2025 H2 FM 9649 P1 Solutions

Singapore–Cambridge Further Mathematics 2025 Paper 1 Solutions © Gerard Sayson 2025 DISCLAIMER. This document is © Gerard Sayson 2025, all rights reserved. These solutions were prepared inde- pendently by the author. While care has been taken to ensure accuracy, errors may still be present, and the solutions should be used as a learning aid rather than an authoritative reference. The author assumes no responsibility or lia- bility for any inaccuracies, misinterpretations, or consequences arising from the use of this document. For corrections or feedback, please email gerard@gsn.bz. This document was last updated on November 17, 2025. 1 Suppose𝑎,𝑏 > 0. Let𝑞1= 2𝑎 and𝑞𝑛+1= 2𝑎+𝑏2/𝑞𝑛. Since(𝑞𝑛)𝑛∈N converges, let𝑞𝑛𝑞𝑛+1= 2𝑎𝑞𝑛+𝑏2. Note that𝐿2−2𝑎𝐿−𝑏2= 0 so 𝐿 = 𝑎± √ 𝑎2+𝑏2. Notice that all terms in the sequence are positive so 𝐿 = 𝑎+ √ 𝑎2+𝑏2, otherwise 𝐿=𝑎− √ 𝑎2+𝑏2 < 0 which is impossible, as𝑎 < √ 𝑎2+𝑏2. 2 (a) Let 𝑧1 = 5+ i and 𝑧2 = 239+ i. Then |𝑧1| = √ 26 and|𝑧2| = √ 57122. Also, arg (𝑧1) = tan−1( 1 5) and arg(𝑧2)= tan−1( 1 239). (b) By considering the properties of 𝑟𝑒 i𝜃, see that |𝑧4| = |𝑧1|4/|𝑧2| = 676√ 57122 = √︃ 456976 57122 = 2 √ 2. Also, arg(𝑧4)= 4 arg(𝑧1)− arg(𝑧2)= 4 tan−1( 1 5)− tan−1( 1 239). (c) We are to show arg(𝑧4)+ arg(𝑧3)= 4 arg(𝑧1), i.e. we must literally find arg(𝑧4). By GC, 𝑧4 1 𝑧2 =(5+ i)4 239+ i = 2+ 2i. Hence arg(𝑧4)= tan−1(1)= 1 4𝜋, and the conclusion follows. 3 (a) 𝑥2+( 𝑦+𝑎)2=𝑟2. (b) Write𝑦= √ 𝑟2−𝑥2−𝑎. Set 𝑦= 0 so𝑥=± √ 𝑟2−𝑎2. Also note that d𝑦 d𝑥 =− 𝑥√ 𝑟2−𝑥2 . Hence 2𝜋 ∫ √ 𝑟2−𝑎2 − √ 𝑟2−𝑎2 ( √︁ 𝑟2−𝑥2−𝑎) √︄ 1+ 𝑥√ 𝑟2−𝑥2 2 d𝑥 = 2𝜋 ∫ √ 𝑟2−𝑎2 − √ 𝑟2−𝑎2 ( √︁ 𝑟2−𝑥2−𝑎) √︂ 𝑟2 𝑟2−𝑥2 d𝑥 = 2𝜋𝑟 ∫ √ 𝑟2−𝑎2 − √ 𝑟2−𝑎2

1− 𝑎√ 𝑟2−𝑥2

d𝑥 = 2𝜋𝑟 h 𝑥−𝑎 sin−1 𝑥 𝑟 i√ 𝑟2−𝑎2 − √ 𝑟2−𝑎2 = 4𝜋𝑟 √︁ 𝑟2−𝑎2−𝑎 sin−1 √ 𝑟2−𝑎2 𝑟 !! . Here,𝑘 = 4𝜋. 4 (a) 𝐿≈𝜋 √ 2+ 3 √ 2 2 𝜋≈ 11.10721≈ 11.1072 (4dp). (b) (i) d𝑦 d𝑥 = sin(𝑥)+ 𝑥 cos(𝑥) so𝐿= ∫ 2𝜋 0 √︄ 1+ d𝑦 d𝑥 2 d𝑥= ∫ 2𝜋 0 √︃ 1+(sin(𝑥)+ 𝑥 cos(𝑥))2 d𝑥. (ii) The size of each interval is 1 4× 2𝜋= 𝜋 2 . Hence, letting 𝑓(𝑥)= √︃ 1+(sin(𝑥)+ 𝑥 cos(𝑥))2, 𝐿≈ 𝜋 4

𝑓(0)+ 2𝑓 𝜋 2

+ 2𝑓(𝜋)+ 2𝑓 3𝜋 2

+ 𝑓(2𝜋)

≈ 15.40396≈ 15.4040 (4dp). LibGSN 9649-PP-2025-01S 1

Singapore–Cambridge Further Mathematics 2025 Paper 1 Solutions © Gerard Sayson 2025 (iii) 𝐿≈ 15.37457≈ 15.3746 (4dp) by GC. (c) The student’s approximation is inaccurate because it is far from the true (GC) value, while the trapezium rule approximation is accurate because it is close to the true value. 5 (a) 𝑧= 𝑓(𝑥,𝑦)=𝑥 tan2(𝑦), so 𝜕𝑧 𝜕𝑥 = tan2(𝑦) and 𝜕𝑧 𝜕𝑦 = 2𝑥 tan(𝑦) sec2(𝑦). Hence 𝑧= 𝑓

2,𝜋 4

+(𝑥− 2)𝑓𝑥

2,𝜋 4

+

𝑦− 𝜋 4

𝑓𝑦

2,𝜋 4

= 2+(𝑥− 2)( 1)+

𝑦− 𝜋 4

(8) =⇒ 𝑧=𝑥+ 8𝑦− 2𝜋 =⇒ 𝑥+ 8𝑦−𝑧= 2𝜋. Then𝑎= 1,𝑏= 8 and𝑐=−1. (b) A vector normal to the plane is ©­ « 1 8 −1 ª® ¬ . (c) Calculate u= i+ j and v= i+ 8k. Then u× v= ©­ « 1 8 −1 ª® ¬ which is normal to the plane. (d) Both u and v are direction vectors on 𝑃. u is the direction vector giving the directional derivative along the𝑥-axis at the point𝑃, while v is that along the𝑦-axis. 6 (a) The annual interest rate is(1.00812− 1)× 100%≈ 1.10% (3sf). 𝐹 is the minimum amount the business repays per month, and 𝐺 is the additional amount that the business pays per month as the number of months passed increases and more interest is charged monthly. (b) Let 𝑀𝑛+1 z }| { 𝐴× 1.008𝑛+1+𝐵(𝑛+ 1)+ 𝐶= 1.008𝑀𝑛 z }| { 1.008(𝐴× 1.008𝑛+𝐵𝑛+𝐶)−𝐹−𝐺𝑛. Then𝐵𝑛+𝐵+𝐶= 1.008𝐵𝑛+ 1.008𝐶−𝐹−𝐺𝑛 =⇒ 0.008𝐵𝑛−𝐵+ 0.008𝐶=𝐹+𝐺𝑛. Hence 𝐵= 125𝐺 and−𝐵+ 0.008𝐶 =−125𝐺+ 0.008𝐶 = 𝐹 =⇒ 𝐶 = 125(𝐹+ 125𝐺). It now remains to determine 𝐴. Since𝑀0= 106= 𝐴+𝐶,𝐴= 106−𝐶= 106− 125(𝐹+ 125𝐺). (c) Set𝑀60= 0. Then (106− 125(𝐹+ 125𝐺))× 1.00860+ 60× 125𝐺+ 125(𝐹+ 125𝐺)= 0 and we solve for𝐺 in terms of𝐹. Indeed, −15625× 1.00860×𝐺+( 106− 125𝐹)× 1.00860+ 23125𝐺+ 125𝐹= 0 𝐺=(106− 125𝐹)× 1.00860+ 125𝐹 15625× 1.00860− 23125 . Since𝐺 varies linearly with𝐹 and is decreasing, max 𝐹≥2500 𝐺≈ 684 (3sf) occurs at min 𝐹≥2500 𝐹= 2500. 7 (a) 𝑇 admits two eigenvalues, one with multiplicity 2. By GC, 𝑝𝑇(𝜆)=

1−𝜆 2 1 3 2 −𝜆 −1 3 −2 3 −𝜆

=(𝜆+ 2)(𝜆− 4)2= 0 implies𝜆=−2 or𝜆= 4. If𝜆= 4, then [𝑇−𝜆· id]𝛽 ©­ « 𝑥 𝑦 𝑧 ª® ¬ = 0 =⇒ ©­ « −3 2 1 3 −2 −1 3 −2 −1 ª® ¬ ©­ « 𝑥 𝑦 𝑧 ª® ¬ = 0 =⇒ ©­ « −3 2 1 0 0 0 0 0 0 ª® ¬ ©­ « 𝑥 𝑦 𝑧 ª® ¬ = 0. LibGSN 9649-PP-2025-01S 2

Singapore–Cambridge Further Mathematics 2025 Paper 1 Solutions © Gerard Sayson 2025 Hence𝑧= 3𝑥− 2𝑦, and the equation ofΠ is Π : r=𝑎 ©­ « 1 0 3 ª® ¬ +𝑏 ©­ « 0 1 −2 ª® ¬ ∀𝑎,𝑏 ∈ R. Else if𝜆=−2, then rref ©­ « 3 2 1 3 4 −1 3 −2 5 ª® ¬ = ©­ « 1 0 1 0 1 −1 0 0 0 ª® ¬ =⇒ 𝑙 : r=𝑎 ©­ « −1 1 1 ª® ¬ ∀𝑎∈ R. (b) (i) For no unique solutions, det(𝑇−𝑡· id)= 0. Hence the values of𝑡 are the eigenvalues of𝑇, i.e.𝑡=−2 and𝑡= 4. (ii) If𝑡=−2 then(𝑎,𝑏,𝑐)∈ R3 must be on𝑙. Hence−𝑎=𝑏=𝑐 for a solution to exist. If𝑡= 4 then(𝑎,𝑏,𝑐) must lie onΠ. Hence 𝑐= 3𝑎− 2𝑏. 8 (a) 𝑢=𝑥 d𝑦 d𝑥+𝑦 =⇒ d𝑢 d𝑥 =𝑥 d2𝑦 d𝑥2+ 2d𝑦 d𝑥 =⇒ d2𝑢 d𝑥2 =𝑥 d3𝑦 d𝑥3+ 3d2𝑦 d𝑥2 . Hence d2𝑢 d𝑥2+𝑢= 2. The characteristic equation is 𝜆2+ 1 = 0 so 𝜆 =±i. Hence a solution to the complementary equation in𝑢 is𝑢 = 𝐶1 sin(𝑥)+ 𝐶2 cos(𝑥) for all𝐶1,𝐶 2∈ R. Thus a solution to the nonhomogenous DE in 𝑢 is 𝑢=𝐶1 sin(𝑥)+ 𝐶2 cos(𝑥)+ 2. Let𝐶1 sin(𝑥)+ 𝐶2 cos(𝑥)+ 2=𝑥 d𝑦 d𝑥+𝑦. Then let 𝑣 =𝑥𝑦 so d𝑣 d𝑥 =𝑥 d𝑦 d𝑥+𝑦=𝐶1 sin(𝑥)+ 𝐶2 cos(𝑥)+ 2. Hence 𝑣=−𝐶1 cos(𝑥)+ 𝐶2 sin(𝑥)+ 2𝑥+𝐶3 =⇒ 𝑦=−𝐶1 cos(𝑥) 𝑥 +𝐶2 sin(𝑥) 𝑥 + 2+𝐶3 𝑥 . d𝑦 d𝑥 =𝐶1×𝑥 sin(𝑥)+ cos(𝑥) 𝑥2 +𝐶2×𝑥 cos(𝑥)− sin(𝑥) 𝑥2 −𝐶3 𝑥2. When𝑥= 𝜋 2 ,𝑦= 2 and d𝑦 d𝑥 =− 2 𝜋 : 2𝐶2 𝜋 + 2+ 2𝐶3 𝜋 = 2 =⇒ 𝐶2=−𝐶3, 2𝐶1 𝜋 − 4𝐶2+ 4𝐶3 𝜋2 =− 2 𝜋 =⇒ 𝐶1=−1. When𝑥=𝜋,𝑦= 2+ 1 𝜋 and d𝑦 d𝑥 = 2 𝜋− 1 𝜋2 . − 1 𝜋+ 2+𝐶3 𝜋 = 2+ 1 𝜋 =⇒ 𝐶3= 2 =⇒ 𝐶2=−2. It follows that 𝑦= cos(𝑥) 𝑥 − 2 sin(𝑥) 𝑥 + 2+ 2 𝑥. (b) By GC, ∫ 3 1 cos(𝑥) 𝑥 − 2 sin(𝑥) 𝑥 + 2+ 2 𝑥

d𝑥≈ 4.17431 and ∫ 3 1 4 𝑥 d𝑥≈ 4.39445, both values to 5dp. The relative error𝜀=

4.17431− 4.39445 4.17431 × 100%≈ 5.27% to 3sf, which is a small error. Considering the graphs: LibGSN 9649-PP-2025-01S 3

Singapore–Cambridge Further Mathematics 2025 Paper 1 Solutions © Gerard Sayson 2025 1 1.5 2 2.5 3 1 2 3 4 5 overestimation underestimation underestimation≈ overestimation 𝑥 𝑦 𝑦= cos𝑥 𝑥 − 2 sin𝑥 𝑥 + 2+ 2 𝑥 𝑦= 4 𝑥 the area of the overestimation is approximately the area of the underestimation, so ∫ 3 1 𝑦 d𝑥= ∫ 3 1 4 𝑥 d𝑥− overestimation+ underestimation≈ ∫ 3 1 4 𝑥 d𝑥 and so ∫ 3 1 4 𝑥 d𝑥 is suitable to approximate ∫ 3 1 𝑦 d𝑥. 9 (a) Write𝑄=Σ(𝑦−𝑎𝑥2−𝑏𝑥)2. Then(𝑦−𝑎𝑥2−𝑏𝑥)2=𝑦2+𝑎2𝑥4+𝑏2𝑥2−2𝑎𝑥2𝑦−2𝑏𝑥𝑦+2𝑎𝑏𝑥 3. Hence𝑄= Σ(𝑦2+𝑎2𝑥4+𝑏2𝑥2−2𝑎𝑥2𝑦−2𝑏𝑥𝑦+2𝑎𝑏𝑥 3)=Σ(𝑦2)+𝑎2Σ(𝑥4)+𝑏2Σ(𝑥2)− 2𝑎Σ(𝑥2𝑦)− 2𝑏Σ(𝑥𝑦)+ 2𝑎𝑏Σ(𝑥3). (b) 𝜕𝑄 𝜕𝑎 = 2𝑎Σ(𝑥4)− 2Σ(𝑥2𝑦)+ 2𝑏Σ(𝑥3) and 𝜕𝑄 𝜕𝑏 = 2𝑏Σ(𝑥2)− 2Σ(𝑥𝑦)+ 2𝑎Σ(𝑥3). (c) (i) Since𝑄 is minimal,∇𝑄= 0. Hence 𝑎Σ(𝑥4)+ 𝑏Σ(𝑥3)=Σ(𝑥2𝑦) 𝑎Σ(𝑥3)+ 𝑏Σ(𝑥2)=Σ(𝑥𝑦) and this is a linear system such that A z }| {Σ(𝑥4) Σ(𝑥3) Σ(𝑥3) Σ(𝑥2) 𝑎 𝑏

= Σ(𝑥2𝑦) Σ(𝑥𝑦)

= u. (ii) Recall that if M= 𝑎 𝑏 𝑐 𝑑

, then M−1= 1 det(M) 𝑑 −𝑏 −𝑐 𝑎

. Thus Σ(𝑥4) Σ(𝑥3) Σ(𝑥3) Σ(𝑥2) −1 = 1 Σ(𝑥4)Σ(𝑥2)−( Σ(𝑥3)) 2 Σ(𝑥2) −Σ(𝑥3) −Σ(𝑥3) Σ(𝑥4)

. Considering A−1A 𝑎 𝑏

= A−1u, 𝑎 𝑏

= 1 Σ(𝑥4)Σ(𝑥2)−( Σ(𝑥3)) 2 Σ(𝑥2) −Σ(𝑥3) −Σ(𝑥3) Σ(𝑥4) Σ(𝑥2𝑦) Σ(𝑥𝑦)

= 1 Σ(𝑥4)Σ(𝑥2)−( Σ(𝑥3)) 2 Σ(𝑥2)Σ(𝑥2𝑦)− Σ(𝑥3)Σ(𝑥𝑦) −Σ(𝑥3)Σ(𝑥2𝑦)+ Σ(𝑥4)Σ(𝑥𝑦)

. Hence 𝑎= Σ(𝑥2)Σ(𝑥2𝑦)− Σ(𝑥3)Σ(𝑥𝑦) Σ(𝑥4)Σ(𝑥2)−( Σ(𝑥3)) 2 𝑏=−Σ(𝑥3)Σ(𝑥2𝑦)+ Σ(𝑥4)Σ(𝑥𝑦) Σ(𝑥4)Σ(𝑥2)−( Σ(𝑥3)) 2 (d) 𝑄 is a sum of squares, so it is bounded below. Hence the values of𝑎 and𝑏 in (c)(i), which give a stationary value of𝑄, must correspond to a minimum value of𝑄. (e) (We are asked to find approximate values of 𝑎 and 𝑏.) Note that for all the points, 𝑥 ≈ 𝑦. Hence Σ(𝑥2𝑦)≈ Σ(𝑥3) andΣ(𝑥𝑦)≈ Σ(𝑥2). Using the result in (c)(ii),𝑎≈ 0 and𝑏≈ 1. LibGSN 9649-PP-2025-01S 4

Singapore–Cambridge Further Mathematics 2025 Paper 1 Solutions © Gerard Sayson 2025 10 (a) (i) 𝑧= exp( 2𝜋𝑘 i 15 ) for𝑘 = 0,..., 14. (ii) 𝑧15− 1=(𝑧5)3− 1=(𝑧5− 1)(𝑧10+𝑧5+ 1). Let 𝑧= cos( 2𝜋 15)+ i sin( 2𝜋 15)= exp( 2𝜋i 15). Clearly exp( 2𝜋i 15) is not a solution to𝑧5− 1= 0 so it must be that𝑧10+𝑧5+ 1= 0. (iii) 𝑧𝑛+𝑧−𝑛= cos(𝑛𝜃)+ i sin(𝑛𝜃)+ cos(𝑛𝜃)− i sin(𝑛𝜃)= 2 cos(𝑛𝜃). (b) Let𝑧= cos( 2𝜋 15)+ i sin( 2𝜋 15). Then 𝑧10+𝑧5+ 1= 0 so𝑧5+𝑧−5+ 1= 0. See that 𝑧𝑛