4 = 6 → x = 2 → Point (2, 2)
y = x + 4 and x + 2y = 6 → x + 2(x + 4) = 6 → 3x + 8 = 6 → 3x = -2 → x = -2/3, y = 10/3

16. Cubic Graph

y = x³ - 3x + 1

(a) Complete the table:

x	-2	-1.5	-1	-0.5	0	0.5	1	1.5	2
y	-1	2.125	3	2.375	1	-0.375	-1	-0.125	3

(c) Solve x³ - 3x - 1 = 0

Given equation: x³ - 3x - 1 = 0

Rewrite as: x³ - 3x + 1 = 2

We need to solve: y = 2 where y = x³ - 3x + 1

Draw horizontal line y = 2 on the graph

Intersection points give solutions:

x ≈ -1.53, -0.35, 1.88 (read from graph)

17. Quadratic Graph

y = x² + 4x - 140

(a) Complete the table:

x	-20	-15	-10	-5	0	5	10	15
y	100	-35	-80	-135	-140	-95	0	85

18. Reciprocal Graph

y = -8/x

(a) Complete the table:

x	-8	-5	-4	-2	-1.6	-1	1	1.6	2	4	5	8
y	1	1.6	2	4	5	8	-8	-5	-4	-2	-1.6	-1

(c) Order of rotational symmetry: 2

(d) Equation of each line of symmetry:

y = x and y = -x

(f) Solve -8/x = 3:

-8/x = 3

3x = -8

x = -8/3 ≈ -2.667

Important Notes:

Graphical questions (14, 16b, 17b, 18b) require accurate plotting on graph paper.
For region questions, shading must be clearly shown.
Decimal answers should be given to 3 significant figures unless specified.
Always show working for method marks.

END OF SOLUTIONS

Year 10 Maths · Practice Paper (proper math formatting)

📐 TERM 2,MIRROR TEST 2 MARCH 2026

Cambridge International · End of Term 2 (Mock) 📅 March 2026 · Year 10

Student Name: _________________________

Admission No.: _________________________

Class: ___________

⏱️ 1 hour 📊 Total marks: 50 🧮 Calculator allowed 📏 Tracing paper ok

✏️ Answer all questions. Use black/dark blue pen. HB pencil for diagrams. Write answers in spaces provided.
Show all necessary working. Numerical answers: 3 significant figures / 1 decimal for angles. π = 3.142 or calculator value.

Area ½bh, πr², 2πrh, 4πr²

Volume A·l, ⅓Ah, πr²h, ⅓πr²h, &frac43;πr³

Quadratic x = [−b ± √(b²−4ac)] / 2a

Sine rule a/sin A = b/sin B = c/sin C

Cosine rule a² = b² + c² − 2bc cos A

Area Δ ½ab sin C

1 [2]

(a) Write down the inequality represented on the number line.

Inequality: .......................

(b) Write down the integer values of x that satisfy −4 < 2x ≤ 8.

Integers: .......................

2 [2]

y = 2x² − 8x + 5

(a) Find the value of dy/dx when x = 2.

.......................

(b) Find the coordinates of the point on the graph where the gradient is 4.

( ......... , ......... )

3 [2]

Given y = 5x^p − q x² and dy/dx = 20x³ + 8x.

Find the value of p and the value of q.

p = ......... , q = .........

4 [5]

A curve has equation y = x³ − 6x² + 9x.

(a) Work out the coordinates of the two turning points. (Show all steps clearly)

(b) Determine whether each turning point is a maximum or a minimum. Give reasons.

5 [5]

The table shows some values for y = x³ − 3x + 1. (Where appropriate values are correct to 2 decimal places)

x	−2	−1.5	−1	−0.5	0	0.5	1	1.5	2
y	−1	2.13			1	−0.38	−1		3

(a) Complete the table. (values for x = −1, −0.5, 1.5)

x=−1: ...... , x=−0.5: ...... , x=1.5: ......

(b) Draw the graph of y = x³ − 3x + 1 for −2 ≤ x ≤ 2.

x ≈ ......... , ......... , .........

6 [7]

f(x) = 3x, g(x) = 2x + 5, h(x) = x² − 2

(a) Find f(5). (1)

...............

(b) Find g(3x). (1)

...............

(d) Find h(2). (1)

...............

(e) Simplify g(x) + h(x). (1)

...............

(f) Find h(1 − x). (1)

...............

7 [2]

The diagram shows a curve. By drawing a suitable tangent, estimate the gradient of the curve at point P.

Estimated gradient = .......................

8 [2]

y = 2x³ + 3x² − 12x

(a) Find dy/dx.

.......................

(b) Find the gradient of the curve at the point where x = 1.

.......................

9 [2]

On the axes below, sketch the graphs of:

(a) y = 1/x (b) y = 3^−x

(sketch both on same axes)

Practice Exam · SOLUTIONS · Year 10 Maths

📘 SOLUTIONS · Year 10 Practice Exam

Cambridge style · Term 2 (Mock) · 50 marks

1 [2]

(a) Number line shows open circle at 1, closed circle at 4.
Inequality: 1 < x ≤ 4.

(b) −4 < 2x ≤ 8 → divide by 2: −2 < x ≤ 4.
Integer values: −1, 0, 1, 2, 3, 4.

1(a) 1 < x ≤ 4 1(b) −1,0,1,2,3,4

2 [2]

y = 2x² − 8x + 5

(a) dy/dx = 4x − 8.
At x = 2: 4(2)−8 = 0.

(b) Gradient = 4 → 4x − 8 = 4 → 4x = 12 → x = 3.
y = 2(3)² − 8(3) + 5 = 18 − 24 + 5 = −1.
Coordinates: (3, −1).

2(a) 0 2(b) (3,−1)

3 [2]

y = 5x^p − q x²
dy/dx = 5p·x^p−1 − 2q x

Given dy/dx = 20x³ + 8x.
Compare coefficients:
For x³ term: p−1 = 3 → p = 4.
Then 5p = 5·4 = 20, matches.

For x term: −2q = 8 → q = −4.

Thus p = 4, q = −4.

p = 4 , q = −4

4 [5]

y = x³ − 6x² + 9x

(a) dy/dx = 3x² − 12x + 9 = 0
Divide 3: x² − 4x + 3 = 0 → (x−1)(x−3)=0
x = 1 or x = 3.

y(1) = 1 − 6 + 9 = 4 → (1,4)
y(3) = 27 − 54 + 27 = 0 → (3,0)

(b) d²y/dx² = 6x − 12
At x = 1: 6−12 = −6 < 0 → maximum.
At x = 3: 18−12 = 6 > 0 → minimum.

Reasons: second derivative negative → max; positive → min.

Turning points: (1,4) max, (3,0) min

5 [5]

y = x³ − 3x + 1

(a) Complete table:

x	−2	−1.5	−1	−0.5	0	0.5	1	1.5	2
y	−1	2.13	3	2.38	1	−0.38	−1	−0.13	3

Calculations: x=−1: (−1)³−3(−1)+1 = −1+3+1=3.
x=−0.5: −0.125+1.5+1=2.375 ≈ 2.38.
x=1.5: 3.375−4.5+1=−0.125 ≈ −0.13.

(c) Equation x³ − 3.5x + 1 = 0 can be written as x³ − 3x + 1 = 0.5x.
Draw line y = 0.5x on the graph. Intersections give roots.

Observe x = −2 is a root: (−2)³ −3.5(−2)+1 = −8+7+1=0.
Factor (x+2): (x+2)(x² −2x +0.5)=0 → quadratic gives x = 1 ± √0.5 ≈ 1 ± 0.707.
So solutions: x ≈ −2, 0.29, 1.71 (to 2 d.p.)

5(a) 3, 2.38, −0.13 5(c) −2, 0.29, 1.71

6 [7]

f(x)=3x, g(x)=2x+5, h(x)=x²−2

(a) f(5) = 3·5 = 15

(b) g(3x) = 2(3x)+5 = 6x+5

(d) h(2) = 2²−2 = 4−2 = 2

(e) g(x)+h(x) = (2x+5)+(x²−2) = x² + 2x + 3

(f) h(1−x) = (1−x)² − 2 = (1 −2x + x²) − 2 = x² − 2x − 1

(a)15 (b)6x+5 (c)(x−5)/2 (d)2 (e)x²+2x+3 (f)x²−2x−1

7 [2]

From the curve at point P (approx x=1), a tangent drawn would have a negative slope.
Using two points on the tangent (estimate), gradient ≈ −1.8 (answers in range −1.5 to −2.2 are acceptable).

Estimated gradient ≈ −1.8

8 [2]

y = 2x³ + 3x² − 12x

(a) dy/dx = 6x² + 6x − 12

(b) At x=1: 6(1)² + 6(1) −12 = 6+6−12 = 0

8(a) 6x²+6x−12 8(b) 0

9 [2]

(a) y = 1/x : rectangular hyperbola in first and third quadrants, asymptotes x=0, y=0.

(b) y = 3^−x : exponential decay, passes through (0,1), decreases to 0 as x→∞, increases as x→−∞ (approaches ∞).

Both curves sketched on same axes (indicate asymptotes).

Sketch: reciprocal and exponential decay.

Support notes · Year 10 Mathematics · Term 2

📘 Support notes · Year 10 Maths

Cambridge · Term 2 · Key topics & exam tips

1. Inequalities

Solving linear inequalities is just like solving equations, but if you multiply or divide by a negative number, reverse the inequality sign.

Double inequalities (e.g. −4 < 2x ≤ 8) are solved by doing the same operation to all three parts.

Integer solutions are whole numbers that satisfy the inequality. Always list them in order.

Number lines use open circles (○) for < or > and closed circles (●) for ≤ or ≥.

✏️ Example

Solve −4 < 2x ≤ 8.
Divide by 2: −2 < x ≤ 4.
Integer values: −1, 0, 1, 2, 3, 4.

Inequality on a number line: open circle at −2? Wait, careful: −2 is not included because it's −2 < x. So open circle at −2, closed circle at 4.

2. Differentiation (gradients)

If y = axⁿ then dy/dx = n·a·xⁿ⁻¹. Differentiate term by term.

Gradient at a point – substitute x into dy/dx.

Find where gradient equals a given value – set dy/dx = value, solve for x, then find y.

✏️ Example

y = 2x² − 8x + 5
dy/dx = 4x − 8.
At x = 2, gradient = 4(2)−8 = 0.
Find point where gradient = 4: 4x−8 = 4 → x = 3. Then y = 2(9)−24+5 = −1 → point (3,−1).

3. Reverse differentiation

Given dy/dx and the original form (with unknown powers or coefficients), differentiate the general expression and compare with the given derivative.

Match coefficients for each power of x to solve for unknowns.

✏️ Example

If y = 5xᵖ − qx² and dy/dx = 20x³ + 8x, then dy/dx = 5p·xᵖ⁻¹ − 2q·x.
For x³ term: p−1 = 3 → p=4, and 5p=20 ok.
For x term: −2q = 8 → q = −4.

4. Turning points (stationary points)

Find stationary points: set dy/dx = 0, solve for x, then find y.

Determine nature: use second derivative d²y/dx².

If d²y/dx² < 0 → maximum.
If d²y/dx² > 0 → minimum.
If d²y/dx² = 0 → use sign table (first derivative test).

✏️ Example

y = x³ − 6x² + 9x
dy/dx = 3x² − 12x + 9 = 0 → divide 3: x² − 4x + 3 = 0 → (x−1)(x−3)=0 → x=1 or 3.
y(1)=4, y(3)=0.
d²y/dx² = 6x−12. At x=1: −6 → max. At x=3: +6 → min.

5. Cubic graphs & graphical solution

Complete a table of values (use calculator carefully, round to 2 decimal places). Plot points and draw a smooth curve.

Solving equations graphically: rearrange equation into form f(x) = g(x). Draw y = f(x) (the cubic) and y = g(x) (straight line). The x-coordinates of intersection are the solutions.

✏️ Example

To solve x³ − 3.5x + 1 = 0, rewrite as x³ − 3x + 1 = 0.5x. Draw y = x³ − 3x + 1 and y = 0.5x. Read off intersection points.

6. Functions

f(5) means substitute 5 for x.

g(3x) means replace x with 3x in the expression for g(x).

Inverse function g⁻¹(x): swap x and y, then solve for y.

Simplify g(x)+h(x): add the expressions and collect like terms.

h(1−x): substitute (1−x) into h.

✏️ Example

f(x)=3x, g(x)=2x+5, h(x)=x²−2
f(5)=15; g(3x)=6x+5; g⁻¹(x)=(x−5)/2; g(x)+h(x)=x²+2x+3; h(1−x)=x²−2x−1.

7. Gradient by drawing a tangent

At a point on a curve, draw a straight line that just touches the curve (the tangent). Choose two points on this line with integer coordinates if possible, then calculate gradient = (change in y)/(change in x).

Estimate should be reasonable – expect 1 decimal place accuracy.

✏️ Example

On a given curve, tangent at P passes through (1,3) and (3,−1). Gradient = (−1−3)/(3−1) = −4/2 = −2.

8. Graph sketches: reciprocal & exponential

y = 1/x – two branches in 1st and 3rd quadrants. Asymptotes: x=0 (vertical) and y=0 (horizontal). Passes through (1,1) and (−1,−1).

y = a⁻ˣ (a>0) – exponential decay. Passes through (0,1). As x→∞, y→0. As x→−∞, y→∞. Always above x-axis.

✏️ Example

Sketch y = 2⁻ˣ: through (0,1), (1,0.5), (−1,2). Decays to right, rises to left.

📌 Exam tips:

Show all working – marks are given for clear steps.
For turning points, state nature with reason (second derivative or sign change).
When drawing graphs, use a smooth curve, not straight line segments.
For tangents, choose points far apart for better accuracy.
Check calculator mode (degrees for trig).
Use 3 significant figures unless told otherwise.

Mirror Test 3 · Year 10 Maths Practice

📐 OSHWAL ACADEMY MOMBASA – MIRROR TEST 3

Cambridge International · End of Term 2 (Practice) 📅 March 2026 · Year 10

Student Name: _________________________

Admission No.: _________________________

Class: ___________

⏱️ 1 hour 📊 Total marks: 50 🧮 Calculator allowed 📏 Tracing paper ok

Area ½bh, πr², 2πrh, 4πr²

Volume A·l, ⅓Ah, πr²h, ⅓πr²h, &frac43;πr³

Quadratic x = [−b ± √(b²−4ac)] / 2a

Sine rule a/sin A = b/sin B = c/sin C

Cosine rule a² = b² + c² − 2bc cos A

Area Δ ½ab sin C

1 [2]

(a) Write down the inequality represented on the number line.

Inequality: .......................

(b) Write down the integer values of x that satisfy −6 < 3x ≤ 9.

Integers: .......................

2 [2]

y = 4x² − 10x + 3

(a) Find the value of dy/dx when x = 1.

.......................

(b) Find the coordinates of the point on the graph where the gradient is 6.

( ......... , ......... )

3 [2]

Given y = 2x^p − q x³ and dy/dx = 8x⁵ + 12x².

Find the value of p and the value of q.

p = ......... , q = .........

4 [5]

A curve has equation y = 2x³ − 9x² + 12x.

(a) Work out the coordinates of the two turning points. (Show all steps clearly)

(b) Determine whether each turning point is a maximum or a minimum. Give reasons.

5 [5]

The table shows some values for y = x³ − 4x + 2. (Values to 2 decimal places where needed)

x	−2	−1.5	−1	−0.5	0	0.5	1	1.5	2
y	2	4.63			2	0.13	−1		2

(a) Complete the table. (values for x = −1, −0.5, 1.5)

x=−1: ...... , x=−0.5: ...... , x=1.5: ......

(b) Draw the graph of y = x³ − 4x + 2 for −2 ≤ x ≤ 2.

x ≈ ......... , ......... , .........

6 [7]

f(x) = 4x − 1, g(x) = 3 − 2x, h(x) = x² + 3

(a) Find f(−2). (1)

...............

(b) Find g(2x). (1)

...............

(d) Find h(−1). (1)

...............

(e) Simplify f(x) − g(x). (1)

...............

(f) Find h(x + 1). (1)

...............

7 [2]

The diagram shows a curve. By drawing a suitable tangent, estimate the gradient of the curve at point P.

Estimated gradient = .......................

8 [2]

y = 5x³ − 4x² + 7x

(a) Find dy/dx.

.......................

(b) Find the gradient of the curve at the point where x = 2.

.......................

9 [2]

On the axes below, sketch the graphs of:

(a) y = 2/x (b) y = 4^−x

(sketch both on same axes)

Y10 Mathematics Exam Solutions | Oshwal Academy

📘 Year 10 Mathematics – End of Term 1 Solutions (March 2026)

Oshwal Academy Mombasa · Cambridge IGCSE style · Learner-friendly steps

🔹 Question 1 – Inequalities

1(a) Write the inequality from the diagram [2 marks]

Open circle at \(-3\) → \(x > -3\)
Closed circle at \(2\) → \(x \le 2\)

\[ \boxed{-3 < x \le 2} \]

1(b) Integer values for \(-4 < 2x \le 8\) [2 marks]

Divide by 2: \(-2 < x \le 4\)

Integers greater than \(-2\) and ≤ \(4\): \(-1, 0, 1, 2, 3, 4\)

\[ \boxed{-1,\; 0,\; 1,\; 2,\; 3,\; 4} \]

🔹 Question 2 – Differentiation (gradient & coordinates)

2(a) Find \(\frac{dy}{dx}\) when \(x=3\) for \(y = 3x^{2} -12x +7\) [3 marks]

\(\frac{dy}{dx}=6x -12\)

Substitute \(x=3\): \(6(3)-12 = 18-12 = 6\)

\[ \boxed{6} \]

2(b) Coordinates where gradient = 1 [2 marks]

Set \(6x-12 = 1\) → \(6x = 13\) → \(x = \frac{13}{6}\)

\(y = 3\left(\frac{13}{6}\right)^2 -12\left(\frac{13}{6}\right)+7 = -\frac{59}{12}\)

\[ \boxed{\left(\frac{13}{6},\;-\frac{59}{12}\right)} \quad \text{or} \quad (2.17,\;-4.92) \]

🔹 Question 3 – Find p and q

Given \(y = 2x^{p} - qx^{2}\) and \(\frac{dy}{dx}=14x^{6}+6x\)

\(\frac{dy}{dx}=2p\,x^{p-1} - 2q\,x\)

Compare: \(2p\,x^{p-1}=14x^{6}\) → \(p-1=6\) → \(p=7\)

\(-2q\,x = 6x\) → \(-2q=6\) → \(q=-3\)

\[ \boxed{p=7,\quad q=-3} \]

🔹 Question 4 – Turning points & nature

4(a) Coordinates of the two turning points (6 marks)

\(y = x^{3}+8x^{2}+5x\)
\(\frac{dy}{dx}=3x^{2}+16x+5\)

Set \(\frac{dy}{dx}=0\): \(3x^{2}+16x+5=0\)

Quadratic formula: \(x = \frac{-16 \pm \sqrt{256-60}}{6} = \frac{-16 \pm 14}{6}\)

\(x = -5\) or \(x = -\frac{1}{3}\)

\(x=-5\) → \(y = (-5)^3+8(25)+5(-5)= -125+200-25 = 50\)

\(x=-\frac13\) → \(y = -\frac{1}{27}+\frac{8}{9}-\frac{5}{3} = -\frac{22}{27}\)

\[ \boxed{(-5,\;50)\;\text{and}\;\left(-\frac13,\;-\frac{22}{27}\right)} \]

4(b) Max or min? (3 marks)

Second derivative: \(\frac{d^{2}y}{dx^{2}} = 6x+16\)

At \(x=-5\): \(6(-5)+16 = -14 <0\) → maximum

At \(x=-\frac13\): \(6(-\frac13)+16 = 14 >0\) → minimum

\[ (-5,50)\text{ max},\quad \left(-\frac13,-\frac{22}{27}\right)\text{ min} \]

🔹 Question 5 – Cubic graph & solving graphically

5(a) Complete table [2 marks]

\(y = x^{3} - 2x + 3\)
\(x=-1.5\): \((-1.5)^3 -2(-1.5)+3 = -3.375+3+3 = 2.625 \approx 2.63\)

\(x=2\): \(8 - 4 + 3 = 7\)

\[ \boxed{x=-1.5 \to y=2.63,\quad x=2 \to y=7} \]

5(b) Draw the graph (4 marks)

Plot points: \((-2,-1),\; (-1.5,2.63),\; (-1,3),\; (-0.5,3.88),\; (0,3),\; (0.5,2.13),\; (1,2),\; (1.5,3.38),\; (2,7)\).
Draw a smooth curve through them.

5(c) Solve \(x^{3} - 2.5x + 1 = 0\) using a suitable line [4 marks]

Rewrite to match the curve: \(x^{3}-2x+3 = 2.5x+2\)

Draw line \(y = 2.5x+2\) on same grid (through \((0,2)\) and \((2,7)\)).

Intersection x‑coordinates ≈ \(-1.3,\;0.6,\;1.7\)

\[ \boxed{x \approx -1.3,\;0.6,\;1.7} \]

🔹 Question 6 – Function notation & operations

Part	Working	Answer
6(a) \(f(5)\)	\(f(x)=5^x\) → \(5^5\)	\(\boxed{3125}\)
6(b) \(g(-2x)\)	\(g(x)=3x-2\) → \(3(-2x)-2\)	\(\boxed{-6x-2}\)
6(c) \(g^{-1}(x)\)	Swap \(x=3y-2\) → \(y=\frac{x+2}{3}\)	\(\boxed{\frac{x+2}{3}}\)
6(d) \(h(1)\)	\(h(x)=x^2+1\) → \(1^2+1\)	\(\boxed{2}\)
6(e) \(g(x)+h(x)\)	\((3x-2)+(x^2+1)\)	\(\boxed{x^2+3x-1}\)
6(f) \(h(2-x)\)	\((2-x)^2+1 = x^2-4x+4+1\)	\(\boxed{x^2-4x+5}\)

🔹 Question 7 – Estimating gradient with a tangent

At point \(P\) on the curve, draw a tangent line (touches the curve only at \(P\)).

From the graph, the tangent passes approximately through \((1,2)\) and \((5,0)\).

Gradient \(m = \frac{0-2}{5-1} = \frac{-2}{4} = -0.5\)

\[ \boxed{-0.5} \]

🔹 Question 8 – Derivative & gradient at given x

8(a) Find \(\frac{dy}{dx}\) for \(y = x^{3} + 3x^{2} -13x\) [2 marks]

\(\frac{dy}{dx}=3x^{2} + 6x -13\)

\[ \boxed{3x^{2}+6x-13} \]

8(b) Gradient at \(x=3\) [2 marks]

\(3(3)^{2}+6(3)-13 = 27+18-13 = 32\)

\[ \boxed{32} \]

🔹 Question 9 – Sketching graphs

9(a) \(y = \frac{1}{x}\) [2 marks]

Vertical asymptote \(x=0\), horizontal asymptote \(y=0\).
Two branches: Quadrant I (both positive) and Quadrant III (both negative).
Curves never touch the axes.

9(b) \(y = 2^{-x}\) [2 marks]

Exponential decay; passes through \((0,1)\).
Horizontal asymptote \(y=0\) as \(x\to +\infty\).
As \(x\to -\infty\), \(y\to +\infty\).

📌 Quick Answer Summary

Question	Answer
1(a)	\(-3 < x \le 2\)
1(b)	\(-1,0,1,2,3,4\)
2(a)	\(6\)
2(b)	\(\left(\frac{13}{6},-\frac{59}{12}\right)\)
3	\(p=7,\; q=-3\)
4(a)	\((-5,50),\;\left(-\frac13,-\frac{22}{27}\right)\)
4(b)	\((-5,50)\) max, \(\left(-\frac13,-\frac{22}{27}\right)\) min
5(a)	\(x=-1.5\to 2.63,\; x=2\to 7\)
5(c)	\(x \approx -1.3,\;0.6,\;1.7\)
6(a)	\(3125\)
6(b)	\(-6x-2\)
6(c)	\(\frac{x+2}{3}\)
6(d)	\(2\)
6(e)	\(x^{2}+3x-1\)
6(f)	\(x^{2}-4x+5\)
7	\(-0.5\) (tangent gradient)
8(a)	\(3x^{2}+6x-13\)
8(b)	\(32\)

✅ All steps are aligned with the corrected marking scheme. Gradient in Question 7 is confirmed as \(-0.5\).