Differentiation: Additional Mathematics Tingkatan 5 questions

Section A: objective questions

11 marks

Choose the correct answer. Each question carries 1 mark.

Find lim_x→3 x² − 9x − 3.

0
6
9
The limit does not exist

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Direct substitution gives 00, so factorise: (x − 3)(x + 3)x − 3 = x + 3. At x = 3 this is 6. The limit exists even though the expression is undefined at x = 3.

For the teacher: Remembering, low difficulty, standard 2.1.1

Textbook p. 31

Which expression defines the first derivative dydx from first principles?

δyδx
δy × δx
lim_δx→∞ δyδx
lim_δx→0 δyδx

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The gradient of the chord δyδx becomes the gradient of the tangent when δx tends to 0. δyδx alone is only the gradient of a chord.

For the teacher: Remembering, low difficulty, standard 2.1.2

Textbook p. 35

Given y = 6√x, find dydx.

3√x
−3x²
−3x√x
−12√x

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y = 6x^−1/2, so dydx = −3x^−3/2 = −3x√x. The power must go down by 1, from −1/2 to −3/2.

For the teacher: Applying, low difficulty, standard 2.2.2

Textbook p. 41

Differentiate (3x − 2)⁵ with respect to x.

5(3x − 2)⁴
3(3x − 2)⁴
15(3x − 2)⁴
15(3x − 2)⁵

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Chain rule: 5(3x − 2)⁴ × 3 = 15(3x − 2)⁴. The first option forgets the derivative of the bracket.

For the teacher: Applying, low difficulty, standard 2.2.3

Textbook p. 43

Given y = x²(2x + 1)³, find dydx.

2x(2x + 1)²(5x + 1)
6x²(2x + 1)²
2x(2x + 1)³
2x(2x + 1)²(x + 1)

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Product rule: 2x(2x + 1)³ + x² · 6(2x + 1)² = 2x(2x + 1)²[(2x + 1) + 3x] = 2x(2x + 1)²(5x + 1).

For the teacher: Applying, low difficulty, standard 2.2.4

Textbook p. 44

Find the gradient of the curve y = 2xx − 1 at x = 3.

−2
12
2
−12

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Quotient rule: dydx = (x − 1)(2) − 2x(1)(x − 1)² = −2(x − 1)². At x = 3 this is −24 = −12.

For the teacher: Applying, medium difficulty, standard 2.2.4

Textbook p. 45

Given y = x⁴ − 3x², find the value of d²ydx² at x = 2.

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dydx = 4x³ − 6x and d²ydx² = 12x² − 6. At x = 2: 12(4) − 6 = 42. 20 is the first derivative 4x³ − 6x evaluated at x = 2.

For the teacher: Applying, low difficulty, standard 2.3.1

Textbook p. 49

The gradient of the normal to the curve y = 2x² − 3 at x = 1 is

−14
4
−4
14

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dydx = 4x, so the tangent gradient at x = 1 is 4. The normal is perpendicular: m = −14. The answer 4 is the tangent's gradient.

For the teacher: Understanding, low difficulty, standard 2.4.2

Textbook p. 53

Find the turning point of the curve y = x² − 6x + 5 and its nature.

(3, 4), maximum
(−3, 32), minimum
(6, 5), minimum
(3, −4), minimum

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dydx = 2x − 6 = 0 gives x = 3 and y = 9 − 18 + 5 = −4. d²ydx² = 2 > 0, so it is a minimum point.

For the teacher: Applying, low difficulty, standard 2.4.4

Textbook p. 58

A curve has equation y = x² + 1. The value of x increases at a rate of 2 units per second. Find the rate of change of y when x = 3.

6 units per second
10 units per second
12 units per second
20 units per second

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dydt = dydx × dxdt = 2x × 2 = 4x. At x = 3 it is 12 units per second. 6 forgets the factor dxdt.

For the teacher: Applying, low difficulty, standard 2.4.6

Textbook p. 66

Given y = x³, find the approximate change in y when x increases from 2 to 2.01.

0.06
0.12
0.24
0.01

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dydx = 3x² = 12 at x = 2 and δx = 0.01, so δy ≈ 12 × 0.01 = 0.12.

For the teacher: Applying, medium difficulty, standard 2.4.8

Textbook p. 71

Section B: structured questions

34 marks

Answer all questions. Show your working.

A curve has equation y = 2x² − 3x.

[3 marks] (a)Use first principles to find dydx.

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δy = 2(x + δx)² − 3(x + δx) − (2x² − 3x) = 4xδx + 2(δx)² − 3δx [1]1
δyδx = 4x + 2δx − 3 [1]1
dydx = lim_δx→0 (4x + 2δx − 3) = 4x − 3 [1]1

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[1 mark] (b)Find the gradient of the tangent at x = 5.

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4(5) − 3 = 17 [1]1

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For the teacher: Understanding, low difficulty, standard 2.1.2

Textbook p. 36

Find the equation of the tangent and the equation of the normal to the curve y = x³ − 6x + 2 at the point where x = 2.

[2 marks] (a)Find the y-coordinate and the gradient of the tangent at x = 2.

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y = 8 − 12 + 2 = −2 [1]1
dydx = 3x² − 6 = 6 [1]1

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[1 mark] (b)Find the equation of the tangent.

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y + 2 = 6(x − 2), so y = 6x − 14 [1]1

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[2 marks] (c)Find the equation of the normal.

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Gradient of normal = −16 [1]1
y + 2 = −16(x − 2), so x + 6y + 10 = 0 [1]1

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For the teacher: Applying, low difficulty, standard 2.4.2

Textbook p. 53

Diagram 1 shows the curve y = x³ − 6x² + 9x. The curve has two turning points, P and Q.

Cubic curve y = x cubed minus 6x squared plus 9x with a turning point P on the left above the x-axis and a turning point Q on the x-axis to the right of P — Diagram 1

[3 marks] (a)Find the coordinates of P and Q.

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dydx = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 [1]1
x = 1, y = 4, so P(1, 4) [1]1
x = 3, y = 0, so Q(3, 0) [1]1

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[2 marks] (b)Use the second derivative to determine whether each point is a maximum or minimum.

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d²ydx² = 6x − 12. At x = 1 it is −6 < 0, so P is a maximum [1]1
At x = 3 it is 6 > 0, so Q is a minimum [1]1

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For the teacher: Analysing, medium difficulty, standard 2.4.4

Textbook p. 59

A closed cylindrical container has base radius r cm and height h cm. Its volume is 128π cm³.

[2 marks] (a)Show that the total surface area is A = 2πr² + 256πr.

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πr²h = 128π, so h = 128r² [1]1
A = 2πr² + 2πrh = 2πr² + 256πr [1]1

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[2 marks] (b)Find the value of r for which A is a minimum.

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dAdr = 4πr − 256πr² = 0 [1]1
r³ = 64, so r = 4 [1]1

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[1 mark] (c)Show that A is a minimum at this value of r.

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d²Adr² = 4π + 512πr³ = 12π > 0 at r = 4 [1]1

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[1 mark] (d)Find the minimum surface area, in terms of π.

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A = 2π(16) + 256π4 = 96π cm² [1]1

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For the teacher: Applying, medium difficulty, standard 2.4.5

Textbook p. 63

A circular oil patch spreads so that its radius, r cm, increases at a constant rate of 0.2 cm s⁻¹. The area of the patch is A cm².

[1 mark] (a)Write down dAdr.

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A = πr², dAdr = 2πr [1]1

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[2 marks] (b)Find the rate of change of the area when r = 5, correct to 4 significant figures.

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dAdt = dAdr × drdt = 2π(5)(0.2) [1]1
= 2π = 6.283 cm² s⁻¹ [1]1

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[1 mark] (c)Find the radius when the area is increasing at 1.2π cm² s⁻¹.

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2πr(0.2) = 1.2π, so r = 3 cm [1]1

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For the teacher: Applying, low difficulty, standard 2.4.7

Textbook p. 69

Use differentiation to answer the following.

[3 marks] (a)Given y = √x, find the approximate value of √25.2.

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dydx = 12√x = 110 at x = 25 [1]1
δx = 0.2, δy ≈ 110 × 0.2 = 0.02 [1]1
√25.2 ≈ 5 + 0.02 = 5.02 [1]1

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[3 marks] (b)Given y = 3x² − 2x + 4, x = 2 increases by 2%. Find the corresponding percentage change in y.

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δx = 0.02 × 2 = 0.04, dydx = 6x − 2 = 10 [1]1
δy ≈ 10 × 0.04 = 0.4; y = 12 − 4 + 4 = 12 [1]1
0.412 × 100 = 3.33% increase [1]1

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[1 mark] (c)Aini says that a 2% change in x always gives a 2% change in y. Is she correct? Use your answer to (b).

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No. y changed by 3.33%, not 2%; the percentage change in y depends on dydx and y at that point [1]1

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For the teacher: Evaluating, high difficulty, standard 2.4.8

Textbook p. 72

You want a quadratic curve with a maximum turning point at x = 2.

[1 mark] (a)Write down a possible equation y = f(x) for such a curve.

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Any quadratic with negative x² coefficient and dydx = 0 at x = 2, e.g. y = −x² + 4x + 1 [1]1

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[2 marks] (b)Show, by differentiation, that your curve has a maximum point at x = 2 and find its y-coordinate.

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For y = −x² + 4x + 1: maximum at (2, 5).

dydx = −2x + 4 = 0 gives x = 2, and d²ydx² = −2 < 0 so it is a maximum [1]1
y = −4 + 8 + 1 = 5 (for the example curve) [1]1

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For the teacher: Creating, medium difficulty, standard 2.4.4

Textbook p. 61

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Section B: structured questions

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18