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IEB mock Paper I (second paper)

A complete 150-mark practice paper: 11 questions in 52 parts, 35 topics, Section A 73 marks and Section B 77. 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 415 real IEB questions — not guessed. None of those questions appears here.

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The paper to print, and its memorandum as a separate file so you can sit it without the answers in front of you. In English and Afrikaans.

The paper

Section A[73 marks]

QUESTION 1

Give answers correct to two decimal places.

  1. (a)
    You deposit R1 900 at the end of every month for 10 years into an account earning 7% per annum compounded monthly. You then stop the deposits, and for the next 3 years the account earns 6% per annum compounded quarterly. Calculate the amount in the account at the end of these 13 years.
    (5)
  2. (b)
    From then on the account earns 5.5% per annum compounded monthly, and you withdraw R3 100 at the end of every month. How many full withdrawals of R3 100 can you make?
    (4)
  3. (c)
    A savings account earns 13% per annum compounded half-yearly. The first deposit of R6 500 is made six months after the account is opened, and the same amount is deposited at the end of every six months after that. How much is in the account 3 years after it was opened, just after the deposit made then?
    (4)

[13]

QUESTION 2

In the diagram below, f(x) = b^x and g, the reflection of f in the line y = x, are drawn. P(−1; 5) lies on f.

xyOy = xP(−1; 5)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) > 2?
    (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) − 25. Write down the equation of the asymptote of h and determine the x-intercept of h.
    (3)

[13]

QUESTION 3

In the diagram below, the graph of f(x) = −x^3 − 3x^2 + 24x + d is drawn, where d is a constant. (2; 0) is a stationary point of f.

xyO(2; 0)f
  1. (a)
    Determine the value of d.
    (2)
  2. (b)
    Determine the coordinates of the other stationary point of f.
    (4)
  3. (c)
    Determine the x-coordinate of the point of inflection of f.
    (2)
  4. (d)
    For which values of k will f(x) = k have exactly one real root?
    (3)
  5. (e)
    Determine the equation of the tangent to f at x = 0.
    (3)

[14]

QUESTION 4
  1. (a)
    Solve for x: (5x − 7)(x^2 + 6x + 9) = 0
    (3)
  2. (b)
    Use the quadratic formula to solve for x, correct to two decimal places, if 4x^2 = 9x + 4. Show all working.
    (4)
  3. (c)
    Hence, solve for x if 4x^2 − 9x − 4 > 0.
    (2)
  4. (d)
    Show that the roots of 3x^2 + px = 2 are real and unequal for all real values of p.
    (3)
  5. (e)
    Determine the values of p for which x^2 − (p − 1)x + 36 = 0 has equal roots.
    (4)

[16]

QUESTION 5

Consider the arithmetic series 7 + 11 + 15 + 19 + … + 223

  1. (a)
    Determine a formula for T_n, the general term of the series.
    (2)
  2. (b)
    Determine the number of terms in the series.
    (3)
  3. (c)
    Calculate the sum of the series.
    (3)
  4. (d)
    Write the series in sigma notation.
    (2)
  5. (e)
    The first five terms of a quadratic sequence are 17; 28; 43; 62; 85; … Determine T_n, the nth term.
    (4)
  6. (f)
    Determine whether 1184 is a term of this quadratic sequence.
    (3)

[17]

Section B[77 marks]

QUESTION 6

The graph of y = f′(x), the derivative of a cubic function f, is drawn below. It cuts the x-axis at (1; 0) and (5; 0) and the y-axis at (0; −5).

xyO(1; 0)(5; 0)(0; −5)f′
  1. (a)
    For which values of x is f increasing?
    (2)
  2. (b)
    Write down the x-coordinates of the turning points of f, and say which one is a local maximum.
    (3)
  3. (c)
    Determine the x-coordinate of the point of inflection of f.
    (2)
  4. (d)
    For which values of x is f concave up?
    (2)
  5. (e)
    Write down the gradient of the tangent to f at x = 0.
    (1)

[10]

QUESTION 7
  1. (a)
    Solve for x and y if x − y = 0 and xy = 1.
    (6)
  2. (b)
    The 288 chairs in a school hall are set out in equal rows. If there were 3 fewer rows, each row would need 8 more chairs. How many rows are there?
    (5)

[11]

QUESTION 8

Stage 1 of a pattern is a shaded equilateral triangle with sides of 36 cm. To make each next stage, every shaded triangle is divided into four equal triangles by joining the midpoints of its sides, and the middle one is removed. Stages 1, 2 and 3 are shown.

123
  1. (a)
    Show that the area of the triangle in stage 1 is 324√3 cm².
    (2)
  2. (b)
    How many shaded triangles are there in stage 6?
    (2)
  3. (c)
    Show that the areas removed in making stages 2, 3, 4, … form a geometric series, and calculate the total area removed in making stages 2 to 7, correct to two decimal places.
    (3)
  4. (d)
    If the pattern could go on for ever, what total area would be removed? What does this tell you about the shaded area?
    (2)
  5. (e)
    In which stage is the total perimeter of the shaded triangles for the first time more than 2550 cm?
    (3)

[12]

QUESTION 9

In the diagram below, the parabola f with turning point T(1; 3) and y-intercept (0; 4), and the straight line g, are drawn. The graphs meet at A and B.

xyOT(1; 3)(0; 4)ABfg
  1. (a)
    Determine the equation of f in the form f(x) = a(x − p)^2 + q.
    (3)
  2. (b)
    g(x) = −2x + 8. Determine the coordinates of A and B, the points where f and g meet.
    (5)
  3. (c)
    For which values of x is f(x) > g(x)?
    (2)
  4. (d)
    Write down the range of f.
    (1)
  5. (e)
    For which values of t will f(x) = t have no real roots?
    (2)

[13]

QUESTION 10
  1. (a)
    A bag holds 4 red and 3 blue balls. Two balls are taken out one after the other, without replacement. Draw a tree diagram to show all the outcomes and their probabilities.
    (3)
  2. (b)
    Calculate the probability that the two balls are the same colour.
    (2)
  3. (c)
    For two events A and B, P(A) = 0.45, P(B′) = 0.55 and P(A′ and B′) = 0.1. Determine P(A or B) and P(A and B).
    (3)
  4. (d)
    Are A and B mutually exclusive? Are they independent? Give reasons.
    (3)
  5. (e)
    In a group of 30 learners, 19 take Life Sciences, 12 take Physical Sciences and 3 take neither. How many take both?
    (2)
  6. (f)
    A learner is chosen at random. Calculate the probability that the learner takes exactly one of the two subjects.
    (2)

[15]

QUESTION 11
  1. (a)
    Given f(x) = −2x^2 − 3x + 1. Determine f′(x) from first principles.
    (5)
  2. (b)
    Determine dy/dx if y = (x^2 − 3x + 2)/√x.
    (4)
  3. (c)
    Determine the equation of the tangent to f at x = −2.
    (4)
  4. (d)
    Determine the average gradient of f between x = −1 and x = 3.
    (3)

[16]

Memorandum

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

QUESTION 1

(a) Answer: R393 192.35

  1. after 120 deposits: 1900[(1 + 7/1200)^120 − 1]/(7/1200) = R328 861.13
  2. then 12 quarters: R328 861.13(1 + 6/400)^12 = R393 192.35

(b) Answer: 190 withdrawals

  1. 393 192.35 = 3100[1 − (1 + 0.004583)^(−n)]/0.004583
  2. (1 + 0.004583)^(−n) = 0.418667
  3. n = 190.40
  4. 190 full withdrawals

(c) Answer: R45 914.23

  1. deposits at 6, 12, …, 36 months: 6 of them
  2. FV = 6500[(1 + 0.065)^6 − 1]/0.065
  3. = R45 914.23

Saving: the future-value annuity

QUESTION 2

(a) Answer: b = 1/5

  1. P lies on f: b^(−1) = 5

(b) Answer: y = log_(1/5) x

  1. g is the reflection of f in y = x: x = (1/5)^y

(c) Answer: x > 0

  1. the domain of g is the range of f

(d) Answer: 0 < x < 1/25

  1. g(x) = 2 when x = (1/5)^2 = 1/25
  2. g is decreasing (0 < b < 1), so the inequality sign is reversed

(e) Answer: AB = 1/5 units

  1. A(1; 1/5): f(1) = (1/5)^1 = 1/5
  2. B(1; 0): g(1) = log_(1/5) 1 = 0
  3. AB = 1/5 − 0 = 1/5

(f) Answer: y = −25; x-intercept (−2; 0)

  1. f's asymptote y = 0 moves down 25: y = −25
  2. (1/5)^x − 25 = 0, so (1/5)^x = 25
  3. x = log_(1/5) 25 = −2

Exponential equations and exponent laws

QUESTION 3

(a) Answer: d = −28

  1. f(2) = 0
  2. 28 + d = 0

(b) Answer: (−4; −108)

  1. f′(x) = −3x^2 − 6x + 24 = 0
  2. −3(x − (−4))(x − (2)) = 0
  3. x = −4
  4. f(−4) = −108

(c) Answer: x = −1

  1. f′′(x) = −6x − 6 = 0

(d) Answer: k < −108 or k > 0

  1. the turning values are −108 and 0
  2. one root when the line y = k passes above the maximum or below the minimum

(e) Answer: y = 24x − 28

  1. f(0) = −28 and f′(0) = 24

Stationary points

QUESTION 4

(a) Answer: x = −3 or x = 7/5

  1. (5x − 7)(x + 3)^2 = 0
  2. 5x − 7 = 0 or x + 3 = 0

(b) Answer: x = −0.38 or x = 2.63

  1. 4x^2 − 9x − 4 = 0
  2. x = (−(−9) ± √((−9)^2 − 4(4)(−4)))/(2(4))
  3. x = (9 ± √(145))/8

(c) Answer: x < −0.38 or x > 2.63

  1. the critical values are x = −0.38 and x = 2.63

(d) Answer: See the proof below.

  1. 3x^2 + px − 2 = 0
  2. Δ = (p)^2 − 4(3)(−2) = p^2 + 24
  3. p^2 ≥ 0, so Δ ≥ 24 > 0
  4. so the roots are real and unequal for all real values of p

(e) Answer: p = −11 or p = 13

  1. Δ = (p − 1)^2 − 4(1)(36) = (p − 1)^2 − 144
  2. Δ = 0 for equal roots: (p − 1)^2 = 144
  3. p − 1 = ±12

Quadratic equations

QUESTION 5

(a) Answer: T_n = 4n + 3

  1. a = 7 and d = 11 − (7) = 4
  2. T_n = 7 + (n − 1)(4) = 4n + 3

(b) Answer: 55 terms

  1. 4n + 3 = 223
  2. 4n = 220
  3. n = 55

(c) Answer: S_(55) = 6325

  1. S_n = n/2(a + l)
  2. S_(55) = 55/2(7 + 223)
  3. = 6325

(d) Answer: Σ_(n=1)^(55) (4n + 3)

  1. the terms are T_n = 4n + 3, from n = 1 to n = 55

(e) Answer: T_n = 2n^2 + 5n + 10

  1. first differences: 11; 15; 19; 23
  2. second difference: 4, so 2a = 4 and a = 2
  3. 3a + b = 11, so b = 5
  4. a + b + c = 17, so c = 10

(f) Answer: No: 1184 is not a term

  1. 2n^2 + 5n + 10 = 1184
  2. 2n^2 + 5n − 1174 = 0
  3. n = 23.01, not a natural number

Arithmetic sequences

QUESTION 6

(a) Answer: 1 < x < 5

  1. f is increasing where f′(x) > 0: where the graph of f′ lies above the x-axis

(b) Answer: x = 5 (local maximum) and x = 1 (local minimum)

  1. f′(x) = 0 at x = 1 and x = 5
  2. f′ changes from positive to negative at x = 5: a maximum

(c) Answer: x = 3

  1. f′′(x) = 0 where f′ has its turning point
  2. x = (1 + 5)/2 = 3

(d) Answer: x < 3

  1. f is concave up where f′′(x) > 0: where f′ is increasing

(e) Answer: −5

  1. the gradient of f at x = 0 is f′(0) = −5

Reading the graph of f'

QUESTION 7

(a) Answer: x = −1, y = −1 or x = 1, y = 1

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

(b) Answer: 12 rows

  1. let there be r rows: 288/(r − 3) − 288/r = 8
  2. 288r − 288(r − 3) = 8r(r − 3)
  3. r^2 − 3r − 108 = 0
  4. (r − 12)(r + 9) = 0
  5. r = 12 (r = −9 is rejected: the number of rows is positive)

Simultaneous equations

QUESTION 8

(a) Answer: 324√3 cm²

  1. area = ½ × 36 × 36 × sin 60°
  2. = ½ × 1296 × √3/2 = 324√3 cm²

(b) Answer: 243

  1. stages 1, 2 and 3 have 1, 3 and 9: every shaded triangle leaves three
  2. stage 6: 3^5 = 243

(c) Answer: 461.31 cm²

  1. stage 2 removes ¼ of 324√3, which is 81√3 cm²
  2. each next stage removes 3 times as many triangles, each ¼ of the size: r = 3/4
  3. S_6 = 81√3(1 − 0.75^6)/(1 − 0.75) = 461.31 cm²

(d) Answer: 324√3 cm², all of stage 1: the shaded area tends to 0

  1. −1 < r = 0.75 < 1, so S∞ = 81√3/(1 − 0.75) = 324√3 cm²
  2. that is the whole area of stage 1, so the shaded area tends to 0

(e) Answer: stage 9

  1. stage 1: 108 cm; each next stage has 3 times as many triangles, each with half the perimeter: r = 1.5
  2. 108(1.5)^(n − 1) > 2550
  3. n − 1 > log(2550/108)/log 1.5 = 7.80
  4. n − 1 = 8, so stage 9

Geometric sequences

QUESTION 9

(a) Answer: f(x) = (x − 1)^2 + 3

  1. f(x) = a(x − 1)^2 + 3
  2. substitute (0; 4): 4 = a(−1)^2 + 3
  3. a = 1

(b) Answer: A(−2; 12) and B(2; 4)

  1. (x − 1)^2 + 3 = −2x + 8
  2. x^2 − 4 = 0
  3. (x + 2)(x − 2) = 0
  4. x = −2 or x = 2

(c) Answer: x < −2 or x > 2

  1. read from the graph at A and B

(d) Answer: y ≥ 3

  1. the turning point is a minimum: y = 3

(e) Answer: t < 3

  1. the line y = t must miss the parabola entirely

The parabola

QUESTION 10

(a) Answer: See the tree below.

4/7R1/2RRR: 2/71/2BRB: 2/73/7B2/3RBR: 2/71/3BBB: 1/7
  1. first draw: red 4/7, blue 3/7
  2. second draw: 6 balls are left, one fewer of the colour taken

(b) Answer: 3/7

  1. P(RR) + P(BB) = 2/7 + 1/7
  2. = 3/7

(c) Answer: P(A or B) = 0.9; P(A and B) = 0

  1. P(A or B) = 1 − P(A′ and B′) = 1 − 0.1 = 0.9
  2. P(B) = 1 − P(B′) = 0.45
  3. P(A and B) = P(A) + P(B) − P(A or B) = 0.45 + 0.45 − 0.9 = 0

(d) Answer: Mutually exclusive, and not independent

  1. P(A and B) = 0: they cannot happen together
  2. P(A) × P(B) = 0.45 × 0.45 = 0.2025 ≠ P(A and B)
  3. mutually exclusive, and not independent

(e) Answer: 4

  1. 19 + 12 − x + 3 = 30
  2. x = 4

(f) Answer: 23/30

  1. only Life Sciences: 15; only Physical Sciences: 8
  2. P = 23/30 = 23/30

The probability rules

QUESTION 11

(a) Answer: f′(x) = −4x − 3

  1. f(x + h) = −2(x + h)^2 − 3(x + h) + 1
  2. = −2x^2 − 4xh − 2h^2 − 3x − 3h + 1
  3. f(x + h) − f(x) = −4xh − 2h^2 − 3h
  4. f′(x) = lim_(h → 0) (−4xh − 2h^2 − 3h)/h
  5. = lim_(h → 0) (−4x − 2h − 3)
  6. = −4x − 3

(b) Answer: dy/dx = (3/2)x^(1/2) − (3/2)x^(−1/2) − x^(−3/2)

  1. y = x^(3/2) − 3x^(1/2) + 2x^(−1/2)

(c) Answer: y = 5x + 9

  1. f(−2) = −1, so the point is (−2; −1)
  2. m = f′(−2) = 5
  3. y + 1 = 5(x + 2)

(d) Answer: −7

  1. f(−1) = 2 and f(3) = −26
  2. (−26 − (2))/(3 − (−1)) = −7

The derivative from first principles