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 t