1a. Show that if log2 3 = 𝑝𝑝

𝑞𝑞, then 2p = 3q. (2)

b. Use proof by contradiction to prove that log2 3 is irrational. (3) __________________________________________________________________________________________

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2. Use proof by contradiction to prove that there are no positive integers, x and y, such that x2 – y2 = 1 (5) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Proof by Contradiction Chapter Reference: Pure 2, Chapter 1

6 minutes

Solutions

1a. log2 3 = 𝑝𝑝

𝑞𝑞

(2𝑝𝑝𝑞𝑞)𝑞𝑞 = 3q

M1

2p = 3q M1 1b.

Assume that log2 3 is rational then, log2 3 = 𝑝𝑝

𝑞𝑞

2p = 3q M1

If 2 and 3 a re co prime, p = q = 0 M1

Therefore, by contradiction, log2 3 is irrational. M1 2.

Assume x2 – y2 = 1 (x + y)(x – y) = 1 M1

x + y = 1 x – y = 1 M1

2x = 2 x = 1 M1

y = 0 M1 Therefore, proof by contradiction there are no positive integer solutions. M1

1. Simplify 𝑥4−5𝑥2+4

𝑥2−𝑥−2(2)

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2. Express as a single fraction in its simplest form, 1

𝑥−3+

3

𝑥2−3𝑥+

𝑥

𝑥2−6𝑥+9(3)

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3a. Express 2

𝑥+5 +

3

(𝑥+2)(𝑥+5) as a single fraction in its simplest form. (2)

b. Hence solve the equation, 2

𝑥+5 +

3

(𝑥+2)(𝑥+5) =

1

3, giving your answers to 2 decimal places. (3)

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A-LevelStarterActivity

Topic: Algebraic Fractions Chapter Reference: Pure 2, Chapter 1

7

minutes

Solutions

1. (𝑥2−1)(𝑥2−4)

(𝑥+1)(𝑥−2)= (𝑥+1)(𝑥−1)(𝑥+2)(𝑥−2)

(𝑥+1)(𝑥−2)M1

= (x – 1)(x + 2) M1

2. 1

𝑥−3+

3

𝑥2−3𝑥+

𝑥

𝑥2−6𝑥+9= 𝑥(𝑥−3)+3(𝑥−3)+𝑥2

𝑥(𝑥−3)2M1

= 2𝑥2−9

𝑥(𝑥−3)2M1

3a. 2

𝑥+5 +

3

(𝑥+2)(𝑥+5) =

2(𝑥+2)+3

(𝑥+2)(𝑥+5)M1

= 2𝑥+7

(𝑥+2)(𝑥+5)M1

3b. 2𝑥+7

(𝑥+2)(𝑥+5)= 1

3M1

3(2x + 7) = (x + 2)(x + 5)

x2 + x – 11 = 0 M1

x = -3.85

x = 2.85 M1

1. Find the value of A, B and C when, (3) 2 – 9x = A(2x – 1)2 + B(x – 3)(2x – 1) + C(x – 3)

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2. Express 9𝑥𝑥2−2𝑥𝑥−12

𝑥𝑥(𝑥𝑥+3)(𝑥𝑥−2) in partial fractions (4)

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3. Express 13−3𝑥𝑥2

(2𝑥𝑥+3)(𝑥𝑥−1)2 in partial fractions (4)

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A-LevelStarterActivity

Topic: Partial Fractions Chapter Reference: Pure 2, Chapter 1

8 minutes

Solutions

1. 2 – 9x = A(2x – 1)2 + B(x – 3)(2x – 1) + C(x – 3) Let x = 1

2

-2.5 = A(0) + B(0) + C(-2.5) C = 1

M1

Let x = 3 -25 = A(25) + B(0) + C(0) A = -1

M1

Let x = 0, A = -1, C = -1 B = 2 M1

A = -1, B = 2, C = 1 2.

9x2 – 2x – 12 = A(x + 3)(x – 2) + Bx(x – 2) + Cx(x + 3) M1 Let x = 0, -12 = A(-6) A = 2

M1

Let x = -3, 75 = 15B B = 5

M1

Let x = 2, 20 = 10C, C = 2

M1

9𝑥𝑥2−2𝑥𝑥−12𝑥𝑥(𝑥𝑥+3)(𝑥𝑥−2)

= 2𝑥𝑥

+ 5𝑥𝑥+3

+ 2𝑥𝑥−2

3.

13 – 3x2 = A(x – 1)2 + B(2x + 3)(x – 1) + C(2x + 3) M1 Let x = -3

2,

254

= 254𝐴𝐴

A = 1

M1

Let x = 1, 10 = 5C C = 2

M1

Coefficients of x2: -3 = A + 2B B = -2 M1

13−3𝑥𝑥2

(2𝑥𝑥+3)(𝑥𝑥−1)2 = 1

2𝑥𝑥+3− 2

𝑥𝑥−1+ 2

(𝑥𝑥 −1)2

1. Express f(x) in partial fractions when f(x) = 2𝑥𝑥−1

(𝑥𝑥−1)(2𝑥𝑥−3) (3)

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2. Express 3𝑥𝑥+2

𝑥𝑥2−2𝑥𝑥−24 in partial fractions (4)

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3. Find the values of the constants, A, B, and C in the identity (4)

3x2 + 17x – 32 = A(x – 1)(x + 3) + B(x – 1)(x – 4) + C(x + 3)(x – 4) __________________________________________________________________________________________

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4. Express in partial fractions 2𝑥𝑥2+4

𝑥𝑥(𝑥𝑥−1)(𝑥𝑥−4) (4)

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A-Level Starter Activity

Topic: Partial Fractions Chapter Reference: Pure 2, Chapter 1

8 minutes

Solutions 1. f(x) = 2𝑥𝑥−1

(𝑥𝑥−1)(2𝑥𝑥−3) = 𝐴𝐴

𝑥𝑥−1+ 𝐵𝐵

2𝑥𝑥−3

2x – 1 = A(2x – 3) + B(x – 1) M1

When x = 1, 1 = A(-1) + B(0) A = 1

M1

When x = 32

2 = A(0) – B(-0.5) B = 4

M1

f(x) = 2𝑥𝑥−1(𝑥𝑥−1)(2𝑥𝑥−3)

= 1𝑥𝑥−1

+ 42𝑥𝑥−3

2.

3𝑥𝑥+2𝑥𝑥2−2𝑥𝑥−24

= 3𝑥𝑥+2(𝑥𝑥−6)(𝑥𝑥+4)

= 𝐴𝐴𝑥𝑥−6

+ 𝐵𝐵𝑥𝑥+4

M1 3x + 2 = A(x + 4) + B(x – 6) M1 When x = -4, -10 = A(0) + B(-10) B = 1

M1

When x = 6 20 = A(10) + B(0) A = 2

M1

3𝑥𝑥+2𝑥𝑥2−2𝑥𝑥−24

= 3𝑥𝑥+2(𝑥𝑥−6)(𝑥𝑥+4)

= 2𝑥𝑥−6

+ 1𝑥𝑥+4

3.

3x2 + 17x – 32 = A(x – 1)(x + 3) + B(x – 1)(x – 4) + C(x + 3)(x – 4) M1 Let x = 1, -12 = A(0) + B(0) + C(-12) C = 1

M1

Let x = -3, -56 = A(0) + B(28) + C(0) B = -2

M1

Let x = 4, 84 = A(21) + B(0) + C(0) A = 4

M1

A = 4, B = -2, C = 1 4.

2𝑥𝑥2+4𝑥𝑥(𝑥𝑥−1)(𝑥𝑥−4)

= 𝐴𝐴𝑥𝑥

+ 𝐵𝐵𝑥𝑥−1

+ 𝐶𝐶𝑥𝑥−4

2𝑥𝑥2 + 4 = A(x – 1)(x – 4) + B(x)(x – 4) + C(x)(x – 1)

M1

Let x = 0, 4 = A(-1)(-4) + B(0) + C(0) A = 1

M1

Let x = 1, 6 = A(0) + B(-3) + C(0) B = 2

M1

Let x = 4, 36 = A(0) + B(0) + C(12) C = 3

M1

2𝑥𝑥2+4𝑥𝑥(𝑥𝑥−1)(𝑥𝑥−4)

= 1𝑥𝑥

+ 2𝑥𝑥−1

+ 3𝑥𝑥−4

1. Find the quotient obtained by dividing (3x3 + 16x2 + 72) by (x + 6) (3) __________________________________________________________________________________________

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2. Find the quotient and the remained obtained by dividing (1 – 22x2 – 6x3) by (x + 2) (4) __________________________________________________________________________________________

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3. f(x) = 6x3

– 7x2 -71x + 12. Given that f(4) = 0, find all solutions to the equation f(x) = 0 (5) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Algebraic Division Chapter Reference: Pure 2, Chapter 1

8 minutes

Solutions

1. Term ‘3x2’ M1 Term ‘-2’ M1 Full correct quotient M1

2.

Term ‘-6x2’ M1 Term ‘-10x’ M1 Full quotient ‘-6x2 – 10x + 20’ M1 Remainder ‘-39’ M1

3.

f(4) = 0, therefore (x – 4) is a factor of f(x) M1

M1 M1

f(x) = (x – 4)(6x2 + 17x – 3) = (x – 4)(6x – 1)(x + 3) M1 x = -3 x = 1

6

x = 4 M1