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IEB practice · Paper I

IEB Differential calculus practice

8 newly generated questions across all 11 differential calculus topics the IEB examines in Paper I, 100 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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The derivative from first principles

2 questions · 29 marks

116 marks
  1. (a)
    Given f(x) = −2x^2 + 4x + 6. Determine f′(x) from first principles.
    (5)
  2. (b)
    Determine dy/dx if y = (x^2 − 6x − 4)/√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 = 2.
    (3)
Show the answers

(a) Answer: f′(x) = −4x + 4

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

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

  1. y = x^(3/2) − 6x^(1/2) − 4x^(−1/2)
  2. dy/dx = (3/2)x^(1/2) − 3x^(−1/2) + 2x^(−3/2)

(c) Answer: y = 12x + 14

  1. f(−2) = −10, so the point is (−2; −10)
  2. m = f′(−2) = 12
  3. y + 10 = 12(x + 2)
  4. y = 12x + 14

(d) Answer: −2

  1. f(1) = 8 and f(2) = 6
  2. (6 − (8))/(2 − (1)) = −2
213 marks
  1. (a)
    Given f(x) = 2x^2 − 2x + 1. 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 = −7x + k is a tangent to the graph of g(x) = x^2 − 3x + 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 = −6x?
    (3)
Show the answers

(a) Answer: 4x + 2h − 2

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

(b) Answer: f′(x) = 4x − 2

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

(c) Answer: (−2; 17); k = 3

  1. g′(x) = 2x − 3 = −7
  2. x = −2
  3. g(−2) = 17, so the point is (−2; 17)
  4. 17 = −7(−2) + k, so k = 3

(d) Answer: (−1; 5)

  1. f′(x) = 4x − 2 = −6
  2. x = −1
  3. f(−1) = 5

Stationary points

2 questions · 31 marks

317 marks

Given: f(x) = x^3 − 9x^2 + 24x − 20

xyO
  1. (a)
    Show that x = 5 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 increasing?
    (2)
  6. (f)
    For which values of k will f(x) = k have three different real roots?
    (2)
Show the answers

(a) Answer: (2; 0) and (5; 0)

  1. f(5) = 0, so (x − 5) is a factor
  2. f(x) = (x − 5)(x − 2)^2
  3. (2; 0) and (5; 0)

(b) Answer: (2; 0) and (4; −4)

  1. f′(x) = 3x^2 − 18x + 24
  2. 3x^2 − 18x + 24 = 0
  3. x = 2 or x = 4
  4. (2; 0) and (4; −4)

(c) Answer: See the sketch below.

xyO(5; 0)(2; 0)(4; −4)(0; −20)f
  1. y-intercept (0; −20); x-intercepts (2; 0), (5; 0)
  2. turning points (2; 0), (4; −4)

(d) Answer: x > 3

  1. f′′(x) = 6x − 18
  2. f′′(x) > 0: 6x − 18 > 0
  3. x > 3

(e) Answer: x < 2 or x > 4

  1. f′(x) > 0
  2. x < 2 or x > 4

(f) Answer: −4 < k < 0

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

In the diagram below, the graph of f(x) = x^3 − 9x^2 + 24x + d is drawn, where d is a constant. (4; 0) is a stationary point of f.

xyO(4; 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)
Show the answers

(a) Answer: d = −16

  1. f(4) = 0
  2. 16 + d = 0
  3. d = −16

(b) Answer: (2; 4)

  1. f′(x) = 3x^2 − 18x + 24 = 0
  2. 3(x − (2))(x − (4)) = 0
  3. x = 2
  4. f(2) = 4

(c) Answer: x = 3

  1. f′′(x) = 6x − 18 = 0
  2. x = 3

(d) Answer: k < 0 or k > 4

  1. the turning values are 0 and 4
  2. one root when the line y = k passes above the maximum or below the minimum
  3. k < 0 or k > 4

(e) Answer: y = 24x − 16

  1. f(0) = −16 and f′(0) = 24
  2. y = 24x − 16

Optimisation

2 questions · 18 marks

59 marks

An open box is made from a square sheet of cardboard 36 cm by 36 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 − 144x^2 + 1296x.
    (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 = 17.
    (2)
Show the answers

(a) Answer: See the working below.

  1. the base is (36 − 2x) by (36 − 2x) and the height is x
  2. V = x(36 − 2x)^2 = 4x^3 − 144x^2 + 1296x

(b) Answer: x = 6 cm; V = 3456 cm³

  1. dV/dx = 12x^2 − 288x + 1296 = 0
  2. (6x − 36)(2x − 36) = 0
  3. x = 6 (x = 18 gives no box)
  4. V = 6(36 − 12)^2 = 3456 cm³

(c) Answer: −132 cm³/cm

  1. dV/dx at x = 17
  2. = −132 cm³/cm
69 marks

A cone has radius r cm and height h cm, and r + h = 12. (V = (1/3)πr^2h)

hr
  1. (a)
    Show that the volume of the cone is V = (π/3)(12r^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 = 6.
    (2)
Show the answers

(a) Answer: See the working below.

  1. h = 12 − r
  2. V = (1/3)πr^2(12 − r) = (π/3)(12r^2 − r^3)

(b) Answer: r = 8 cm; V = 256π/3 cm³

  1. dV/dr = (π/3)(24r − 3r^2) = 0
  2. r(24 − 3r) = 0
  3. r = 8 (r = 0 gives no cone)
  4. V = (π/3)(12(8)^2 − (8)^3) = 256π/3 cm³

(c) Answer: 12π cm³/cm

  1. dV/dr at r = 6
  2. = 12π cm³/cm

Reading the graph of f'

2 questions · 22 marks

710 marks

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

xyO(−5; 0)(1; 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)
Show the answers

(a) Answer: −5 < x < 1

  1. f is increasing where f′(x) > 0: where the graph of f′ lies above the x-axis
  2. −5 < x < 1

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

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

(c) Answer: x = −2

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

(d) Answer: x < −2

  1. f is concave up where f′′(x) > 0: where f′ is increasing
  2. x < −2

(e) Answer: 5

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

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; 3).

xyOC(−1; 0)(0; 3)f′
  1. (a)
    Determine the equation of f′.
    (3)
  2. (b)
    Given that f(0) = −1, 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 12.
    (3)
Show the answers

(a) Answer: f′(x) = 3x^2 + 6x + 3

  1. f′(x) = a(x − (−1))^2, touching the x-axis at C
  2. 3 = a(−1)^2, so a = 3
  3. f′(x) = 3x^2 + 6x + 3

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

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

(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 = −3 or x = 1

  1. 3(x − (−1))^2 = 12
  2. (x − (−1))^2 = 4
  3. x = −3 or x = 1

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