Mathbench

IEB mock Paper I (third paper)

A complete 150-mark practice paper: 11 questions in 53 parts, 32 topics, Section A 76 marks and Section B 74. 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[76 marks]

1[9 marks]

A closed cylindrical can must hold 250 cm³. Its radius is r cm and its height h cm. (V = πr^2h and the total surface area is S = 2πr^2 + 2πrh)

rh
  1. (a)
    Show that S = 2πr^2 + 500/r.
    (3)
  2. (b)
    Determine, correct to two decimal places, the radius for which the surface area is a minimum.
    (4)
  3. (c)
    Calculate the minimum surface area, correct to two decimal places.
    (2)
2[12 marks]

You want to have R570 000 in 12 years' time. Your savings account earns 9.5% per annum compounded monthly. Give answers correct to two decimal places.

  1. (a)
    Calculate the equal monthly deposit that reaches the goal, if the first deposit is made one month from now and the last one at the end of the 12 years.
    (4)
  2. (b)
    Instead, you deposit a single amount of R230 000 now and nothing more. After how many months will the account first hold at least R570 000?
    (4)
  3. (c)
    Instead, you deposit R230 000 now and R800 at the end of every month. What will the account hold at the end of the 12 years?
    (4)
3[13 marks]

The sum of the first n terms of a sequence is given by S_n = −2n^2 − 3n.

  1. (a)
    Determine the first three terms of the sequence.
    (3)
  2. (b)
    Show that the sequence is arithmetic, and determine T_n.
    (3)
  3. (c)
    Which term of the sequence is equal to −77?
    (2)
  4. (d)
    In another arithmetic series the common difference is −3. The sum of the first n terms is −234 and the sum of the first 2n terms is −900. Determine n and the first term of this series.
    (5)
4[14 marks]
  1. (a)
    If 2^x = k, write 8^x in terms of k.
    (2)
  2. (b)
    Write 2^(2x − 1) in terms of k.
    (2)
  3. (c)
    If log 2 = p and log 3 = q, determine log 18 in terms of p and q.
    (2)
  4. (d)
    Determine log 4.5 in terms of p and q.
    (2)
  5. (e)
    Solve for x: log_x 27 = 3
    (2)
  6. (f)
    Solve for x: log_5 (x − 2) + log_5 (x + 22) = 2
    (4)
5[14 marks]
  1. (a)
    Solve for x in terms of p: x^2 + 12p^2 = 7px
    (3)
  2. (b)
    If x^2 + 2x − 6 = (x − m)^2 + p, determine the values of m and p.
    (3)
  3. (c)
    Hence write down the minimum value of x^2 + 2x − 6.
    (1)
  4. (d)
    Solve for x: x^2 ≤ 3x + 10
    (3)
  5. (e)
    Determine the value(s) of p for which the roots of x(x − 8) = 4p are real and equal.
    (4)
6[14 marks]

In the diagram below, the graph of f(x) = a/(x + p) + q is drawn with its asymptotes. A(3; 2) lies on f.

xyOx = 2y = −1A(3; 2)f
  1. (a)
    Write down the values of p and q.
    (2)
  2. (b)
    Determine the value of a.
    (2)
  3. (c)
    Determine the equations of the axes of symmetry of f.
    (2)
  4. (d)
    The axis of symmetry y = x − 3 meets the graph of f at B and D. Determine the coordinates of B and D, leaving your answers in surd form where necessary.
    (3)
  5. (e)
    Write down the range of f.
    (1)
  6. (f)
    For which values of x is f(x) > 0?
    (2)
  7. (g)
    The graph of f is shifted 1 unit to the right and 1 unit down to form g. Write down the equation of g.
    (2)

Section B[74 marks]

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(2; 0) and cuts the y-axis at (0; 4).

xyOC(2; 0)(0; 4)f′
  1. (a)
    Determine the equation of f′.
    (3)
  2. (b)
    Given that f(0) = −6, determine f(x).
    (4)
  3. (c)
    Is x = 2 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 1.
    (3)
8[13 marks]
  1. (a)
    Given f(x) = 2x^2 + 4x − 6. 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 = 2x + k is a tangent to the graph of g(x) = x^2 − 4x − 7. 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 = 12x?
    (3)
9[15 marks]
  1. (a)
    Calculate Σ_(n=1)^(16) (2n + 7).
    (3)
  2. (b)
    Determine the value of p if Σ_(n=1)^(p) (4n + 5) = 774.
    (4)
  3. (c)
    Calculate Σ_(n=1)^(∞) (−6(1/4)^(n − 1)).
    (3)
  4. (d)
    Determine the value of p if Σ_(n=1)^(∞) ((4/5)^n) + Σ_(n=1)^(p) (6n + 4) = 990.
    (5)
10[16 marks]
  1. (a)
    In how many different ways can the letters of the word PEPPERS be arranged?
    (3)
  2. (b)
    The letters of PEPPERS are arranged at random. What is the probability that the arrangement starts with S and ends with R?
    (3)
  3. (c)
    8 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 4 particular cars must be next to each other, in any order?
    (3)
  5. (e)
    The digits 1; 3; 4; 5; 6; 9 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 90 000?
    (4)
11[18 marks]
  1. (a)
    P(A) = 0.45 and P(B) = 0.1. Determine P(A or B) if A and B are mutually exclusive.
    (2)
  2. (b)
    Determine P(A or B) if A and B are independent.
    (3)
  3. (c)
    A spinner lands on red (R), blue (B) or green (G) with probabilities 0.45, 0.35 and 0.2. It is spun twice, and the spins are independent. Draw a tree diagram to show all the outcomes and their probabilities.
    (3)
  4. (d)
    Calculate the probability of one blue and one green, in any order.
    (3)
  5. (e)
    Calculate the probability of at least one red.
    (2)
  6. (f)
    In a group of 50 learners, 25 take chess, 23 take drama and 13 take both. A learner is chosen at random. Calculate the probability that the learner takes chess or drama.
    (2)
  7. (g)
    Are the events 'takes chess' and 'takes drama' independent? Show your calculation.
    (3)

Memorandum

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

1

(a) Answer: See the working below.

  1. h = 250/(πr^2)
  2. S = 2πr^2 + 2πr(250/(πr^2))
  3. S = 2πr^2 + 500/r

(b) Answer: r = 3.41 cm

  1. dS/dr = 4πr − 500/r^2 = 0
  2. r^3 = 250/(2π)
  3. r = 3.41 cm

(c) Answer: S = 219.69 cm²

  1. S = 2π(3.4139)^2 + 500/3.4139
  2. S = 219.69 cm²

Optimisation

2

(a) Answer: R2 135.83

  1. i = 9.5%/12 and n = 144
  2. 570 000 = x[(1 + i)^144 − 1]/i
  3. x = R2 135.83

(b) Answer: 116 months

  1. 230 000(1 + i)^n ≥ 570 000
  2. n ≥ log(570 000/230 000)/log(1 + i) = 115.09
  3. 116 months

(c) Answer: R929 436.18

  1. 230 000(1 + i)^144 + 800[(1 + i)^144 − 1]/i
  2. = R929 436.18

Saving: the future-value annuity

3

(a) Answer: −5; −9; −13

  1. S_1 = −5, S_2 = −14, S_3 = −27
  2. T_1 = S_1 = −5; T_2 = S_2 − S_1 = −9; T_3 = S_3 − S_2 = −13

(b) Answer: T_n = −4n − 1

  1. T_2 − T_1 = −4 and T_3 − T_2 = −4: a constant difference, so arithmetic
  2. T_n = −5 + (n − 1)(−4) = −4n − 1

(c) Answer: T_(19)

  1. −4n − 1 = −77
  2. n = 19

(d) Answer: n = 12 and a = −3

  1. S_n = n/2[2a + (n − 1)(−3)] = −234 and S_(2n) = n[2a + (2n − 1)(−3)] = −900
  2. S_(2n) − 2S_n = −3n^2 = −432
  3. n^2 = 144, so n = 12
  4. 12/2[2a − 33] = −234, so a = −3

Arithmetic sequences

4

(a) Answer: 8^x = k^3

  1. 8^x = (2^x)^3
  2. = k^3

(b) Answer: 2^(2x − 1) = k^2/2

  1. 2^(2x − 1) = (2^x)^2 ÷ 2
  2. = k^2/2

(c) Answer: log 18 = p + 2q

  1. log 18 = log(2 × 3^2)
  2. = p + 2q

(d) Answer: log 4.5 = −p + 2q

  1. log 4.5 = log(3^2/2)
  2. = −p + 2q

(e) Answer: x = 3

  1. x^3 = 27
  2. x = 3 (a base is positive and not 1)

(f) Answer: x = 3

  1. log_5 [(x − 2)(x + 22)] = 2
  2. (x − 2)(x + 22) = 25
  3. x^2 + 20x − 69 = 0
  4. x = 3 or x = −23
  5. x = −23 gives the log of a negative number, so x = 3

Exponential equations and exponent laws

5

(a) Answer: x = 3p or x = 4p

  1. x^2 − 7px + 12p^2 = 0
  2. (x − 3p)(x − 4p) = 0
  3. x = 3p or x = 4p

(b) Answer: m = −1 and p = −7

  1. (x − m)^2 + p = x^2 − 2mx + m^2 + p
  2. −2m = 2, so m = −1
  3. m^2 + p = −6, so p = −7

(c) Answer: −7

  1. (x − m)^2 ≥ 0, so the least value is p = −7

(d) Answer: −2 ≤ x ≤ 5

  1. x^2 − 3x − 10 ≤ 0
  2. (x + 2)(x − 5) ≤ 0
  3. critical values: x = −2 and x = 5
  4. −2 ≤ x ≤ 5

(e) Answer: p = −4

  1. x^2 − 8x − 4p = 0
  2. Δ = (−8)^2 − 4(1)(−4p) = 64 + 16p
  3. Δ = 0 for equal roots: 64 + 16p = 0
  4. p = −4

Quadratic equations

6

(a) Answer: p = −2 and q = −1

  1. the asymptotes are x = 2 and y = −1
  2. x + p = 0 at x = 2, so p = −2; q = −1

(b) Answer: a = 3

  1. 2 = a/(3 − 2) − 1
  2. a = (2 + 1)(3 − 2) = 3

(c) Answer: y = x − 3 and y = −x + 1

  1. both pass through the point where the asymptotes meet, (2; −1)
  2. y = (x − 2) − 1 and y = −(x − 2) − 1

(d) Answer: (2 − √3; −1 − √3) and (2 + √3; −1 + √3)

  1. 3/(x − 2) − 1 = x − 3
  2. (x − 2)^2 = 3
  3. x = 2 ± √3
  4. (2 − √3; −1 − √3) and (2 + √3; −1 + √3)

(e) Answer: y ∈ ℝ, y ≠ −1

  1. f never takes the value of its horizontal asymptote, y = −1

(f) Answer: 2 < x < 5

  1. read from the graph: the x-intercept is x = 5 and the asymptote x = 2 is never included
  2. 2 < x < 5

(g) Answer: g(x) = 3/(x − 3) − 2

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

The hyperbola

7

(a) Answer: f′(x) = x^2 − 4x + 4

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

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

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

(c) Answer: A point of inflection

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

(d) Answer: x = 1 or x = 3

  1. 1(x − (2))^2 = 1
  2. (x − (2))^2 = 1
  3. x = 1 or x = 3

Reading the graph of f'

8

(a) Answer: 4x + 2h + 4

  1. f(x + h) = 2(x + h)^2 + 4(x + h) − 6
  2. f(x + h) − f(x) = 4xh + 2h^2 + 4h
  3. (4xh + 2h^2 + 4h)/h = 4x + 2h + 4

(b) Answer: f′(x) = 4x + 4

  1. f′(x) = lim_(h → 0) (4x + 2h + 4)
  2. = 4x + 4

(c) Answer: (3; −10); k = −16

  1. g′(x) = 2x − 4 = 2
  2. x = 3
  3. g(3) = −10, so the point is (3; −10)
  4. −10 = 2(3) + k, so k = −16

(d) Answer: (2; 10)

  1. f′(x) = 4x + 4 = 12
  2. x = 2
  3. f(2) = 10

The derivative from first principles

9

(a) Answer: 384

  1. an arithmetic series: a = 9, d = 2, n = 16
  2. S_(16) = 16/2[2(9) + (15)(2)] = 384

(b) Answer: p = 18

  1. a = 9, d = 4: p/2[2(9) + (p − 1)(4)] = 774
  2. 4p^2 + 14p − 1548 = 0
  3. p = 18 (p is a natural number)

(c) Answer: −8

  1. a geometric series with a = −6 and r = 1/4; −1 < r < 1
  2. S_(∞) = −6/(1 − (1/4)) = −8

(d) Answer: p = 17

  1. the first sum: a = r = 4/5, so S_(∞) = (4/5)/(1 − 4/5) = 4
  2. the second: p/2[2(10) + (p − 1)(6)] = 986
  3. 6p^2 + 14p − 1972 = 0
  4. p = 17

Sigma notation

10

(a) Answer: 420

  1. 7 letters, with E 2 times, P 3 times
  2. 7!/(2! × 3!) = 420

(b) Answer: 1/42

  1. fix S first and R last: 10 arrangements of the other 5 letters
  2. P = 10/420 = 1/42

(c) Answer: 40 320

  1. 8! = 40 320

(d) Answer: 2880

  1. treat the 4 as one block: 5 units, 5! ways
  2. the block itself in 4! ways
  3. 5! × 4! = 2880

(e) Answer: 720

  1. 6 × 5 × 4 × 3 × 2
  2. = 720

(f) Answer: 48

  1. the first digit is at least 9, the last digit even
  2. count each choice of first and last digit, then fill the middle
  3. = 48

The fundamental counting principle

11

(a) Answer: 0.55

  1. mutually exclusive: P(A and B) = 0
  2. P(A or B) = 0.45 + 0.1 = 0.55

(b) Answer: 0.505

  1. independent: P(A and B) = 0.45 × 0.1 = 0.045
  2. P(A or B) = 0.45 + 0.1 − 0.045 = 0.505

(c) Answer: See the tree below.

0.45R0.45RRR: 0.20250.35BRB: 0.15750.2GRG: 0.090.35B0.45RBR: 0.15750.35BBB: 0.12250.2GBG: 0.070.2G0.45RGR: 0.090.35BGB: 0.070.2GGG: 0.04
  1. three branches for the first spin, three from each for the second
  2. each outcome's probability is the product along its branches

(d) Answer: 0.14

  1. P(BG) + P(GB) = 0.35 × 0.2 + 0.2 × 0.35
  2. = 0.14

(e) Answer: 0.6975

  1. 1 − P(no red) = 1 − (0.55)^2
  2. = 0.6975

(f) Answer: 7/10

  1. n(chess or drama) = 25 + 23 − 13 = 35
  2. P = 35/50 = 7/10

(g) Answer: No, they are not independent

  1. P(chess) × P(drama) = 1/2 × 23/50 = 23/100
  2. P(both) = 13/50
  3. not equal: not independent

The probability rules