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IEB practice · Paper I

IEB Differential calculus practice

21 newly generated questions across all 11 differential calculus topics the IEB examines in Paper I, 259 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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Tangents to a circle

2 questions · 28 marks

Paper I · Section A · 14 marks

QUESTION 1
  1. (a)
    Given f(x) = x^3 − 4x^2 − x + 3. Determine the equation of the tangent to f at the point where x = 2.
    (4)
  2. (b)
    The tangent meets the graph of f again at the point Q. Determine the coordinates of Q.
    (5)
  3. (c)
    Determine the gradient of the graph of f at Q.
    (2)
  4. (d)
    Determine the x-coordinate of the point of inflection of f, and show that it lies one third of the way from x = 2 to x = 0.
    (3)

[14]

Show the answers

(a) Answer: y = −5x + 3

  1. f′(x) = 3x^2 − 8x − 1
  2. m = f′(2) = −5; f(2) = −7
  3. y + 7 = −5(x − 2)

(b) Answer: (0; 3)

  1. x^3 − 4x^2 − x + 3 = −5x + 3
  2. x^3 − 4x^2 + 4x = 0
  3. the tangent touches at x = 2, so (x − 2)^2 is a factor
  4. x(x − 2)^2 = 0
  5. x = 0; y = f(0) = 3

(c) Answer: −1

  1. f′(0) = −1

(d) Answer: x = 4/3

  1. f′′(x) = 6x − 8 = 0: x = 4/3
  2. 2 + (0 − 2)/3 = 4/3

Paper I · Section A · 14 marks

QUESTION 2
  1. (a)
    Given f(x) = 6/x. Determine the equation of the tangent to f at the point where x = 3.
    (4)
  2. (b)
    The tangent cuts the x-axis at A and the y-axis at B. Calculate the area of triangle AOB.
    (3)
  3. (c)
    Show that the tangent at any point (p; 6/p) on the graph, p > 0, cuts off a triangle of the same area.
    (4)
  4. (d)
    Determine the coordinates of the points on the graph of f where the gradient of the tangent is −6.
    (3)

[14]

Show the answers

(a) Answer: y = −(2/3)x + 4

  1. f(x) = 6x^(−1), so f′(x) = −6x^(−2) = −6/x^2
  2. m = f′(3) = −2/3; f(3) = 2
  3. y − 2 = (−2/3)(x − 3)

(b) Answer: 12 square units

  1. A(6; 0) and B(0; 4)
  2. area = (1/2)(6)(4) = 12

(c) Answer: the area is 12 for every p

  1. m = −6/p^2: y − 6/p = (−6/p^2)(x − p)
  2. y = 0: x = 2p; x = 0: y = 12/p
  3. area = (1/2)(2p)(12/p) = 12, whatever p is

(d) Answer: (−1; −6) and (1; 6)

  1. −6/x^2 = −6
  2. x^2 = 1
  3. x = −1 or x = 1

Differentiating by rule

1 questions · 15 marks

Paper I · Section A · 15 marks

QUESTION 3
  1. (a)
    Determine dy/dx if y = (x^3 − 9)/x^2.
    (4)
  2. (b)
    Given f(x) = 2x^3 + 9x^2 − 6. Determine the equation of the tangent to f at the point where x = 0.
    (4)
  3. (c)
    Determine the x-coordinate of the other point on f at which the tangent is parallel to the tangent at x = 0.
    (3)
  4. (d)
    The line y = −12x + k is a tangent to f. Determine the possible values of k.
    (4)

[15]

Show the answers

(a) Answer: dy/dx = 1 + 18x^(−3)

  1. y = x − 9x^(−2)

(b) Answer: y = −6

  1. f′(x) = 6x^2 + 18x
  2. m = f′(0) = 0; f(0) = −6
  3. y + 6 = 0(x)

(c) Answer: x = −3

  1. f′(x) = 0
  2. 6x^2 + 18x = 0
  3. x = 0 or x = −3; the other point: x = −3

(d) Answer: k = −11 or k = −10

  1. f′(x) = −12: 6x^2 + 18x + 12 = 0
  2. x = −1 or x = −2
  3. k = f(x) − (−12)x: 1 − (12) = −11 and 14 − (24) = −10

The derivative from first principles

2 questions · 28 marks

Paper I · Section A · 15 marks

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

[15]

Show the answers

(a) Answer: f′(x) = −6x + 3

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

(b) Answer: dy/dx = 8x^3 + 12x^(−4)

  1. y = 2x^4 − 4x^(−3)

(c) Answer: y = 21x + 29

  1. f(−3) = −34, so the point is (−3; −34)
  2. m = f′(−3) = 21
  3. y + 34 = 21(x + 3)

(d) Answer: −12

  1. f(1) = 2 and f(4) = −34
  2. (−34 − (2))/(4 − (1)) = −12

Paper I · Section A · 13 marks

QUESTION 5
  1. (a)
    Given f(x) = x^2 + 4x − 5. P and Q are the points on the graph of f with x-coordinates x and x + h. Determine the gradient of PQ in terms of x and h, in its simplest form.
    (3)
  2. (b)
    Hence determine f′(x).
    (2)
  3. (c)
    The line y = 6x + k is a tangent to the graph of g(x) = x^2 + 4x − 6. 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 = 0 (the x-axis)?
    (3)

[13]

Show the answers

(a) Answer: 2x + h + 4

  1. gradient of PQ = (f(x + h) − f(x))/((x + h) − x)
  2. f(x + h) = (x + h)^2 + 4(x + h) − 5
  3. f(x + h) − f(x) = 2xh + h^2 + 4h
  4. (2xh + h^2 + 4h)/h = 2x + h + 4

(b) Answer: f′(x) = 2x + 4

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

(c) Answer: (1; −1); k = −7

  1. g′(x) = 2x + 4 = 6
  2. x = 1
  3. g(1) = −1, so the point is (1; −1)
  4. −1 = 6(1) + k, so k = −7

(d) Answer: (−2; −9)

  1. f′(x) = 2x + 4 = 0
  2. x = −2
  3. f(−2) = −9

Stationary points

3 questions · 43 marks

Paper I · Section A · 14 marks

QUESTION 6

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]

Show the answers

(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 − (−2))(x − (4)) = 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

Paper I · Section A · 12 marks

QUESTION 7

f(x) = ax^3 + bx^2 + cx + d. The graph of f cuts the y-axis at (0; 2), has a turning point at (−1; 6) and a point of inflection where x = −2.

  1. (a)
    Write down the value of d.
    (1)
  2. (b)
    Determine the values of a, b and c.
    (6)
  3. (c)
    Determine the coordinates of the other turning point of f.
    (3)
  4. (d)
    For which values of x is the graph of f concave down?
    (2)

[12]

Show the answers

(a) Answer: d = 2

  1. f(0) = d = 2

(b) Answer: a = −1, b = −6 and c = −9

  1. f′′(−2) = 0: 6a(−2) + 2b = 0, so b = 6a
  2. f′(−1) = 0: 3a(−1)^2 + 2b(−1) + c = 0, so c = 9a
  3. f(−1) = 6: a(−1)^3 + b(−1)^2 + c(−1) + 2 = 6
  4. −4a = 4, so a = −1
  5. b = −6 and c = −9

(c) Answer: (−3; 2)

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

(d) Answer: x > −2

  1. f′′(x) = −6x − 12 < 0

Paper I · Section A · 17 marks

QUESTION 8

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)

[17]

Show the answers

(a) Answer: (1; 0) and (4; 0)

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

(b) Answer: (1; 0) and (3; 4)

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

(c) Answer: See the sketch below.

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

(d) Answer: x < 2

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

(e) Answer: x < 1 or x > 3

  1. f′(x) < 0

(f) Answer: 0 < k < 4

  1. the line y = k must cut the graph three times: between the turning values

Sketching a cubic

1 questions · 13 marks

Paper I · Section A · 13 marks

QUESTION 9

Given: f(x) = 2x^3 + ax^2 + bx − 4, where a and b are constants. (x − 2) is a factor of f(x), and when f(x) is divided by (x − 1) the remainder is −9.

  1. (a)
    Determine the values of a and b.
    (5)
  2. (b)
    Hence, factorise f(x) fully.
    (3)
  3. (c)
    Determine the remainder when f(x) is divided by (2x + 3).
    (2)
  4. (d)
    Solve for x if f(x) ≥ 0.
    (3)

[13]

Show the answers

(a) Answer: a = 1 and b = −8

  1. f(2) = 0: 4a + 2b = −12
  2. f(1) = −9: a + b = −7
  3. solving the two together: a = 1 and b = −8

(b) Answer: f(x) = (x − 2)(x + 2)(2x + 1)

  1. f(x) = 2x^3 + x^2 − 8x − 4
  2. = (x − 2)(2x^2 + 5x + 2)
  3. = (x − 2)(x + 2)(2x + 1)

(c) Answer: 7/2

  1. 2x + 3 = 0 at x = −3/2
  2. remainder = f(−3/2) = 7/2

(d) Answer: −2 ≤ x ≤ −1/2 or x ≥ 2

  1. f(x) = 0 at x = −2, −1/2, 2
  2. f changes sign at each root, and f(x) > 0 for large x

Optimisation

9 questions · 98 marks

Paper I · Section B · 9 marks

QUESTION 10

A cone has radius r cm and height h cm, and r + h = 21. (V = (1/3)πr^2h)

hr
  1. (a)
    Show that the volume of the cone is V = (π/3)(21r^2 − r^3).
    (2)
  2. (b)
    Determine the radius that gives the largest volume, and that volume in terms of π.
    (5)
  3. (c)
    Determine the rate at which the volume changes with respect to r when r = 18.
    (2)

[9]

Show the answers

(a) Answer: See the working below.

  1. h = 21 − r
  2. V = (1/3)πr^2(21 − r) = (π/3)(21r^2 − r^3)

(b) Answer: r = 14 cm; V = 1372π/3 cm³

  1. dV/dr = (π/3)(42r − 3r^2) = 0
  2. r(42 − 3r) = 0
  3. r = 14 (r = 0 gives no cone)
  4. V = (π/3)(21(14)^2 − (14)^3) = 1372π/3 cm³

(c) Answer: −72π cm³/cm

  1. dV/dr at r = 18
  2. = −72π cm³/cm

Paper I · Section B · 9 marks

QUESTION 11

A rectangular garden is built against a straight wall. The other three sides are fenced with 80 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 = 80x − 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 = 36.
    (2)

[9]

Show the answers

(a) Answer: See the working below.

  1. the side along the wall is 80 − 2x
  2. A = x(80 − 2x) = 80x − 2x^2

(b) Answer: 20 m by 40 m; area 800 m²

  1. dA/dx = 80 − 4x = 0
  2. x = 20 m
  3. the side along the wall is 40 m
  4. A = 20 × 40 = 800 m²

(c) Answer: −64 m²/m

  1. dA/dx at x = 36
  2. = −64 m²/m

Paper I · Section B · 11 marks

QUESTION 12

f(x) = 24 − 2x^2. P(t; 24 − 2t^2) is a point on the graph of f in the first quadrant, and O is the origin.

xyPfOAB
  1. (a)
    Show that the equation of the tangent to f at P is y = −4tx + 2t^2 + 24.
    (3)
  2. (b)
    The tangent at P cuts the x-axis at A and the y-axis at B. Show that the area of ΔAOB is (2t^2 + 24)^2/(8t).
    (3)
  3. (c)
    Determine the value of t for which the area of ΔAOB is a minimum, and the minimum area.
    (5)

[11]

Show the answers

(a) Answer: See the working below.

  1. f′(x) = −4x, so the gradient at P is −4t
  2. y − (24 − 2t^2) = −4t(x − t)
  3. y = −4tx + 2t^2 + 24

(b) Answer: See the working below.

  1. y = 0: x = (2t^2 + 24)/(4t), so OA = (2t^2 + 24)/(4t)
  2. x = 0: y = 2t^2 + 24, so OB = 2t^2 + 24
  3. area = ½(OA)(OB) = (2t^2 + 24)^2/(8t)

(c) Answer: t = 2; area 64 square units

  1. area = (1/2)t^3 + 12t + 72t^(−1)
  2. d/dt: (3/2)t^2 + 12 − 72t^(−2) = 0
  3. multiply by 8t^2: 12t^4 + 96t^2 − 576 = 0
  4. (6t^2 − 24)(2t^2 + 24) = 0, so t^2 = 4 and t = 2 (t > 0)
  5. the derivative changes from negative to positive at t = 2: a minimum
  6. area = (8 + 24)^2/16 = 64 square units

Paper I · Section B · 9 marks

QUESTION 13

An open box is made from a square sheet of cardboard 18 cm by 18 cm by cutting a square of side x cm from each corner and folding up the sides.

xx
  1. (a)
    Show that the volume of the box is V = 4x^3 − 72x^2 + 324x.
    (2)
  2. (b)
    Determine the value of x for which the volume is a maximum, and the maximum volume.
    (5)
  3. (c)
    Determine the rate at which the volume changes with respect to x when x = 6.
    (2)

[9]

Show the answers

(a) Answer: See the working below.

  1. the base is (18 − 2x) by (18 − 2x) and the height is x
  2. V = x(18 − 2x)^2 = 4x^3 − 72x^2 + 324x

(b) Answer: x = 3 cm; V = 432 cm³

  1. dV/dx = 12x^2 − 144x + 324 = 0
  2. (6x − 18)(2x − 18) = 0
  3. x = 3 (x = 9 gives no box)
  4. V = 3(18 − 6)^2 = 432 cm³

(c) Answer: −108 cm³/cm

  1. dV/dx at x = 6
  2. = −108 cm³/cm

Paper I · Section B · 9 marks

QUESTION 14

A closed cylindrical can must hold 1250 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 + 2500/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)

[9]

Show the answers

(a) Answer: See the working below.

  1. h = 1250/(πr^2)
  2. S = 2πr^2 + 2πr(1250/(πr^2))
  3. S = 2πr^2 + 2500/r

(b) Answer: r = 5.84 cm

  1. dS/dr = 4πr − 2500/r^2 = 0
  2. r^3 = 1250/(2π)

(c) Answer: S = 642.37 cm²

  1. S = 2π(5.8377)^2 + 2500/5.8377

Paper I · Section B · 12 marks

QUESTION 15

A camera on a pole at P watches a road that follows the curve y = x^2 + 3, where the units are hundreds of metres. P is the point (−7; 1). To find where the curve is nearest to P, it is enough to make the SQUARE of the distance as small as possible.

  1. (a)
    Q(x; x^2 + 3) is any point on the curve. Show that the square of the distance PQ is D(x) = x^4 + 5x^2 + 14x + 53.
    (3)
  2. (b)
    Determine the coordinates of the point on the curve nearest to P.
    (5)
  3. (c)
    Hence determine the shortest distance from P to the curve, correct to two decimal places.
    (2)
  4. (d)
    Use D′′(x) to show that x = −1 gives the least distance.
    (2)

[12]

Show the answers

(a) Answer: D(x) = x^4 + 5x^2 + 14x + 53

  1. PQ^2 = (x + 7)^2 + (x^2 + 3 − 1)^2
  2. = x^2 + 14x + 49 + x^4 + 4x^2 + 4
  3. = x^4 + 5x^2 + 14x + 53

(b) Answer: (−1; 4)

  1. D′(x) = 4x^3 + 10x + 14 = 0
  2. D′(−1) = 0, so (x + 1) is a factor
  3. (x + 1)(4x^2 − 4x + 14) = 0, and 4x^2 − 4x + 14 = 0 has no real roots
  4. x = −1, so the point is (−1; 4)

(c) Answer: 6.71 units

  1. D(−1) = 45
  2. PQ = 3√5 = 6.71

(d) Answer: D′′(−1) = 22 > 0, so a minimum

  1. D′′(x) = 12x^2 + 10
  2. D′′(−1) = 22 > 0: the graph of D is concave up there

Paper I · Section B · 13 marks

QUESTION 16

A grain silo is a cylinder of radius r metres and height h metres, with a flat circular floor and a hemispherical roof. It must hold 300 m³. The floor costs R150 per square metre to build, the curved wall R300 per square metre and the roof R450 per square metre. (Volume of a sphere = (4/3)πr^3; surface area of a sphere = 4πr^2)

rh
  1. (a)
    Show that h = 300/(πr^2) − (2/3)r.
    (3)
  2. (b)
    Show that the cost of building the silo is C = 180 000/r + 650πr^2 rand.
    (3)
  3. (c)
    Determine, correct to two decimal places, the radius that makes the cost a minimum.
    (4)
  4. (d)
    Calculate the minimum cost, to the nearest rand, and the height h of the cylinder for it, correct to two decimal places.
    (3)

[13]

Show the answers

(a) Answer: See the working below.

  1. the roof is half a sphere
  2. πr^2h + (2/3)πr^3 = 300
  3. h = 300/(πr^2) − (2/3)r

(b) Answer: See the working below.

  1. floor 150(πr^2); wall 300(2πrh); roof 450(2πr^2)
  2. C = 150πr^2 + 600πr(300/(πr^2) − (2/3)r) + 900πr^2
  3. = 150πr^2 + 180 000/r − 400πr^2 + 900πr^2
  4. = 180 000/r + 650πr^2

(c) Answer: r = 3.53 m

  1. dC/dr = −180 000/r^2 + 1300πr = 0
  2. r^3 = 90 000/(650π)

(d) Answer: C = R76 437; h = 5.30 m

  1. C = 180 000/3.5323 + 650π(3.5323)^2 = R76 437.07
  2. h = 300/(π(3.5323)^2) − (2/3)(3.5323) = 5.30 m

Paper I · Section B · 13 marks

QUESTION 17

P(x; y) is a point on the curve y = −x^2 + 12 in the first quadrant, and Q is the point on the x-axis directly below P. O is the origin.

OPQxy
  1. (a)
    Show that the area of triangle OPQ is A(x) = −(1/2)x^3 + 6x.
    (2)
  2. (b)
    Determine the coordinates of P for which the area of triangle OPQ is a maximum, and that maximum area.
    (6)
  3. (c)
    For which values of x does the area of triangle OPQ increase as x increases?
    (2)
  4. (d)
    Determine the equation of the tangent to the curve at this point P.
    (3)

[13]

Show the answers

(a) Answer: A(x) = −(1/2)x^3 + 6x

  1. OQ = x and PQ = −x^2 + 12
  2. A = (1/2)x(−x^2 + 12) = −(1/2)x^3 + 6x

(b) Answer: P(2; 8); the area is 8 square units

  1. A′(x) = −(3/2)x^2 + 6 = 0
  2. x = 2 (x > 0)
  3. y = 8
  4. A(2) = 8

(c) Answer: 0 < x < 2

  1. A′(x) > 0 between x = 0 and x = 2

(d) Answer: y = −4x + 16

  1. dy/dx = −2x = −4 at x = 2
  2. y − 8 = −4(x − 2)

Paper I · Section B · 13 marks

QUESTION 18

Two toy trains run on parallel straight tracks. t seconds after a stopwatch is started, train A is s_A(t) = t^3 − 13t^2 + 62t centimetres past a marker, and train B is s_B(t) = 18t + 32 centimetres past it, for 0 ≤ t ≤ 9.

  1. (a)
    Calculate the velocity of train A when t = 6.
    (2)
  2. (b)
    At what time is train A moving at its slowest, and what is its velocity then? Give your answers correct to two decimal places where necessary.
    (3)
  3. (c)
    Show that the trains are side by side when t = 1, and determine the other times when they are side by side.
    (4)
  4. (d)
    Train B is ahead again between t = 4 and t = 8. Determine, correct to two decimal places, when in this time train B's lead is greatest, and that lead.
    (4)

[13]

Show the answers

(a) Answer: 14 cm/s

  1. v_A(t) = s_A′(t) = 3t^2 − 26t + 62
  2. v_A(6) = 14 cm/s

(b) Answer: t = 4.33 s; 5.67 cm/s

  1. a_A(t) = v_A′(t) = 6t − 26 = 0 when t = 13/3 ≈ 4.33
  2. a_A changes from negative to positive there, so the velocity is least
  3. v_A(13/3) = 17/3 ≈ 5.67 cm/s

(c) Answer: t = 4 s and t = 8 s

  1. s_A(t) − s_B(t) = t^3 − 13t^2 + 44t − 32
  2. at t = 1 this is 0, so (t − 1) is a factor
  3. (t − 1)(t^2 − 12t + 32) = 0
  4. t = 4 or t = 8

(d) Answer: t = 6.36 s; a lead of 20.75 cm

  1. B's lead = s_B(t) − s_A(t) = −t^3 + 13t^2 − 44t + 32
  2. its derivative: −3t^2 + 26t − 44 = 0
  3. t = 6.36 (the root between 4 and 8)
  4. lead = 20.75 cm

Reading the graph of f'

2 questions · 22 marks

Paper I · Section B · 12 marks

QUESTION 19

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

xyOC(1; 0)(0; −1)f′
  1. (a)
    Determine the equation of f′.
    (3)
  2. (b)
    Given that f(0) = −7, determine f(x).
    (4)
  3. (c)
    Is x = 1 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)

[12]

Show the answers

(a) Answer: f′(x) = −x^2 + 2x − 1

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

(b) Answer: f(x) = −(1/3)x^3 + x^2 − x − 7

  1. f(x) = ax^3 + bx^2 + cx + d with f′(x) = 3ax^2 + 2bx + c = −x^2 + 2x − 1
  2. a = −1/3, b = 1, c = −1
  3. f(0) = d = −7

(c) Answer: A point of inflection

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

(d) Answer: x = 0 or x = 2

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

Paper I · Section B · 10 marks

QUESTION 20

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 (1; 0) and the y-axis at (0; 1/2).

xyO(−1; 0)(1; 0)(0; 1/2)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]

Show the answers

(a) Answer: −1 < x < 1

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

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

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

(c) Answer: x = 0

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

(d) Answer: x < 0

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

(e) Answer: 1/2

  1. the gradient of f at x = 0 is f′(0) = 1/2

Average gradient and the limit

1 questions · 12 marks

Paper I · Section A · 12 marks

QUESTION 21
  1. (a)
    A water rocket is launched straight up from a platform. Its height above the ground t seconds later is h(t) = −5t^2 + 15t + 20 metres, until it lands. From what height above the ground does it start?
    (1)
  2. (b)
    Calculate the average rate of change of its height over the first 2 seconds.
    (2)
  3. (c)
    Calculate the rate at which its height is changing at the moment t = 3 s, and say whether it is rising or falling then.
    (3)
  4. (d)
    Determine the greatest height it reaches, and when.
    (3)
  5. (e)
    At what speed does it hit the ground?
    (3)

[12]

Show the answers

(a) Answer: 20 m

  1. h(0) = 20

(b) Answer: 5 m/s

  1. (h(2) − h(0))/(2 − 0) = (30 − 20)/2
  2. = 5 m/s

(c) Answer: −15 m/s: falling

  1. h′(t) = −10t + 15
  2. h′(3) = −15
  3. negative: falling

(d) Answer: 31.25 m, after 1.5 s

  1. h′(t) = −10t + 15 = 0
  2. t = 1.5
  3. h(1.5) = 31.25

(e) Answer: 25 m/s

  1. h(t) = 0: −5(t − 4)(t + 1) = 0, so t = 4 (t > 0)
  2. h′(4) = −25
  3. the speed is 25 m/s

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