EJC 9649 2025 Prelim P2 Solutions

2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [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 2 9649/02 15 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 2 Section A: Pure Mathematics [50 marks] 1 Solution Let f ( ) ln( 2)xx= + ( )( ) 1.5 1 ln( 2) d 1 0.5 f ( 1) 2(f ( 0.5) f (0) f (0.5) f (1)) f (1.5)2 2.02 (3sf) xx − + ≈ −+ − + + + + = ∫ The graph is concave downwards over the interval 1 1.5x−≤ ≤ . Each The total area of the 5 rectangles is less than the actual area under the curve. Hence, the approximation by the trapezium rule is an underestimate of the area. 2 Solution (a) 22 zxx xz xy ∂ = =∂ + , 22 zyy yz xy ∂ = =∂ + Let 22 0 00z xy= + Equation of tangent plane at 00(,)xy : x y

3 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over 00 00 0 00 22 0 0 00 0 00 0 00 00 00 00 00 ( )( )xyzz xx yy zz x y xyz xy zz z xyz x yzzz xy xyzz = + −+ − += ++− = ++− = + When 0x= and 0y= , we have 0z= So the tangent plane passes through the origin. (b) Normal of tangent plane 0 0 0 x y z   − Angle between the normal and the z-axis 1 0 2 22 0 00 1 0 2 0 1 cos cos 2 1cos 2 45 z xyz z z − − −  =  ++  =  =  = ° Alternative Let the angle between the tangent plane and xy-plane be θ , then 22 2 00 0 22 2 00 0 tan f 1xy z zz z θ = ∇= + = = . Hence the angle between the tangent plane and z-axis is also 45° 3 Solution (a) 22 2 S( , ) 0 3 20 23 (2 3 ) xy xy x y xx yxx = +−= = ±− = ±− So points correspond to perfectly balanced spots must have (2 3 ) 0 20 3 xx x −≥ ≤≤ Alternatively,

4 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 22 2 2 2 2 3 20 113 33 19 31 3 xy x xy xy +−=  − +=  −+= Points correspond to perfectly balanced spots are on an ellipse centred 1,03   with semi-major 1 3 and semi- minor 1 3 . (b) S62x x= − , S2y y= 1S0 3 x x=⇒= S0 0y y=⇒= 1,03 ∴ is a stationary point S6xx = , S2yy = , SS0xy yx= = ( ) 2 S S S 12 0xx yy xy−= > Since S0xx > , S has a local minimum at 1 ,03   (c) On the boundary, 22 2 2 44xy y x+= ⇒= − 22 2 2 S( ) 3 4 2 2( 2) 172 2222 xx xx xx xx = +− − = −+ = − + −≤≤ S( )x∴ is a quadratic function and has a minimum at 1 2x= and 17S 22  = At endpoints 2x=± : S(2) 8= S( 2) 16−= Minimum on the boundary: 1 15 7,,2 22  ± Maximum on the boundary: ( )2,0,16− Hence, absolute maximum is at ( )2,0,16− and absolute minimum is at 11,0,33  − . 4 Solution

5 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over (a) 01 10 3 = A Consider 0λ−=AI 2 1 010 3 (3 ) 10 0 3 10 0 λ λ λλ λλ − =− − −−= −−= 2λ =− or 5λ = For 2λ =− : 21 0 10 5 0 x y      =         An eigenvector is 1 2  − For 5λ = : 51 0 10 2 0 x y −     =    −     An eigenvector is 1 5   (b) Let 1 n n n u u + = v Then 11 01 10 3 n nn −− = = v v Av and 0 0 1 1 1 u u  = =  v By (a), 1 1 11 2 0 11 25 0 5 25 − − −   = =    −−    A QDQ

6 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 1 2 2 0 1 0 1 11 11 11 2 0 11 1 25 0 5 25 1 1 1 5 11( 2) 0 1 25 2 1 1705 4 ( 2) 5 7 3( 2) 5 7 43( 2) (5)77 43( 2) (5)77 nn n n n n n n nn nn nn nn − − − − ++ ++ = = = = −    =    −−     −−    = ×    −       −= −    −+ = −+  v Av Av Av QD Q v       Hence 43( 2) (5)77 nn nu =−+ Alternative Let 12 111 1 25 cc   = +   −    By GC, 12 43,77cc= = 1 11 1 1 1143 2577 1143 2577 1143( 2) (5) 2577 43( 2) (5)77 43( 2) (5)77 nn n n nn nn nn nn − ++ = =     = +    −     = +   −     = ⋅− + ⋅   −    −+ =  −+ v Av A A AA Hence 43( 2) (5)77 nn nu =−+ (c) Let λ be the corresponding eigenvalue for 0 1 u u  ′   ′

7 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over i.e. 00 11 uu uu λ  ′′  = ′′  A 00 1 11 nn nn uu uu λ−  ′′  ∴= == =  ′′  v Av A  and this means 0 n nuu λ= which a geometric progression with common ratio the eigenvalue λ. 5 Solution (a) Volume of required solid of revolution, 2π d a a V yx − =∫ . 2 2 2 1yx b += ( ) 22 2 1yb x⇒= − ( ) ( ) 1 22 1 122 0 133 22 2 0 π 1 d 2π 1 d 142π 2π 1 0 π3 33 V b xx b xx xbx b b − = − = −  = − = −−=     ∫ ∫ (b) 2 2 2 2 2 1 2d d2 0, dd yx b y y y bxx x xyb += +== − Curved surface area of required solid of revolution, ( ) ( ) 1 2 available from MF27 1 1 42 due to reflective symmetry about 02 0 1 2 42 0 1 2 2 42 0 1 2 22 0 1 22 0 d2π 1 d , d 4π 1 d , 4πd 4π1 d 4π 1 d 4 1 1 d (shown x ySy x x bxyx y y bx x b x bx x b x bx x S b b xxπ − = = +  = + = + = −+ = −+ = +−∴ ⌠ ⌡ ⌠ ⌡ ⌠⌡ ⌠⌡ ∫ ∫ ) (c) (i) When 1b= , 2 10b −= .

8 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 ( ) [ ] 1 22 0 1 0 1 0 4π 1 1d 4π(1) 1d 4π 4 π x x S b b xx x x S = = = +− = = ∴ = ⌠⌡ ∫ (c) (ii) When 1b< , 210 b−> . 22 2 2 22 2 2 sin d cos , andd11 sin( 1) ( 1) 1 sin xx bb b xb b θθ θ θ θ = ⇒= −−  −= −  − =− ( ) [ ] ( ) 12 12 12 12 12 1 22 0 sin 1 2 0 sin 1 20 sin 1 2 0 sin 1 2 0 sin 1 02 12 2 2 4π 1 1d d4π 1 sin d d cos4π cos d 1 4π 1 cos 2d21 2π1 sin 221 2π sin cos 1 2π sin 1 1 1 b b b b b S b b xx xb b b b b b b b b b b bb b θθ θ θθθ θ θ θθ θ θθ − − − − − − − − − − − = +− = − = − += − = + − = + − = −+− − ⋅ ⌠⌡ ⌠⌡ ⌠ ⌡ ⌠⌡ 122 2 1 cos 2π sin 1 2 π b bS bb b θ − = = −+∴ −  (d) (i) Consolidating from the results of parts (a) and (c), 2 12 2 22 2 2π2π sin 1 , if 1 1 4π , if 1 2π2π ln 1 , if 1 1 bb bb b Sb bb bb b b − + −< −= =   + +− >  − and 24 π3Vb= ,

9 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over 2 3 12 3 2 3 2 3 2 9π sin 1 , if 12 1 36π , if 1 ln 19π , if Required 12 rati 1 o bbbb b S b V bb bbb b −  −  +<  −= =   +− + ∴ > − Obtaining the graph of 2 3 S V versus b (using the GC) : (d) (ii) When this (isoperimetric) ratio is minimised, the shape of the solid of revolution is a sphere. b S V 2 3 36∙π( ) 1 3 = 4.836 O 1

10 2025 JC2 H2 Further Mathematics Preliminary Examination Paper 2 Section B: Probability and Statistics [50 marks] 6 Solution (a) 0H: The number of absentees is independent of the day of the week. If the number of absentees is independent of the day of the week, then we’d expect the total of 500 to be uniformly spre