a-level topic: proof by contradiction 6 starter activity ...β¬Β¦Β Β· 1a. show that if log 2 3 =...
TRANSCRIPT
1a. Show that if log2 3 = ππ
ππ, 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 β 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 = ππ
ππ
(2ππππ)ππ = 3q
M1
2p = 3q M1 1b.
Assume that log2 3 is rational then, log2 3 = ππ
ππ
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 β y2 = 1 (x + y)(x β y) = 1 M1
x + y = 1 x β y = 1 M1
2x = 2 x = 1 M1
y = 0 M1 Therefore, proof by contradiction there are no positive integer solutions. M1
1. Simplify π₯4β5π₯2+4
π₯2βπ₯β2(2)
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2. Express as a single fraction in its simplest form, 1
π₯β3+
3
π₯2β3π₯+
π₯
π₯2β6π₯+9(3)
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3a. Express 2
π₯+5 +
3
(π₯+2)(π₯+5) as a single fraction in its simplest form. (2)
b. Hence solve the equation, 2
π₯+5 +
3
(π₯+2)(π₯+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. (π₯2β1)(π₯2β4)
(π₯+1)(π₯β2)= (π₯+1)(π₯β1)(π₯+2)(π₯β2)
(π₯+1)(π₯β2)M1
= (x β 1)(x + 2) M1
2. 1
π₯β3+
3
π₯2β3π₯+
π₯
π₯2β6π₯+9= π₯(π₯β3)+3(π₯β3)+π₯2
π₯(π₯β3)2M1
= 2π₯2β9
π₯(π₯β3)2M1
3a. 2
π₯+5 +
3
(π₯+2)(π₯+5) =
2(π₯+2)+3
(π₯+2)(π₯+5)M1
= 2π₯+7
(π₯+2)(π₯+5)M1
3b. 2π₯+7
(π₯+2)(π₯+5)= 1
3M1
3(2x + 7) = (x + 2)(x + 5)
x2 + x β 11 = 0 M1
x = -3.85
x = 2.85 M1
1. Find the value of A, B and C when, (3) 2 β 9x = A(2x β 1)2 + B(x β 3)(2x β 1) + C(x β 3)
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2. Express 9π₯π₯2β2π₯π₯β12
π₯π₯(π₯π₯+3)(π₯π₯β2) in partial fractions (4)
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3. Express 13β3π₯π₯2
(2π₯π₯+3)(π₯π₯β1)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 β 9x = A(2x β 1)2 + B(x β 3)(2x β 1) + C(x β 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 β 2x β 12 = A(x + 3)(x β 2) + Bx(x β 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
9π₯π₯2β2π₯π₯β12π₯π₯(π₯π₯+3)(π₯π₯β2)
= 2π₯π₯
+ 5π₯π₯+3
+ 2π₯π₯β2
3.
13 β 3x2 = A(x β 1)2 + B(2x + 3)(x β 1) + C(2x + 3) M1 Let x = -3
2,
254
= 254π΄π΄
A = 1
M1
Let x = 1, 10 = 5C C = 2
M1
Coefficients of x2: -3 = A + 2B B = -2 M1
13β3π₯π₯2
(2π₯π₯+3)(π₯π₯β1)2 = 1
2π₯π₯+3β 2
π₯π₯β1+ 2
(π₯π₯ β1)2
1. Express f(x) in partial fractions when f(x) = 2π₯π₯β1
(π₯π₯β1)(2π₯π₯β3) (3)
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2. Express 3π₯π₯+2
π₯π₯2β2π₯π₯β24 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 β 32 = A(x β 1)(x + 3) + B(x β 1)(x β 4) + C(x + 3)(x β 4) __________________________________________________________________________________________
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4. Express in partial fractions 2π₯π₯2+4
π₯π₯(π₯π₯β1)(π₯π₯β4) (4)
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A-Level Starter Activity
Topic: Partial Fractions Chapter Reference: Pure 2, Chapter 1
8 minutes
Solutions 1. f(x) = 2π₯π₯β1
(π₯π₯β1)(2π₯π₯β3) = π΄π΄
π₯π₯β1+ π΅π΅
2π₯π₯β3
2x β 1 = A(2x β 3) + B(x β 1) M1
When x = 1, 1 = A(-1) + B(0) A = 1
M1
When x = 32
2 = A(0) β B(-0.5) B = 4
M1
f(x) = 2π₯π₯β1(π₯π₯β1)(2π₯π₯β3)
= 1π₯π₯β1
+ 42π₯π₯β3
2.
3π₯π₯+2π₯π₯2β2π₯π₯β24
= 3π₯π₯+2(π₯π₯β6)(π₯π₯+4)
= π΄π΄π₯π₯β6
+ π΅π΅π₯π₯+4
M1 3x + 2 = A(x + 4) + B(x β 6) M1 When x = -4, -10 = A(0) + B(-10) B = 1
M1
When x = 6 20 = A(10) + B(0) A = 2
M1
3π₯π₯+2π₯π₯2β2π₯π₯β24
= 3π₯π₯+2(π₯π₯β6)(π₯π₯+4)
= 2π₯π₯β6
+ 1π₯π₯+4
3.
3x2 + 17x β 32 = A(x β 1)(x + 3) + B(x β 1)(x β 4) + C(x + 3)(x β 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.
2π₯π₯2+4π₯π₯(π₯π₯β1)(π₯π₯β4)
= π΄π΄π₯π₯
+ π΅π΅π₯π₯β1
+ πΆπΆπ₯π₯β4
2π₯π₯2 + 4 = A(x β 1)(x β 4) + B(x)(x β 4) + C(x)(x β 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
2π₯π₯2+4π₯π₯(π₯π₯β1)(π₯π₯β4)
= 1π₯π₯
+ 2π₯π₯β1
+ 3π₯π₯β4
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 β 22x2 β 6x3) by (x + 2) (4) __________________________________________________________________________________________
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3. f(x) = 6x3
β 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 β3x2β M1 Term β-2β M1 Full correct quotient M1
2.
Term β-6x2β M1 Term β-10xβ M1 Full quotient β-6x2 β 10x + 20β M1 Remainder β-39β M1
3.
f(4) = 0, therefore (x β 4) is a factor of f(x) M1
M1 M1
f(x) = (x β 4)(6x2 + 17x β 3) = (x β 4)(6x β 1)(x + 3) M1 x = -3 x = 1
6
x = 4 M1