A question will appear on the screen along with a student’s solution.

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Can you find all of the mistakes and correct them?

Year 2 Pure

PURE MATHS Contents

• Partial Fractions – SLIDE 3

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• Proof by contradiction – SLIDE 5

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• Functions – SLIDE 7

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• Arcs and sectors – SLIDE 9

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• Trigonometry: Using identities/solving equations – SLIDE 11

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• Trigonometry: Harmonic form – SLIDE 13

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• Trigonometry: Compound angle formulae – SLIDE 15

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• Trigonometry: Inverse functions – SLIDE 17

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• Differentiation: Chain rule – SLIDE 19

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• Differentiation: Product rule – SLIDE 21

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• Differentiation: Implicit – SLIDE 23

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• Differentiation: Parametric equations – SLIDE 25

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• Integration: By substitution – SLIDE 27

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• Integration: Parametric equations – SLIDE 29

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• Integration: By parts – SLIDE 31

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• Differential equations – SLIDE 33

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• Numerical methods: Trapezium rule – SLIDE 35

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• Numerical methods: Iteration & roots – SLIDE 37

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• Numerical methods: Newton-Raphson – SLIDE 39

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• Binomial expansion – SLIDE 41

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• Sequences and series - SLIDE 43

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• Vectors - SLIDE 45

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1. Arc length = rθ = 5 x 28 = 140 m

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• Area of sector = ½ r ² θ = = ½ x 25 x 28 = 350

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• MC = 2.5 m , therefore BM = 2.5 m

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So perimeter = 2.5 + 2.5 + 140 = 145

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d) Area of triangle ABM = ½ absinC

= ½ x 5 x 2.5 x sin28 = 2.93

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So shaded area = area of sector – area of triangle

= 350 – 2.93 = 347.07

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CORRECT SOLUTION

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b) Area of sector =

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c) AM = MC = 2.5 m

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Using cosine rule,

BM² = 5² + 2.5² - 2 x 5 x 2.5 x cos28°

BM = 3.03 m (3.s.f)

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So perimeter = BM + MC + BC = 3.03 + 2.5 + 2.44 = 7.97 m (3.s.f)

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d)

Area of triangle ABM = ½absinC = ½ x 5 x 2.5 x sin28°

= 2.93 m² (3.s.f)

Area of shaded region = area of sector – area of triangle

= 6.11 – 2.93 = 3.18 m² (3.s.f.)

Arc length =

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1. 1+tan²x = sec²x

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So = 1 +

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Therefore LHS = 1 + + = 1 + 2

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b) 1 + 2tan²x = 3

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2tan²x = 2

tan²x = 1

So tanx = 1

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x = 45°

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a)

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If tanx = 1, x = 45°, 225°

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If tanx = -1, x = 135°, 315°

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b) Max points occur where dy/dx =0

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Using chain rule,

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Smallest positive value of x is 0.983 (3.d.p)

c) Divide by cosx,

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2tanx - 3 = 1

2tanx = 4

tanx = 2

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x = 1.107, 4.249 (3.d.p)

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CORRECT SOLUTION

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a)

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Hence 2sinx – 3cosx =

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b) The greatest value of = 1

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Hence the greatest value of =

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This maximum occurs when

Therefore,

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c) = 1

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Give your answer in the form ax + by = c ,

where a, b and c are integer values.

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Give your answer in the form ax + by = c ,

where a, b and c are integer values.

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(your answer should be in terms of x only)

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(your answer should be in terms of x only)

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Using product rule

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Give your answer in its simplest form.

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