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Cambridge Maths Academy
P2 §1.6 Mixed exercise for chapter 1 본문
P2 §1.6 Mixed exercise for chapter 1
Cambridge Maths Academy 2022. 6. 28. 21:35Pure mathematics Year 2
Table of contents
- 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.
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.
Question 3. Simplify: (a)x−46×2x+8x2−16(b)x2−3x−103x2−21×6x2+24x2+6x+8(c)4x2+12x+9x2+6x÷4x2−92x2+9x−18
(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q3.)
Solution.
(a) x−46×2x+8x2−16=x−46×2(x+4)(x+4)(x−4)=13✓
(b) x2−3x−103x2−21×6x2+24x2+6x+8=(x−5)(x+2)3(x2−7)×6(x2+4)(x+2)(x+4)=(x−5)(x2−7)×2(x2+4)(x+4)=2(x2+4)(x−5)(x2−7)(x+4)✓
(c) 4x2+12x+9x2+6x÷4x2−92x2+9x−18=(2x+3)2x(x+6)×(2x−3)(x+6)(2x+3)(2x−3)=2x+3x✓
Question 4. (E/P)(a) Simplify fully 4x2−8xx2−3x−4×x2+6x+52x2+10x [3 marks]
(b) Given that ln[(4x2−8x)(x2+6x+5)]=6+ln[(x2−3x−4)(2x2+10x)], find x in terms of e. [4 marks]
(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q4.)
Solution.
(a) 4x2−8xx2−3x−4×x2+6x+52x2+10x=4x(x−2)(x−4)(x+1)×(x+1)(x+5)2x(x+5)=2(x−2)x−4✓
(b) ln[(4x2−8x)(x2+6x+5)]=6+ln[(x2−3x−4)(2x2+10x)]⇒ln[(4x2−8x)(x2+6x+5)(x2−3x−4)(2x2+10x)]=6⇒ln[2(x−2)x−4]=6⇒2(x−2)x−4=e6⇒2x−4=e6(x−4)⇒(2−e6)x=4(1−e6)⇒x=4(1−e6)(2−e6)=4(e6−1)(e6−2)✓
Note: A similar question: P2 Ch1.2 Exercise 1B Q5.
Question 5. (E/P) g(x)=4x3−9x2−9x32x+24÷x2+6x+56x2−13x−5(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)=4x3−9x2−9x32x+24÷x2−3x6x2−13x−5=x(4x+3)(x−3)8(4x+3)×(3x+1)(2x−5)x(x−3)=(3x+1)(2x−5)8=6x2−13x−58=34x2−138x−58✓
(b) g(x)=34x2−138x−58⇒g′(x)=32x−138⇒g′(−2)=32(−2)−138=−3−138=−378✓
Note: A similar question: P2 Ch1.2 Exercise 1B Q6.
Question 6. (E) Express 6x+1x−5+5x+3x2−3x−10 as a single fraction in its simplest form. [4 marks]
(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q6.)
Solution.
6x+1x−5+5x+3x2−3x−10=6x+1x−5+5x+3(x−5)(x+2)=(6x+1)(x+2)+5x+3(x−5)(x+2)=(6x2+13x+2)+5x+3(x−5)(x+2)=6x2+18x+5(x−5)(x+2)✓=6x2+18x+5x2−3x−10✓
Question 7. (E) f(x)=x+3x−1−12x2+2x−3,x∈R,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+3x−1−12x2+2x−3=x+3x−1−12(x+3)(x−1)=x(x+3)(x−1)+3(x+3)−12(x+3)(x−1)=(x3+2x2−3x)+(3x+9)−12(x+3)(x−1)=x3+2x2−3(x+3)(x−1) For the numerator, let g(x)=x3+2x2−3 and note that: g(1)=1+2−3=0 By the factor theorem, we find: g(x)=x3+2x2−3=(x−1)(x2+3x+3) This gives ⇒f(x)=g(x)(x+3)(x−1)=(x−1)(x2+3x+3)(x+3)(x−1)=x2+3x+3x+3✓
Question 8. (E) f(x)=x−3x(x−1) Show that f(x) can be written in the form Ax+Bx−1 where A and B are constants to be found. [3 marks]
(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q8.)
Solution.
Let f(x)=x−3x(x−1)≡Ax+Bx−1≡A(x−1)+Bxx(x−1) Compare the numerators on both sides: A(x−1)+Bx≡x−3. By the method of substitution: (i)x=0⇒−A=−3⇒A=3(ii)x=1⇒B=−2 and we find: ⇒f(x)=x−3x(x−1)≡3x−2x−1✓ where A=3,B=−2✓
Question 9. (E) −15x+21(x−2)(x+1)(x−5)≡Px−2+Qx+1+Rx−5. 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(x−2)(x+1)(x−5)≡Px−2+Qx+1+Rx−5≡P(x+1)(x−5)+Q(x−2)(x−5)+R(x−2)(x+1)(x−2)(x+1)(x−5) Compare the numerators on both sides: P(x+1)(x−5)+Q(x−2)(x−5)+R(x−2)(x+1)≡−15x+21. By the method of substitution: (i)x=2⇒(3)(−3)P=−30+21⇒−9P=−9⇒P=1(ii)x=−1⇒(−3)(−6)Q=15+21⇒18Q=36⇒Q=2(iii)x=5⇒(3)(6)R=−75+21⇒18R=−54⇒R=−3 and we find: ⇒−15x+21(x−2)(x+1)(x−5)≡1x−2+2x+1−3x−5✓ where P=1,Q=2,R=−3✓
Question 10. (E) Show that 16x−1(3x+2)(2x−1) can be written in the form D3x+2+E2x−1 and find the values of the constants D and E. [4 marks]
(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q10.)
Solution.
Let 16x−1(3x+2)(2x−1)≡D3x+2+E2x−1≡D(2x−1)+E(3x+2)(3x+2)(2x−1) Compare the numerators on both sides: D(2x−1)+E(3x+2)≡16x−1. By the method of substitution: (i)x=−23⇒(−43−1)D=−323−1⇒−73D=−353⇒D=5(ii)x=12⇒(32+2)E=8−1⇒72E=7⇒E=2 and we find: ⇒16x−1(3x+2)(2x−1)≡53x+2+22x−1✓ where D=5,E=2✓
Question 11. (E) 7x2+2x−2x2(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+2x−2x2(x+1)≡Ax+Bx2+Cx+1≡Ax(x+1)+B(x+1)+Cx2x2(x+1) Compare the numerators on both sides: Ax(x+1)+B(x+1)+Cx2≡7x2+2x−2. By the method of substitution: (i)x=0⇒B=−2(ii)x=−1⇒C=7(−1)2+2(−1)−2=3(iii)x=1⇒2A+2B+C=7+2−2=7⇒2A=7−2B−C=7+4−3=8⇒A=4 and we find: ⇒7x2+2x−2x2(x+1)≡4x−2x2+3x+1✓ where A=4,B=−2,C=3✓
Question 12. (E) h(x)=21x2−13(x+5)(3x−1)2 Show that h(x) can be written in the form Dx+5+E(3x−1)+F(3x−1)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)=21x2−13(x+5)(3x−1)2≡Dx+5+E(3x−1)+F(3x−1)2≡D(3x−1)2+E(x+5)(3x−1)+F(x+5)(x+5)(3x−1)2 Compare the numerators on both sides: D(3x−1)2+E(x+5)(3x−1)+F(x+5)≡21x2−13. By the method of substitution: (i)x=−5⇒(−16)2D=21(−5)2−13⇒256D=512⇒D=2(ii)x=13⇒(13+5)F=21(13)2−13⇒163F=73−13=7−393=−323⇒F=−2(iii)x=0⇒D−5E+5F=−13⇒−5E=−13−D−5F=−13−2+10=−5⇒E=1 and we find: ⇒h(x)=21x2−13(x+5)(3x−1)2≡2x+5+1(3x−1)−2(3x−1)2✓ where D=2,E=1,F=−2✓
Question 13. (E) Find the values of the constants A,B,C and D in the following identity: x3−6x2+11x+2≡(x−2)(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), x3−6x2+11x+2≡(x−2)(x2−4x+3)+7
Question 14. (E) Show that 4x3−6x2+8x−52x+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), 4x3−6x2+8x−5≡(2x2−4x+6)(2x+1)−11⇔4x3−6x2+8x−52x+1≡2x2−4x+6−112x+1
Question 15. (E) Show that x4+2x2−1≡Ax2+Bx+C+Dx2−1, 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)(x2−1)+3⇔x4+2x2−1≡x2+1+3x2−1
Question 16. (E) x4x2−2x+1≡Ax2+Bx+C+Dx−1+E(x−1)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)(x2−2x+1)+4x−3⇔x4x2−2x+1≡x2+2x+3+4x−3x2−2x+1
(ii) Step 2: Partial fractions.
Factorise the denominator first: x2−2x+1=(x−1)2 and this gives partial fractions with repeated factors. Let 4x−3x2−2x+1≡4x−3(x−1)2≡Dx−1+E(x−1)2≡D(x−1)+E2x−1 Compare the numerators on both sides: D(x−1)+E≡4x−3 By the method of equating coefficients: {D=4−D+E=−3⇒{D=4E=1 and we obtain: ⇒4x−3x2−2x+1≡4x−3(x−1)2≡4x−1+1(x−1)2
Finally, we find: ⇒x4x2−2x+1≡x2+2x+3+4x−3x2−2x+1≡x2+2x+3+4x−1+1(x−1)2✓ and thus: A=1B=2C=3D=4E=1
Question 17. (E) h(x)=2x2+2x−3x2+2x−3 Show that h(x) can be written in the form A+Bx+3+Cx−1, 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+2x−3≡2(x2+2x−3)−2x+3⇔h(x)=2x2+2x−3x2+2x−3≡2+−2x+3x2+2x−3
(ii) Step 2: Partial fractions.
Factorise the denominator first: x2+2x−3=(x+3)(x−1) Let −2x+3x2+2x−3≡−2x+3(x+3)(x−1)≡Bx+3+Cx−1≡B(x−1)+C(x+3)(x+3)(x−1) Compare the numerators on both sides: B(x−1)+C(x+3)≡−2x+3 By the method of substitution: (i)x=−3⇒−4B=6+3=9⇒B=−94(ii)x=1⇒4C=−2+3=1⇒C=14 and we obtain: ⇒−2x+3x2+2x−3≡−2x+3(x+3)(x−1)≡−94x+3+14x−1≡14(−9x+3+1x−1)
Finally, we find: ⇒h(x)=2x2+2x−3x2+2x−3≡2+−2x+3x2+2x−3≡2+14(−9x+3+1x−1)✓ and thus: A=2B=−94C=14
Question 18. (E) Given that x2+1x(x−2)≡P+Qx+Rx−2, 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+1≡x2−2x+2x+1≡x(x−2)+2x+1⇔x2+1x(x−2)≡1+2x+1x(x−2)
(ii) Step 2: Partial fractions.
Let 2x+1x(x−2)≡Qx+Rx−2≡Q(x−2)+Rxx(x−2) Compare the numerators on both sides: (Q+R)x−2Q≡2x+1 By the method of equating coefficients: {Q+R=2−2Q=1⇒{Q=−12R=52 and we obtain: ⇒2x+1x(x−2)≡Qx+Rx−2≡−12x+52(x−2)
Finally, we find: ⇒x2+1x(x−2)≡1+2x+1x(x−2)≡1−12x+52(x−2)✓ and thus: P=1Q=−12R=52
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+81−30+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+1≡A(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=10⇒A=1(ii)x=−12⇒(52)(12)B=10⇒B=8(iii)x=−1⇒(2)(−1)C=10⇒C=−5 and we obtain: ⇒10f(x)=10(x+3)(2x+1)(x+1)≡1x+3+82x+1−5x+1✓
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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