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P2 §1.6 Mixed exercise for chapter 1 본문

A-level Mathematics/Pure Mathematics 2

P2 §1.6 Mixed exercise for chapter 1

Cambridge Maths Academy 2022. 6. 28. 21:35
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Pure mathematics Year 2

Table of contents

  1. Edexcel P2 Ch1 Mixed exercise 1

1. Edexcel P2 Ch1 Mixed exercise 1

Question 1. (E/P) Prove by contradiction that 12 is an irrational number. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q1.)

 

Solution.

 

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Question 2. (P) Prove that if q2 is an irrational number then q is an irrational number.

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q2.)

 

Solution.

 

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Question 3. Simplify: (a)x46×2x+8x216(b)x23x103x221×6x2+24x2+6x+8(c)4x2+12x+9x2+6x÷4x292x2+9x18

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q3.)

 

Solution.

 

(a) x46×2x+8x216=x46×2(x+4)(x+4)(x4)=13

 

(b) x23x103x221×6x2+24x2+6x+8=(x5)(x+2)3(x27)×6(x2+4)(x+2)(x+4)=(x5)(x27)×2(x2+4)(x+4)=2(x2+4)(x5)(x27)(x+4)

 

(c) 4x2+12x+9x2+6x÷4x292x2+9x18=(2x+3)2x(x+6)×(2x3)(x+6)(2x+3)(2x3)=2x+3x

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Question 4. (E/P)

(a) Simplify fully 4x28xx23x4×x2+6x+52x2+10x [3 marks]

 

(b) Given that ln[(4x28x)(x2+6x+5)]=6+ln[(x23x4)(2x2+10x)], find x in terms of e. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q4.)

 

Solution.

(a) 4x28xx23x4×x2+6x+52x2+10x=4x(x2)(x4)(x+1)×(x+1)(x+5)2x(x+5)=2(x2)x4

 

(b) ln[(4x28x)(x2+6x+5)]=6+ln[(x23x4)(2x2+10x)]ln[(4x28x)(x2+6x+5)(x23x4)(2x2+10x)]=6ln[2(x2)x4]=62(x2)x4=e62x4=e6(x4)(2e6)x=4(1e6)x=4(1e6)(2e6)=4(e61)(e62)

 

Note: A similar question: P2 Ch1.2 Exercise 1B Q5.

 

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Question 5. (E/P) g(x)=4x39x29x32x+24÷x2+6x+56x213x5

(a) Show that g(x) can be written in the form ax2+bx+c, where a,b and c are constants to be found. [4 marks]

 

(b) Hence differentiate g(x) and find g(2). [3 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q5.)

 

Solution.

(a) g(x)=4x39x29x32x+24÷x23x6x213x5=x(4x+3)(x3)8(4x+3)×(3x+1)(2x5)x(x3)=(3x+1)(2x5)8=6x213x58=34x2138x58

 

(b) g(x)=34x2138x58g(x)=32x138g(2)=32(2)138=3138=378

 

Note: A similar question: P2 Ch1.2 Exercise 1B Q6.

 

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Question 6. (E) Express 6x+1x5+5x+3x23x10 as a single fraction in its simplest form. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q6.)

 

Solution.

6x+1x5+5x+3x23x10=6x+1x5+5x+3(x5)(x+2)=(6x+1)(x+2)+5x+3(x5)(x+2)=(6x2+13x+2)+5x+3(x5)(x+2)=6x2+18x+5(x5)(x+2)=6x2+18x+5x23x10

 

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Question 7. (E) f(x)=x+3x112x2+2x3,xR,x>1 Show that f(x)=x2+3x+3x+3 [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q7.)

 

Solution.

f(x)=x+3x112x2+2x3=x+3x112(x+3)(x1)=x(x+3)(x1)+3(x+3)12(x+3)(x1)=(x3+2x23x)+(3x+9)12(x+3)(x1)=x3+2x23(x+3)(x1) For the numerator, let g(x)=x3+2x23 and note that: g(1)=1+23=0 By the factor theorem, we find: g(x)=x3+2x23=(x1)(x2+3x+3) This gives f(x)=g(x)(x+3)(x1)=(x1)(x2+3x+3)(x+3)(x1)=x2+3x+3x+3

 

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Question 8. (E) f(x)=x3x(x1) Show that f(x) can be written in the form Ax+Bx1 where A and B are constants to be found. [3 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q8.)

 

Solution.

Let f(x)=x3x(x1)Ax+Bx1A(x1)+Bxx(x1) Compare the numerators on both sides: A(x1)+Bxx3. By the method of substitution: (i)x=0A=3A=3(ii)x=1B=2 and we find: f(x)=x3x(x1)3x2x1 where A=3,B=2

 

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Question 9. (E) 15x+21(x2)(x+1)(x5)Px2+Qx+1+Rx5. Find the values of the constants P,Q and R. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q9.)

 

Solution.

Let 15x+21(x2)(x+1)(x5)Px2+Qx+1+Rx5P(x+1)(x5)+Q(x2)(x5)+R(x2)(x+1)(x2)(x+1)(x5) Compare the numerators on both sides: P(x+1)(x5)+Q(x2)(x5)+R(x2)(x+1)15x+21. By the method of substitution: (i)x=2(3)(3)P=30+219P=9P=1(ii)x=1(3)(6)Q=15+2118Q=36Q=2(iii)x=5(3)(6)R=75+2118R=54R=3 and we find: 15x+21(x2)(x+1)(x5)1x2+2x+13x5 where P=1,Q=2,R=3

 

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Question 10. (E) Show that 16x1(3x+2)(2x1) can be written in the form D3x+2+E2x1 and find the values of the constants D and E. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q10.)

 

Solution.

Let 16x1(3x+2)(2x1)D3x+2+E2x1D(2x1)+E(3x+2)(3x+2)(2x1) Compare the numerators on both sides: D(2x1)+E(3x+2)16x1. By the method of substitution: (i)x=23(431)D=323173D=353D=5(ii)x=12(32+2)E=8172E=7E=2 and we find: 16x1(3x+2)(2x1)53x+2+22x1 where D=5,E=2

 

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Question 11. (E) 7x2+2x2x2(x+1)Ax+Bx2+Cx+1 Find the values of the constants [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q11.)

 

Solution.

We have: 7x2+2x2x2(x+1)Ax+Bx2+Cx+1Ax(x+1)+B(x+1)+Cx2x2(x+1) Compare the numerators on both sides: Ax(x+1)+B(x+1)+Cx27x2+2x2. By the method of substitution: (i)x=0B=2(ii)x=1C=7(1)2+2(1)2=3(iii)x=12A+2B+C=7+22=72A=72BC=7+43=8A=4 and we find: 7x2+2x2x2(x+1)4x2x2+3x+1 where A=4,B=2,C=3

 

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Question 12. (E) h(x)=21x213(x+5)(3x1)2 Show that h(x) can be written in the form Dx+5+E(3x1)+F(3x1)2, where D,E and F are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q12.)

 

Solution.

Let h(x)=21x213(x+5)(3x1)2Dx+5+E(3x1)+F(3x1)2D(3x1)2+E(x+5)(3x1)+F(x+5)(x+5)(3x1)2 Compare the numerators on both sides: D(3x1)2+E(x+5)(3x1)+F(x+5)21x213. By the method of substitution: (i)x=5(16)2D=21(5)213256D=512D=2(ii)x=13(13+5)F=21(13)213163F=7313=7393=323F=2(iii)x=0D5E+5F=135E=13D5F=132+10=5E=1 and we find: h(x)=21x213(x+5)(3x1)22x+5+1(3x1)2(3x1)2 where D=2,E=1,F=2

 

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Question 13. (E) Find the values of the constants A,B,C and D in the following identity: x36x2+11x+2(x2)(Ax2+Bx+C)+D [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q13.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), x36x2+11x+2(x2)(x24x+3)+7

 

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Question 14. (E) Show that 4x36x2+8x52x+1 can be put in the form Ax2+Bx+C+D2x+1. Find the values of the constants A,B,C and D. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q14.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), 4x36x2+8x5(2x24x+6)(2x+1)114x36x2+8x52x+12x24x+6112x+1

 

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Question 15. (E) Show that x4+2x21Ax2+Bx+C+Dx21, where A,B,C and D are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q15.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), x4+2(x2+1)(x21)+3x4+2x21x2+1+3x21

 

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Question 16. (E) x4x22x+1Ax2+Bx+C+Dx1+E(x1)2. Find the values of the constants A,B,C,D and E. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q16.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), x4(x2+2x+3)(x22x+1)+4x3x4x22x+1x2+2x+3+4x3x22x+1

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: x22x+1=(x1)2 and this gives partial fractions with repeated factors. Let 4x3x22x+14x3(x1)2Dx1+E(x1)2D(x1)+E2x1 Compare the numerators on both sides: D(x1)+E4x3 By the method of equating coefficients: {D=4D+E=3{D=4E=1 and we obtain: 4x3x22x+14x3(x1)24x1+1(x1)2

 

Finally, we find: x4x22x+1x2+2x+3+4x3x22x+1x2+2x+3+4x1+1(x1)2 and thus: A=1B=2C=3D=4E=1

 

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Question 17. (E) h(x)=2x2+2x3x2+2x3 Show that h(x) can be written in the form A+Bx+3+Cx1, where A,B and C are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q17.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), 2x2+2x32(x2+2x3)2x+3h(x)=2x2+2x3x2+2x32+2x+3x2+2x3

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: x2+2x3=(x+3)(x1) Let 2x+3x2+2x32x+3(x+3)(x1)Bx+3+Cx1B(x1)+C(x+3)(x+3)(x1) Compare the numerators on both sides: B(x1)+C(x+3)2x+3 By the method of substitution: (i)x=34B=6+3=9B=94(ii)x=14C=2+3=1C=14 and we obtain: 2x+3x2+2x32x+3(x+3)(x1)94x+3+14x114(9x+3+1x1)

 

Finally, we find: h(x)=2x2+2x3x2+2x32+2x+3x2+2x32+14(9x+3+1x1) and thus: A=2B=94C=14

 

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Question 18. (E) Given that x2+1x(x2)P+Qx+Rx2, find the values of the constants P,Q and R. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q18.)

 

Solution.

(i) Step 1: Algebraic division.

x2+1x22x+2x+1x(x2)+2x+1x2+1x(x2)1+2x+1x(x2)

 

(ii) Step 2: Partial fractions.

Let 2x+1x(x2)Qx+Rx2Q(x2)+Rxx(x2) Compare the numerators on both sides: (Q+R)x2Q2x+1 By the method of equating coefficients: {Q+R=22Q=1{Q=12R=52 and we obtain: 2x+1x(x2)Qx+Rx212x+52(x2)

 

Finally, we find: x2+1x(x2)1+2x+1x(x2)112x+52(x2) and thus: P=1Q=12R=52

 

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Question 19. (P) Given that f(x)=2x3+9x2+10x+3,

(a) show that 3 is a root of f(x);

 

(b) express 10f(x) as partial fractions.

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q19.)

 

Solution.

(a) f(3)=2(3)3+9(3)2+10(3)+3=54+8130+3=0.

 

(b) We factorise the denominator f(x) first. By the factor theorem, (x+3) is a factor. f(x)=2x3+9x2+10x+3=(x+3)(2x2+3x+1)=(x+3)(2x+1)(x+1) For partial fractions, let 10f(x)=10(x+3)(2x+1)(x+1)Ax+3+B2x+1+Cx+1A(2x+1)(x+1)+B(x+3)(x+1)+C(x+3)(2x+1)(x+3)(2x+1)(x+1) Compare the numerators on both sides: A(2x+1)(x+1)+B(x+3)(x+1)+C(x+3)(2x+1)10 By the method of substitution: (i)x=3(5)(2)A=10A=1(ii)x=12(52)(12)B=10B=8(iii)x=1(2)(1)C=10C=5 and we obtain: 10f(x)=10(x+3)(2x+1)(x+1)1x+3+82x+15x+1

 

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Challenge. The line L meets the circle C with centre O at exactly one point, A.

 

Prove by contradiction that the line L is perpendicular to the radius OA. [Hint: In a right-angled triangle, the side opposite the right-angle is always the longest side.]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Challenge.)

 

Solution.

 

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