EJC 9649 2025 Prelim P1 Solutions

2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CIVICS GROUP INDEX NO. FURTHER MATHEMATICS Paper 1 9649/01 02 September 2025 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF27) READ THESE INSTRUCTIONS FIRST Answer all questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 6 printed pages and 2 blank pages.

2 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 1 Solution (a) Cofactor expansion along the first column: 2 2 d e t ( ) () ( ) () () () () () () ( ) ( )( ) ( )( ( ) ( )) ( )( )( ) β αβ γ γα α αβ γ βγ α γα β βγ α β αβ γαβ αβ α γβ γ α β αβ γ αγ βγ αβγ βγ α βγ αββγγα = −− ++− = − −− −+− + =−−−++ =− −− − = − −− A (b) det 3( )( )( )αββγγα= −− − −B 2 Solution (a) sin 4 z y xyx π∂ = −+ ∂  , sin 4 z x xyy π∂ = −+ ∂  ddd ddd 1sin e sin 4 41 t z zx zy t xt yt y xy x xy t ππ ∂∂= +∂∂  = − +⋅ − +⋅  +  When 0t = , 1, 0xy= = Therefore d2 sind 42 z t π= −= − (b) f (, ) 0xy xy = implies fx is independent of y f (, ) 0yx xy = implies fy is independent of x So f must take the form of f ( , ) g( ) h( )xy x y= + for some functions g and h

3 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over 3 Solution (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 21 11 4 44 2 4 42 2d 1 d4 22 44 2222 11 11 24 2422 4 22 11 442 4 21 42 4 ,1 when 2 4 ,42 4 1 24 2 Length of arc 1 d 1 d 1 d d 2d d d , ln x xx y xx xx xx x x xx xx x PQ x xx x xx x xx x xx xx x + = += + ≤≤ = + = +− = + −+ = ++ = ++ = + = +  = + ⌠ ⌠ ⌡⌡ ⌠⌠⌡⌡ ⌠⌡ ⌠⌡ ∫  ( ) ( ) 4 2 221 1 14 1 2 4 42 4 4 2 ln4 ln2 6 (ln ) 6 ln2. (shown)  = −+ − = + = + (bi) ( ) 4 2d d 2 4 2 2 Area of surface of revolution generated by arc about the -axis 2π1 d 121 d 4 120.4277... 120.43 (2 d.p.) y x PQ y xx xx x xπ = + = +−  = = ⌠⌡ ⌠ ⌡ (bii) ( ) 2 4 2 4 ln 242 Volume of solid of revolution generated by shaded region abt. -axis 2 πd 2 π d, 177.9648... 177.96 (2 d.p.) xx y xy x xx = = − = = ∫ ∫

4 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 4 Solution By de Moivre’s Theorem, 5(cos isin ) cos5 isin 5θθ θ θ+= + 5 54 3 2 23 4 5 5 32 4 4 23 5 (cos isin ) cos 5cos (isin ) 10cos (isin ) 10cos (isin ) 5cos (isin ) (isin ) cos 10cos sin 5cos sin i(5cos sin 10cos sin sin ) θθ θ θθ θθ θθ θθ θ θ θθ θθ θθ θ θ θ + = ++ + ++ = −+ +− + Hence ( ) ( ) 5 32 4 4 22 4 4 2 2 22 4 2 4 24 42 cos5 cos 10cos sin 5cos sin cos (cos 10cos sin 5sin ) cos cos 10cos (1 cos ) 5(1 cos ) cos cos 10cos 10cos 5 10cos 5cos cos (16cos 20cos 5) θ θ θθ θθ θ θ θθ θ θθ θ θ θ θθ θ θ θ θ θθ θ = −+ = −+ = − − +− = −++ −+ = −+ Substitute 10 πθ = : 42 42 42 cos cos 16cos 20cos 52 10 10 10 0 cos 16cos 20cos 510 10 10 16cos 20cos 5 0 since cos 010 10 10 ππ π π ππ π ππ π = −+  = −+  − += ≠ 2 20 80 5 5cos 10 32 8 π ±±= = Note that the cosine function is decreasing from 0 to 2 π , so 22 3cos cos10 6 4 ππ>= . Since 5 53 84 − < , 2 55cos 10 8 π += 2 55 15cos 2cos 1 2 15 10 8 4 ππ  ++= −= −=  (shown)

5 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over 5(a)(i) Solution (a)(ii) The circle’s tangents through the origin have gradients 5 12± , both smaller in magnitude than the half lines’ gradients 1 2± . So the points satisfying 13 5z−≤ also satisfy 1111tan arg( ) tan22 z−−−≤≤ . Hence the inequality 1111tan arg( ) tan22 z−−−≤≤ is implied by 13 5z−≤ . (b) Maximum distance of z from the origin 18= Minimum distance of z from the origin 225 13 194 = + = 6 Solution (a) Consider λ=Ax x where λ is an eigenvalue and x is the corresponding non-zero eigenvector. Pre-multiplying both sides by Tx : TT TT 2T ( ) () λ λ λ = = = x Ax x x x Ax x x x Ax x Since A is positive definite, T 0>x Ax for all non-zero x. Also, 2 0>x since x is a non-zero. Therefore, T 2 0λ = >x Ax x (shown) 5

6 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 (b) ( ) ( ) T 22 22 22 2 2 22 2 2 2 2 2 ( 0) ab xxy bc y xax by bx cy y ax bxy bxy cy ax bxy cy ba x xy cy aa bba x y y cyaa b ac bax y yaa   =      = ++   =+++ = ++ = ++ ≠ = +−+  −= ++    x Ax For A to be positive definite, we need 0a> and 2 0ac b−> . (c) 21 13 = A Since 20a= > and 2 50ac b−= > , A is positive definite Tf( , ) 0xy∴=> x Ax for all (, )xy except the origin where f (0,0) 0= Hence the origin must be the global minimum.

7 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over 7 Solution (a) 1 32nnuu+ = + for 0n≥ with 0uk= 1 2 2 2 2 3 32 3 12 0 32 3(3 2) 2 3 32 2 3 (3 2) 3 2 2 3 3 2 32 2 3 3 2 3 2 32 2 2(1 3 )3 13 ( 1) 3 1 nn n n n n nn n n n n uu u u u u u k k − − − − − −− = + = ++ = +⋅+ = + +⋅+ = + ⋅+⋅+ = + ⋅+ ⋅+ +⋅+ −= ⋅+ − = +⋅−   Alternatively 1 1 2 2 0 32 1 3( 1) 3 ( 1) 3 ( 1) nn nn n n uu uu u u − − − = + += + = + = +  3 ( 1) 1n nuk∴= +− (b) 1 32 n nnvv+ = + Multiplying 1nx + to both sides, 1 11 1 111 1 000 32 32 n n nn nn n n nn nn nnn v x vx x v x vx x + ++ + ∞∞∞ +++ + = = = = + = +∑∑∑ 0 1f( ) 3 f( ) 12x v xx x x− = +⋅ − (1 3 )f ( ) 1 12 xxx x−= + − 1(1 3 )f ( ) 12 xxx x −−= − 1f( ) (1 3 )(1 2 ) xx xx −= −− 1 (1 3 )(1 2 ) 1 3 1 2 x AB xx x x − = +− − − − Let 1 3x= , 11 3 2112 3 A − = = −  Let 1 2x= , 11 2 1113 2 B − = =− − 

8 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 Hence, 21f( ) 13 12x xx −= +−− (c) 00 0 21f( ) 13 12 2 (3 ) (2 ) (2 3 2 ) nn nn n nn n x xx xx x ∞∞ = = ∞ = −= +−− = − = ⋅− ∑∑ ∑ Hence 23 2nn nv = ⋅− 22233 n n n v = −→ as n→∞ Further note: 1 1 1 32 32 n nn n n n vv vv + − − = + = + By eliminating 2n , ( )11 11 133 2 56 nn n n n nn vv v v v vv −+ +− = +− = − Therefore, the sequence is also a solution of another second order linear recurrence relation. 8 Solution (a) Given that 22 2 cos sin 0rr θθ− += , 22 22 22 ( 2cos ) ( 2cos ) 4(1)(sin ) 2(1) 2cos 4(cos sin ) 2 cos cos sin cos co 2 s r r θθ θ θ θθ θ θθ θθ ± ± ± ⇒= ± −− − −= −= = −

9 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over Curve C is a combination of an outer branch 1C & an inner branch 2C . (b) y x θ = 0C2 : r = cosθ − cos2θ C1 : r = cosθ + cos2θ θ = − π 4 θ = π 4 2O y x θ = 0C2 : r = cosθ − cos2θ C1 : r = cosθ + cos2θ θ = − π 4 θ = π 4 2O

10 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 1 due to reflective symmetry Required area of region enclosed by (shaded) 2 a nd above horizontal axis 0 Area of region enclose d b y C C θ⋅= = about 0 2 θ= = 1 2 ⋅ π 4 2 outer0 1d 2 r θ −∫ ( ) ( ) ( ) ( ) π 4 π 4 π 4 π 4 π 4 2 inner0 22 0 2 2 0 0 2 0 d cos cos 2 cos cos 2 d cos cos 2 d cos cos 2 4 cos 2 d = 4 d (shown). 2cos cos 2 2cos cos 2 cos cos 1 2sin r θ θ θ θ θ θθ θθ θ θθ θθ θθ θ θ θθ θ   = + −−  =  − ++ −+ −  = ⌠⌡ ⌠  ⌡ ∫ ∫ ∫ π 4