Mathbench

IEB mock Paper I

A complete 150-mark practice paper: 10 questions in 53 parts, 31 topics, Section A 75 marks and Section B 75. 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[75 marks]

1[13 marks]
  1. (a)
    Solve for x in terms of p: x^2 − 3p^2 = −2px
    (3)
  2. (b)
    If x^2 + 12x + 34 = (x − m)^2 + p, determine the values of m and p.
    (3)
  3. (c)
    Hence write down the minimum value of x^2 + 12x + 34.
    (1)
  4. (d)
    Solve for x: x^2 ≤ −4x − 3
    (3)
  5. (e)
    Show that the roots of 2x^2 − px = 3 are real and unequal for all real values of p.
    (3)
2[14 marks]

In the diagram below, the graph of f(x) = a/(x + p) + q is drawn with its asymptotes. A(−1; −5) lies on f.

xyOx = −2y = 3A(−1; −5)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 + 1 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 domain 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 2 units up to form g. Write down the equation of g.
    (2)
3[16 marks]
  1. (a)
    Solve for x in terms of y: 3^(x + 3) · 81^y = 3^y
    (3)
  2. (b)
    Solve for x: √(x + 53) = x − 3
    (5)
  3. (c)
    Solve for x: log_2 (x^2 − 3x − 16) = 1
    (4)
  4. (d)
    Solve for x: 2^(2x) + 1(2^x) − 6 = 0
    (4)
4[16 marks]
  1. (a)
    Given f(x) = −x^2 − 6x − 6. Determine f′(x) from first principles.
    (5)
  2. (b)
    Determine dy/dx if y = (x^2 + 4x + 5)/√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 = 2 and x = 4.
    (3)
5[16 marks]

You can afford to repay R5 500 per month on a home loan. Give answers correct to two decimal places.

  1. (a)
    The bank quotes an effective annual interest rate of 11.57%. Determine the equivalent nominal annual rate, compounded monthly, correct to two decimal places.
    (3)
  2. (b)
    Using this nominal rate, calculate the largest loan that repayments of R5 500 per month can pay off in 30 years. The first payment is made one month after the loan is granted.
    (4)
  3. (c)
    The house you want costs R670 000. How much must you pay as a deposit?
    (2)
  4. (d)
    At the end of year 9, immediately after the payment then, you pay a lump sum of R100 000 into the loan. Calculate the balance outstanding immediately after the lump sum.
    (4)
  5. (e)
    You keep paying R5 500 per month. How many more payments will pay off the loan? (The last payment may be smaller.)
    (3)

Section B[75 marks]

6[13 marks]

In the diagram below, the parabola f with turning point T(−2; 6) and y-intercept (0; 2), and the straight line g, are drawn. The graphs meet at A and B.

xyOT(−2; 6)(0; 2)ABfg
  1. (a)
    Determine the equation of f in the form f(x) = a(x − p)^2 + q.
    (3)
  2. (b)
    g(x) = −4x − 7. 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)
7[14 marks]

In the diagram below, the graph of f(x) = x^3 − 6x^2 + d is drawn, where d is a constant. (0; 0) is a stationary point of f.

xyO(0; 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)
8[14 marks]
  1. (a)
    A PIN is made of 5 digits chosen from 0 to 9. How many different PINs can be made if digits may be repeated?
    (2)
  2. (b)
    How many different PINs can be made if no digit may be used more than once?
    (2)
  3. (c)
    A PIN is chosen at random, digits allowed to repeat. What is the probability that at least one digit appears more than once? Give your answer as a decimal, correct to four decimal places where needed.
    (3)
  4. (d)
    How many PINs, digits allowed to repeat, start with a digit greater than 4 and end with an odd digit?
    (3)
  5. (e)
    How many 3-letter arrangements can be made from the letters of the word GARDEN, if no letter is used more than once?
    (2)
  6. (f)
    How many of these arrangements start with a vowel?
    (2)
9[17 marks]

Given: f(x) = −x^3 − 6x^2 − 9x − 4

xyO
  1. (a)
    Show that x = −4 is a root of f(x) = 0, and hence determine the x-intercepts of f.
    (3)
  2. (b)
    Determine the coordinates of the turning points of f.
    (4)
  3. (c)
    Sketch the graph of f on the axes provided. Show all intercepts with the axes and the turning points.
    (4)
  4. (d)
    For which values of x is the graph of f concave up?
    (2)
  5. (e)
    For which values of x is f decreasing?
    (2)
  6. (f)
    For which values of k will f(x) = k have three different real roots?
    (2)
10[17 marks]

Consider the arithmetic series −1 − 4 − 7 − 10 + … − 91

  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 6; 21; 42; 69; 102; … Determine T_n, the nth term.
    (4)
  6. (f)
    Determine whether 2182 is a term of this quadratic sequence.
    (3)

Memorandum

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

1

(a) Answer: x = −3p or x = p

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

(b) Answer: m = −6 and p = −2

  1. (x − m)^2 + p = x^2 − 2mx + m^2 + p
  2. −2m = 12, so m = −6
  3. m^2 + p = 34, so p = −2

(c) Answer: −2

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

(d) Answer: −3 ≤ x ≤ −1

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

(e) Answer: See the proof below.

  1. 2x^2 − px − 3 = 0
  2. Δ = (−p)^2 − 4(2)(−3) = 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

Quadratic equations

2

(a) Answer: p = 2 and q = 3

  1. the asymptotes are x = −2 and y = 3
  2. x + p = 0 at x = −2, so p = 2; q = 3

(b) Answer: a = −8

  1. −5 = a/(−1 + 2) + 3
  2. a = (−5 − 3)(−1 + 2) = −8

(c) Answer: y = x + 5 and y = −x + 1

  1. both pass through the point where the asymptotes meet, (−2; 3)
  2. y = (x + 2) + 3 and y = −(x + 2) + 3

(d) Answer: (−2 − √8; 3 + √8) and (−2 + √8; 3 − √8)

  1. −8/(x + 2) + 3 = −x + 1
  2. (x + 2)^2 = 8
  3. x = −2 ± √8
  4. (−2 − √8; 3 + √8) and (−2 + √8; 3 − √8)

(e) Answer: x ∈ ℝ, x ≠ −2

  1. f is undefined only at x = −2

(f) Answer: −2 < x < 2/3

  1. read from the graph: the x-intercept is x = 2/3 and the asymptote x = −2 is never included
  2. −2 < x < 2/3

(g) Answer: g(x) = −8/(x + 1) + 5

  1. g(x) = f(x − 1) + 2
  2. g(x) = −8/(x + 1) + 5

The hyperbola

3

(a) Answer: x = −3y − 3

  1. 3^(x + 3) · 3^(4y) = 3^(y)
  2. x + 3 + 4y = y
  3. x = −3y − 3

(b) Answer: x = 11

  1. x + 53 = x^2 − 6x + 9
  2. x^2 − 7x − 44 = 0
  3. (x − 11)(x + 4) = 0
  4. check x = −4: √(−4 + 53) = 7 but −4 − 3 = −7, so x = −4 is rejected
  5. x = 11

(c) Answer: x = −3 or x = 6

  1. x^2 − 3x − 16 = 2^1
  2. x^2 − 3x − 18 = 0
  3. (x + 3)(x − 6) = 0
  4. x = −3 or x = 6

(d) Answer: x = 1

  1. let t = 2^x: t^2 + t − 6 = 0
  2. (t − 2)(t + 3) = 0
  3. 2^x = −3 has no solution
  4. 2^x = 2, so x = 1

Exponential equations and exponent laws

4

(a) Answer: f′(x) = −2x − 6

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

(b) Answer: dy/dx = (3/2)x^(1/2) + 2x^(−1/2) − (5/2)x^(−3/2)

  1. y = x^(3/2) + 4x^(1/2) + 5x^(−1/2)
  2. dy/dx = (3/2)x^(1/2) + 2x^(−1/2) − (5/2)x^(−3/2)

(c) Answer: y = −10x − 2

  1. f(2) = −22, so the point is (2; −22)
  2. m = f′(2) = −10
  3. y + 22 = −10(x − 2)
  4. y = −10x − 2

(d) Answer: −12

  1. f(2) = −22 and f(4) = −46
  2. (−46 − (−22))/(4 − (2)) = −12

The derivative from first principles

5

(a) Answer: 11.00%

  1. 1 + 0.1157 = (1 + i/12)^12
  2. i = 12[(1 + 0.1157)^(1/12) − 1]
  3. i = 11.00%

(b) Answer: R577 534.90

  1. i = 11.00%/12 and n = 360
  2. PV = 5500[1 − (1 + i)^(−360)]/i
  3. = R577 534.90

(c) Answer: R92 465.10

  1. deposit = R670 000 − R577 534.90
  2. = R92 465.10

(d) Answer: R439 813.20

  1. after 108 payments: R577 534.90(1 + i)^108 − 5500[(1 + i)^108 − 1]/i = R539 813.20
  2. R539 813.20 − R100 000 = R439 813.20

(e) Answer: 145 payments

  1. 439813.2 = 5500[1 − (1 + i)^(−n)]/i
  2. n = 144.72
  3. 145 payments

Nominal and effective rates

6

(a) Answer: f(x) = −(x + 2)^2 + 6

  1. f(x) = a(x + 2)^2 + 6
  2. substitute (0; 2): 2 = a(2)^2 + 6
  3. a = −1
  4. f(x) = −(x + 2)^2 + 6

(b) Answer: A(−3; 5) and B(3; −19)

  1. −(x + 2)^2 + 6 = −4x − 7
  2. −x^2 + 9 = 0
  3. −(x + 3)(x − 3) = 0
  4. x = −3 or x = 3
  5. A(−3; 5) and B(3; −19)

(c) Answer: x < −3 or x > 3

  1. read from the graph at A and B
  2. x < −3 or x > 3

(d) Answer: y ≤ 6

  1. the turning point is a maximum: y = 6

(e) Answer: t > 6

  1. the line y = t must miss the parabola entirely
  2. t > 6

The parabola

7

(a) Answer: d = 0

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

(b) Answer: (4; −32)

  1. f′(x) = 3x^2 − 12x = 0
  2. 3(x − (0))(x − (4)) = 0
  3. x = 4
  4. f(4) = −32

(c) Answer: x = 2

  1. f′′(x) = 6x − 12 = 0
  2. x = 2

(d) Answer: k < −32 or k > 0

  1. the turning values are −32 and 0
  2. one root when the line y = k passes above the maximum or below the minimum
  3. k < −32 or k > 0

(e) Answer: y = 0

  1. f(0) = 0 and f′(0) = 0
  2. y = 0

Stationary points

8

(a) Answer: 100000

  1. 10 × 10 × 10 × 10 × 10
  2. = 100000

(b) Answer: 30240

  1. 10 × 9 × 8 × 7 × 6
  2. = 30240

(c) Answer: 0.6976

  1. P(nothing repeated) = 30240/100000
  2. P(at least one repeated) = 1 − 30240/100000 = 0.6976

(d) Answer: 25000

  1. first: 5 choices; last: 5 choices; the middle 3: 10 each
  2. 5 × 10 × 10 × 10 × 5 = 25000

(e) Answer: 120

  1. 6 × 5 × 4
  2. = 120

(f) Answer: 40

  1. 2 vowels for the first letter, then 5 × 4
  2. 2 × 20 = 40

The fundamental counting principle

9

(a) Answer: (−4; 0) and (−1; 0)

  1. f(−4) = 0, so (x + 4) is a factor
  2. f(x) = −(x + 4)(x + 1)^2
  3. (−4; 0) and (−1; 0)

(b) Answer: (−3; −4) and (−1; 0)

  1. f′(x) = −3x^2 − 12x − 9
  2. −3x^2 − 12x − 9 = 0
  3. x = −3 or x = −1
  4. (−3; −4) and (−1; 0)

(c) Answer: See the sketch below.

xyO(−4; 0)(−3; −4)(−1; 0)(0; −4)f
  1. y-intercept (0; −4); x-intercepts (−4; 0), (−1; 0)
  2. turning points (−3; −4), (−1; 0)

(d) Answer: x < −2

  1. f′′(x) = −6x − 12
  2. f′′(x) > 0: −6x − 12 > 0
  3. x < −2

(e) Answer: x < −3 or x > −1

  1. f′(x) < 0
  2. x < −3 or x > −1

(f) Answer: −4 < k < 0

  1. the line y = k must cut the graph three times: between the turning values
  2. −4 < k < 0

Stationary points

10

(a) Answer: T_n = −3n + 2

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

(b) Answer: 31 terms

  1. −3n + 2 = −91
  2. −3n = −93
  3. n = 31

(c) Answer: S_(31) = −1426

  1. S_n = n/2(a + l)
  2. S_(31) = 31/2(−1 − 91)
  3. = −1426

(d) Answer: Σ_(n=1)^(31) (−3n + 2)

  1. the terms are T_n = −3n + 2, from n = 1 to n = 31
  2. Σ_(n=1)^(31) (−3n + 2)

(e) Answer: T_n = 3n^2 + 6n − 3

  1. first differences: 15; 21; 27; 33
  2. second difference: 6, so 2a = 6 and a = 3
  3. 3a + b = 15, so b = 6
  4. a + b + c = 6, so c = −3
  5. T_n = 3n^2 + 6n − 3

(f) Answer: No: 2182 is not a term

  1. 3n^2 + 6n − 3 = 2182
  2. 3n^2 + 6n − 2185 = 0
  3. n = 26.01, not a natural number

Arithmetic sequences