Mathbench

IEB mock Paper I (second paper)

A complete 150-mark practice paper: 11 questions in 49 parts, 30 topics, Section A 70 marks and Section B 80. Every question was generated for this page and every answer was independently verified.

The split between the sections, and how much of the paper each topic is worth, were measured from 282 real IEB questions — not guessed. None of those questions appears here.

All IEB practice · Sit this paper under time · Afrikaans

The paper

Section A[70 marks]

1[13 marks]
  1. (a)
    Given f(x) = x^2 − 2x − 5. Determine (f(x + h) − f(x))/h, the average rate of change of f over the interval [x; x + h], in its simplest form.
    (3)
  2. (b)
    Hence determine f′(x).
    (2)
  3. (c)
    The line y = 4x + k is a tangent to the graph of g(x) = −x^2 + 2x − 1. Determine the coordinates of the point where the line touches g, and the value of k.
    (5)
  4. (d)
    At which point on f is the tangent parallel to the line y = −8x?
    (3)
2[13 marks]

Given: f(x) = −4 − 8/(x − 4)

xyO
  1. (a)
    Determine f(0).
    (1)
  2. (b)
    Determine the value of x for which f(x) = 0.
    (2)
  3. (c)
    Sketch the graph of f on the axes provided. Label clearly all asymptotes and intercepts with the axes.
    (5)
  4. (d)
    Write down the domain of f.
    (1)
  5. (e)
    For which values of x is f(x) > 0?
    (2)
  6. (f)
    The graph of f is shifted 3 units to the right and 2 units up to form g. Write down the equation of g.
    (2)
3[14 marks]
  1. (a)
    Solve for x: 3^(2x) − 10(3^x) − 24 = 0
    (4)
  2. (b)
    Solve for x: 3^(x + 1) + 3^(x − 1) = 30
    (3)
  3. (c)
    If log 2 = p and log 3 = q, determine log 6 in terms of p and q.
    (2)
  4. (d)
    Solve for x: x − √(x + 14) = −2
    (5)
4[15 marks]
  1. (a)
    Solve for x and y if 2^(x + y) = 2 and 3^(x − y) = 1/27.
    (5)
  2. (b)
    For which values of k will the line y = −2x + k never meet the parabola y = x^2 + 3x − 4?
    (4)
  3. (c)
    A bus and a car travel the same route of 75 km. The car travels 25 km/h faster than the bus and arrives half an hour earlier. Determine the average speed of the bus.
    (6)
5[15 marks]

A home loan of R1 250 000 is repaid in equal monthly payments over 20 years. The interest rate is 11% per annum compounded monthly, and the first payment is made one month after the loan is granted. Give answers correct to two decimal places.

  1. (a)
    Calculate the minimum monthly repayment needed to pay off the loan.
    (4)
  2. (b)
    Calculate the balance outstanding at the end of year 14, immediately after the payment at that time.
    (4)
  3. (c)
    Determine the effective annual interest rate of the loan.
    (3)
  4. (d)
    If you pay R16 800 per month instead, how many payments will it take to pay off the loan? (The last payment may be smaller.)
    (4)

Section B[80 marks]

6[9 marks]

A rectangular garden is built against a straight wall. The other three sides are fenced with 56 m of fencing. The two sides at right angles to the wall are each x m long.

xx
  1. (a)
    Show that the area of the garden is A = 56x − 2x^2.
    (2)
  2. (b)
    Determine the dimensions of the garden with the largest area, and that area.
    (5)
  3. (c)
    Determine the rate at which the area changes with respect to x when x = 19.
    (2)
7[12 marks]

The graph of y = f′(x), the derivative of a cubic function f, is drawn below. It touches the x-axis at C(4; 0) and cuts the y-axis at (0; 16).

xyOC(4; 0)(0; 16)f′
  1. (a)
    Determine the equation of f′.
    (3)
  2. (b)
    Given that f(0) = 6, determine f(x).
    (4)
  3. (c)
    Is x = 4 a turning point of f, or a point of inflection? Give a reason.
    (2)
  4. (d)
    Determine the x-values at which the tangent to f has a gradient of 9.
    (3)
8[13 marks]
  1. (a)
    Solve for x: (3x − 5)(x^2 + 8x + 16) = 0
    (3)
  2. (b)
    Use the quadratic formula to solve for x, correct to two decimal places, if 3x^2 + 5x = 1. Show all working.
    (4)
  3. (c)
    Hence, solve for x if 3x^2 + 5x − 1 > 0.
    (2)
  4. (d)
    Determine the value(s) of p for which the roots of x(x − 10) = −3p are real and equal.
    (4)
9[13 marks]

In the diagram below, f(x) = b^x and g, the reflection of f in the line y = x, are drawn. P(1; 1/3) lies on f.

xyOy = xP(1; 1/3)ABfg
  1. (a)
    Determine the value of b.
    (2)
  2. (b)
    Write down the equation of g in the form y = …
    (2)
  3. (c)
    Write down the domain of g.
    (1)
  4. (d)
    For which values of x is g(x) ≥ 1?
    (2)
  5. (e)
    The line x = 1 meets f at A and g at B. Determine the length of AB.
    (3)
  6. (f)
    h(x) = f(x) − 2. Write down the equation of the asymptote of h and determine the x-intercept of h, correct to two decimal places.
    (3)
10[16 marks]
  1. (a)
    In how many different ways can the letters of the word LETTERS be arranged?
    (3)
  2. (b)
    The letters of LETTERS are arranged at random. What is the probability that the arrangement starts with T and ends with S?
    (3)
  3. (c)
    6 different cars are arranged in a row of parking bays. In how many ways can this be done?
    (1)
  4. (d)
    In how many ways can this be done if 2 particular cars must be next to each other, in any order?
    (3)
  5. (e)
    The digits 0; 4; 5; 7; 8 are used to form 5-digit numbers, without repetition (a number cannot start with 0). How many such numbers can be formed?
    (2)
  6. (f)
    How many of these numbers are even and greater than 80000?
    (4)
11[17 marks]
  1. (a)
    3x − 14; 5x − 24 and 2x + 6 are the first three terms of an arithmetic sequence. Determine the value of x and the three terms.
    (3)
  2. (b)
    Which term of this sequence is equal to 220?
    (3)
  3. (c)
    How many terms of this sequence must be added to give a sum of 598?
    (4)
  4. (d)
    A pattern of dots is built picture by picture. Pictures 1 to 4 have 7; 17; 33; 55 dots, and the numbers of dots form a quadratic sequence. How many dots are in picture 20?
    (4)
  5. (e)
    Which picture has 6123 dots?
    (3)

Memorandum

Every answer below was confirmed by an independent verifier at the moment the question was made.

1

(a) Answer: 2x + h − 2

  1. f(x + h) = (x + h)^2 − 2(x + h) − 5
  2. f(x + h) − f(x) = 2xh + h^2 − 2h
  3. (2xh + h^2 − 2h)/h = 2x + h − 2

(b) Answer: f′(x) = 2x − 2

  1. f′(x) = lim_(h → 0) (2x + h − 2)
  2. = 2x − 2

(c) Answer: (−1; −4); k = 0

  1. g′(x) = −2x + 2 = 4
  2. x = −1
  3. g(−1) = −4, so the point is (−1; −4)
  4. −4 = 4(−1) + k, so k = 0

(d) Answer: (−3; 10)

  1. f′(x) = 2x − 2 = −8
  2. x = −3
  3. f(−3) = 10

The derivative from first principles

2

(a) Answer: f(0) = −2

  1. f(0) = −8/(−4) − 4 = −2

(b) Answer: x = 2

  1. −8/(x − 4) = 4
  2. −8 = 4(x − 4)
  3. x = 2

(c) Answer: See the sketch below.

xyOx = 4y = −4(0; −2)(2; 0)f
  1. asymptotes: x = 4 and y = −4
  2. y-intercept (0; −2); x-intercept (2; 0)

(d) Answer: x ∈ ℝ, x ≠ 4

  1. f is undefined only at x = 4

(e) Answer: 2 < x < 4

  1. read from the graph: the x-intercept is x = 2 and the asymptote x = 4 is never included
  2. 2 < x < 4

(f) Answer: g(x) = −8/(x − 7) − 2

  1. g(x) = f(x − 3) + 2
  2. g(x) = −8/(x − 7) − 2

The hyperbola

3

(a) Answer: x = 2.26

  1. let t = 3^x: t^2 − 10t − 24 = 0
  2. (t − 12)(t + 2) = 0
  3. 3^x = −2 has no solution
  4. 3^x = 12, so x = log_3 12 = 2.26

(b) Answer: x = 2

  1. 3^x · (3 + 1/3) = 30
  2. 3^x · 10/3 = 30
  3. 3^x = 9
  4. x = 2

(c) Answer: log 6 = p + q

  1. log 6 = log(2 × 3)
  2. = p + q

(d) Answer: x = 2

  1. √(x + 14) = x + 2
  2. x + 14 = x^2 + 4x + 4
  3. x^2 + 3x − 10 = 0
  4. (x − 2)(x + 5) = 0
  5. check x = −5: √(−5 + 14) = 3 but −5 + 2 = −3, so x = −5 is rejected
  6. x = 2

Exponential equations and exponent laws

4

(a) Answer: x = −1, y = 2

  1. 2^(x + y) = 2^(1), so x + y = 1
  2. 3^(x − y) = 3^(−3), so x − y = −3
  3. adding: 2x = −2, so x = −1
  4. y = 1 − (−1) = 2

(b) Answer: k < −41/4

  1. x^2 + 3x − 4 = −2x + k
  2. x^2 + 5x − 4 − k = 0
  3. no point of intersection: Δ < 0
  4. (5)^2 − 4(1)(−4 − k) < 0
  5. k < −41/4

(c) Answer: 50 km/h

  1. let the bus travel at v km/h: 75/v − 75/(v + 25) = 1/2
  2. 75(v + 25) − 75v = (1/2)v(v + 25)
  3. v^2 + 25v − 3750 = 0
  4. (v − 50)(v + 75) = 0
  5. v = 50 (v = −75 is rejected: a speed is positive)

Simultaneous equations

5

(a) Answer: R12 902.35

  1. i = 11%/12 and n = 240
  2. 1250000 = x[1 − (1 + 0.009167)^(−240)]/0.009167
  3. x = R12 902.35

(b) Answer: R677 857.43

  1. after 168 payments
  2. balance = 1250000(1 + i)^168 − 12902.35[(1 + i)^168 − 1]/i
  3. = R677 857.43

(c) Answer: 11.57%

  1. 1 + i_eff = (1 + 0.11/12)^12
  2. i_eff = 11.57%

(d) Answer: 126 payments

  1. 1250000 = 16800[1 − (1 + i)^(−n)]/i
  2. (1 + i)^(−n) = 0.317956
  3. n = 125.57
  4. 126 payments

Borrowing: the present-value annuity

6

(a) Answer: See the working below.

  1. the side along the wall is 56 − 2x
  2. A = x(56 − 2x) = 56x − 2x^2

(b) Answer: 14 m by 28 m; area 392 m²

  1. dA/dx = 56 − 4x = 0
  2. x = 14 m
  3. the side along the wall is 28 m
  4. A = 14 × 28 = 392 m²

(c) Answer: −20 m²/m

  1. dA/dx at x = 19
  2. = −20 m²/m

Optimisation

7

(a) Answer: f′(x) = x^2 − 8x + 16

  1. f′(x) = a(x − (4))^2, touching the x-axis at C
  2. 16 = a(4)^2, so a = 1
  3. f′(x) = x^2 − 8x + 16

(b) Answer: f(x) = (1/3)x^3 − 4x^2 + 16x + 6

  1. f(x) = ax^3 + bx^2 + cx + d with f′(x) = 3ax^2 + 2bx + c = x^2 − 8x + 16
  2. a = 1/3, b = −4, c = 16
  3. f(0) = d = 6
  4. f(x) = (1/3)x^3 − 4x^2 + 16x + 6

(c) Answer: A point of inflection

  1. f′(4) = 0, but f′ does not change sign at x = 4
  2. so f has a horizontal point of inflection there

(d) Answer: x = 1 or x = 7

  1. 1(x − (4))^2 = 9
  2. (x − (4))^2 = 9
  3. x = 1 or x = 7

Reading the graph of f'

8

(a) Answer: x = −4 or x = 5/3

  1. (3x − 5)(x + 4)^2 = 0
  2. 3x − 5 = 0 or x + 4 = 0
  3. x = −4 or x = 5/3

(b) Answer: x = −1.85 or x = 0.18

  1. 3x^2 + 5x − 1 = 0
  2. x = (−(5) ± √((5)^2 − 4(3)(−1)))/(2(3))
  3. x = (−5 ± √(37))/6
  4. x = −1.85 or x = 0.18

(c) Answer: x < −1.85 or x > 0.18

  1. the critical values are x = −1.85 and x = 0.18
  2. x < −1.85 or x > 0.18

(d) Answer: p = 25/3

  1. x^2 − 10x + 3p = 0
  2. Δ = (−10)^2 − 4(1)(3p) = 100 − 12p
  3. Δ = 0 for equal roots: 100 − 12p = 0
  4. p = 25/3

Quadratic equations

9

(a) Answer: b = 1/3

  1. P lies on f: b^1 = 1/3
  2. b = 1/3

(b) Answer: y = log_(1/3) x

  1. g is the reflection of f in y = x: x = (1/3)^y
  2. y = log_(1/3) x

(c) Answer: x > 0

  1. the domain of g is the range of f

(d) Answer: 0 < x ≤ 1/3

  1. g(x) = 1 when x = (1/3)^1 = 1/3
  2. g is decreasing (0 < b < 1), so the inequality sign is reversed
  3. 0 < x ≤ 1/3

(e) Answer: AB = 1/3 units

  1. A(1; 1/3): f(1) = (1/3)^1 = 1/3
  2. B(1; 0): g(1) = log_(1/3) 1 = 0
  3. AB = 1/3 − 0 = 1/3

(f) Answer: y = −2; x-intercept (−0.63; 0)

  1. f's asymptote y = 0 moves down 2: y = −2
  2. (1/3)^x − 2 = 0, so (1/3)^x = 2
  3. x = log_(1/3) 2 = −0.63

Exponential equations and exponent laws

10

(a) Answer: 1260

  1. 7 letters, with E 2 times, T 2 times
  2. 7!/(2! × 2!) = 1260

(b) Answer: 1/21

  1. fix T first and S last: 60 arrangements of the other 5 letters
  2. P = 60/1260 = 1/21

(c) Answer: 720

  1. 6! = 720

(d) Answer: 240

  1. treat the 2 as one block: 5 units, 5! ways
  2. the block itself in 2! ways
  3. 5! × 2! = 240

(e) Answer: 96

  1. 4 × 4 × 3 × 2 × 1
  2. = 96

(f) Answer: 12

  1. the first digit is at least 8, the last digit even
  2. count each choice of first and last digit, then fill the middle
  3. = 12

The fundamental counting principle

11

(a) Answer: x = 8; the terms are 10; 16; 22

  1. T_2 − T_1 = T_3 − T_2
  2. 2(5x − 24) = (3x − 14) + (2x + 6)
  3. 5x − 40 = 0
  4. x = 8

(b) Answer: T_(36)

  1. 10 + (n − 1)(6) = 220
  2. n − 1 = 35
  3. n = 36

(c) Answer: 13 terms

  1. n/2[2(10) + (n − 1)(6)] = 598
  2. 6n^2 + 14n − 1196 = 0
  3. n = 13 (n is a natural number)

(d) Answer: 1223 dots

  1. second difference 6: a = 3
  2. T_n = 3n^2 + n + 3
  3. T_(20) = 1223

(e) Answer: picture 45

  1. 3n^2 + n + 3 = 6123
  2. 3n^2 + n − 6120 = 0
  3. n = 45 (n > 0)

Arithmetic sequences