Scholar Tests · Sample paper
Secondary 3 Additional Mathematics, Paper 1
25 questions · 90 marks · 135 minutes · no negative marking
Name ______________________ Date ____________
Paper 1: multiple choice (calculator allowed)
Questions 1 to 10, 3 marks each. Circle the letter of the best answer.
- 1.
Express x2 + 12x + 42 in the form a(x − h)2 + k.
- A(x + 6)2 + 6
- B(x + 6)2 + 78
- C(x + 6)2 − 6
- D(x − 6)2 + 6
- 2.
Given that (−4x − 9)/((x + 2)(x + 3)) = A/(x + 2) + B/(x + 3), find the value of A.
- A1
- B−3
- C3
- D−1
- 3.
The vertices of triangle ABC are A(−5, −4), B(4, 2) and C(6, 7). Find the area of ABC, in square units.
- A8.25
- B16.5
- C121
- D33
- 4.
Find the coefficient of x2 in the expansion of (1 − 2x)4.
- A−12
- B−24
- C24
- D−32
- 5.
Solve the equation x3 − 6x2 + 5x = 0.
- Ax = −5, −1, 0
- Bx = 1, 5
- Cx = 0, 1
- Dx = 0, 1, 5
- 6.
The population, P, of a town after t years is given by P = 62505e0.1t. Find the time taken for P to rise to 312525, correct to 3 significant figures.
- A50.0 years
- B6.99 years
- C1.61 years
- D16.1 years
- 7.
Solve x2 + 4x < 0.
- A−4 < x < 0
- Bx < −4 or x > 0
- C−4 ≤ x ≤ 0
- D0 < x < 4
- 8.
Find the radius of the circle x2 + y2 + 8x + 4y + 10 = 0, correct to 3 significant figures.
- A5.48
- B10
- C8.37
- D3.16
- 9.
Solve the equation 4x − 24(2x) + 128 = 0.
- Ax = 3 or x = 4
- Bx = 4 only
- Cx = 8 or x = 16
- Dx = −3 or x = −4
- 10.
Find the range of values of k for which the line y = −x + k does not meet the curve y = x2 − 7x + 6.
- Ak < −3
- Bk > −3
- Ck ≤ −3
- Dk < 15
Paper 1: short answers
Questions 11 to 25, 4 marks each. Write your answer on the line.
- 11.
Find the term independent of x in the expansion of (x2 − 2/x2)4.
Answer: ______________
- 12.
The line y = −4x + 34 meets the curve x2 + y2 = 85 at the points A and B. Find the length of AB, correct to 3 significant figures.
Answer: ______________
- 13.
Find the remainder when x3 − 8x2 + 19x − 9 is divided by 2x + 3.
Answer: ______________
- 14.
The variables x and y are related by y = kbx, where k and b are constants. When lg y is plotted against x, a straight line passes through the points (1, 0.75) and (7, 2.25). Find the value of b, correct to 3 significant figures.
Answer: ______________
- 15.
The point P(−5, 2) lies on the circle (x + 4)2 + (y + 1)2 = 10. The tangent to the circle at P meets the y-axis at Q. Find the y-coordinate of Q.
Answer: ______________
- 16.
By completing the square, find the maximum value of y = −3x2 − 30x − 63.
Answer: ______________
- 17.
The line y = −3x + c is a tangent to the curve y = x2 − 9x + 4. Find the value of c.
Answer: ______________
- 18.
The vertices of triangle ABC are A(−1, −5), B(−1, 1) and C(−5, 3). Find the area of ABC, in square units.
Answer: ______________
- 19.
Given that (−x2 + 3)/((x + 1)x2) = A/(x + 1) + B/x + C/x2, find the value of C.
Answer: ______________
- 20.
Find the number of integer values of x for which x2 − 5x + 6 ≤ 0.
Answer: ______________
- 21.
The value, V dollars, of a machine after t years is given by V = 43148e−0.03t. Find the time taken for V to fall to 8630, correct to 3 significant figures. Give your answer in years.
Answer: ______________
- 22.
Solve the equation √(−3x + 40) = x.
Answer: ______________
- 23.
Find the term independent of x in the expansion of (x2 − 1/x)9.
Answer: ______________
- 24.
Solve the equation 2x + 3 = 59, giving your answer correct to 3 significant figures.
Answer: ______________
- 25.
The line y = 4x + c is a tangent to the curve y = x2 + 6x + 5. Find the value of c.
Answer: ______________
Answers and working
Paper 1: multiple choice (calculator allowed)
- 1.A ((x + 6)2 + 6). Halve the x-coefficient: x2 + 12x = (x + 6)2 − 36. Complete the square inside and tidy the constant: (x + 6)2 + 6. The answer is (x + 6)2 + 6.
- 2.D (−1). Multiply up: −4x − 9 = A(x + 3) + B(x + 2). Put x = −2: −1 = A(1), so A = −1. The answer is −1.
- 3.B (16.5). List the vertices in order, back to the first, and use the shoelace formula: ½|Σ(x1y2 − x2y1)|. The sum is 33, so the area is 16.5. The answer is 16.5.
- 4.C (24). The term in x2 is 4C2 × 12 × (−2x)2. 4C2 = 6, so the coefficient is 24. The answer is 24.
- 5.D (x = 0, 1, 5). Try small whole numbers: f(1) = 0, so (x − 1) is a factor. Divide to leave a quadratic, then factorise it: the roots are 0, 1, 5. The answer is x = 0, 1, 5.
- 6.D (16.1 years). 312525 = 62505e0.1t, so e0.1t = 5.000. Take ln: 0.1t = ln 5.000, so t = 16.1. The answer is 16.1 years.
- 7.A (−4 < x < 0). Factorise: x2 + 4x = (x + 4)x, with roots −4 and 0. The curve is U-shaped, so it is below the x-axis between the roots: −4 < x < 0. The answer is −4 < x < 0.
- 8.D (3.16). g = 4, f = 2 and c = 10. r = √(g2 + f2 − c) = √10. The answer is 3.16.
- 9.A (x = 3 or x = 4). Let y = 2x, so 4x = y2: y2 − 24y + 128 = 0. (y − 8)(y − 16) = 0, so 2x = 8 or 16. x = 3 or x = 4. The answer is x = 3 or x = 4.
- 10.A (k < −3). Put the line into the curve: x2 − 6x + (6 − k) = 0. b2 − 4ac = 36 − 4(6 − k), which must be negative: k < −3. The answer is k < −3.
Paper 1: short answers
- 11.24. The general term is 4C r × (x2)n − r × (−2/x2)r. The powers of x cancel when 2(4 − r) = 2r, so r = 2. The term is 4C2 × 12 × (−2)2 = 24. The answer is 24.
- 12.8.25. Solving the two equations gives A(7, 6) and B(9, −2). AB = √(22 + (−8)2) = 8.25. The answer is 8.25.
- 13.−471/8. By the remainder theorem, the remainder is f(−3/2), where 2x + 3 = 0. f(−3/2) = −471/8. The answer is −471/8.
- 14.1.78. Take lg: lg y = (lg b)x + lg k, a straight line with gradient lg b and intercept lg k. Gradient lg b = 0.25 and intercept lg k = 0.5; undo each lg with 10 to the power. The answer is 1.78.
- 15.11/3. The radius to P has gradient −3, so the tangent has gradient 1/3. The tangent is x − 3y = −11; put x = 0: y = 11/3. The answer is 11/3.
- 16.12. −3x2 − 30x − 63 = −3(x + 5)2 + 12. The squared bracket is never negative, and it is subtracted, so y is greatest when x = −5, where y = 12. The answer is 12.
- 17.−5. Put the line into the curve: x2 − 6x + (4 − c) = 0. A tangent gives equal roots: (−6)2 − 4(4 − c) = 0, so c = −5. The answer is −5.
- 18.12. List the vertices in order, back to the first, and use the shoelace formula: ½|Σ(x1y2 − x2y1)|. The sum is 24, so the area is 12. The answer is 12.
- 19.3. Multiply up: −x2 + 3 = Ax2 + B(x + 1)x + C(x + 1). Put x = 0: the first two terms vanish, leaving C = 3. The answer is 3.
- 20.2. Factorise: the roots are x = 2 and x = 3. The curve is U-shaped, so the inequality holds for 2 ≤ x ≤ 3. Count the whole numbers in that range: 2. The answer is 2.
- 21.53.6. 8630 = 43148e−0.03t, so e−0.03t = 0.2000. Take ln: −0.03t = ln 0.2000, so t = 53.6. The answer is 53.6 years.
- 22.5. Square both sides: −3x + 40 = x2, which gives x = 5 or x = −8. Check each in the original: x = −8 makes the right side negative, so only x = 5 works. The answer is 5.
- 23.84. The general term is 9C r × (x2)n − r × (−1/x)r. The powers of x cancel when 2(9 − r) = r, so r = 6. The term is 9C6 × 13 × (−1)6 = 84. The answer is 84.
- 24.2.88. Take lg of both sides: (x + 3) lg 2 = lg 59. x = lg 59 ÷ lg 2 − 3 = 2.88. The answer is 2.88.
- 25.4. Put the line into the curve: x2 + 2x + (5 − c) = 0. A tangent gives equal roots: (2)2 − 4(5 − c) = 0, so c = 4. The answer is 4.
Written by Scholar Tests to SEAB's SEC G3 Additional Mathematics syllabus. Not a school paper. Scholar Tests is not connected with MOE, SEAB or any school.