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%≈