Fu Hua 2025 4E5N AM P1 - Solutions

Candidate Name: Class: I n d e x No.: A n s w e r Scheme. FUHUA SECONDARY FUHUA SECONDARY SCHOOL Secondary 4 Express/ 5 Normal (Academic) PRELIMINARY EXAMINATION 2025 4E5N Fufua S e c o n d a r y School Fuhua S e c o n d a r y School Fuhua S e c o n d a r y S c h o o l Fuhua S e c o n d a r y School Fufra S e c o n d a r y School Fuhua S e c o n d a r y S c h o o f Fufua S e c o n d a r y School Fufua S e c o n d a r y S c h o o f Fuhua S e c o n d a r y School Fuhua S e c o n d a r y School Fuhua S e c o n d a r y School Fuhua S e c o n d a r y School Fultua S e c o n d a r y S c h o o l Fuhua S e c o n d a r y S c h o o l F u f u a S e c o n d a r y School Fuhua S e c o n d a r y S c h o o f Fufua S e c o n d a r y School Fufua S e c o n d a r y School Fufua Secondary School Fuhua S e c o n d a r y S c h o o f Fuhua Secondary School Fuhua S e c o n d a r y School Fuhua S e c o n d a r y School Fuhua Secondary School Fuhua Secondary School Fufua S e c o n d a r y School Fuhua Secondary School Fufua Secondary School Fuhua Secondary School Fuhua Secondary School 4049/01 ADDITIONAL MATHEMATICS Paper 1 DATE TIME DURATION 26 August 2025 1050 - 1305 2 hours 1 5 minutes READ THESE INSTRUCTIONS FIRST Write y o u r class, i n d e x n u m b e r a n d name i n the spaces at the t o p o f this page. Write i n dark blue o r black pen. Y o u m a y use an H B pencil f o r any diagrams o r graphs. D o not use staples, paper c l i p s , glue o r correction f l u i d . Answer all the questions. Give non-exact numerical answers correct t o 3 significant f i g u r e s , o r 1 d e c i m a l place i n the case o f angles i n d e g r e e s , unless a d i f f e r e n t level o f a c c u r a c y is s p e c i f i e d i n the q u e s t i o n . The use o f a n approved scientific calculator is expected, where appropriate. Y o u are r e m i n d e d o f the n e e d for clear presentation i n y o u r a n s w e r s . The n u m b e r o f marks is g i v e n i n brackets [ ] at the e n d o f each question o r part question. The total number o f marks f o r this paper is 9 0 . F O R EXAMINER'S USE Statements/Workings A c c u r a c y ( i n c l . U n i t s ) Presentation Setter: Ms Gao Conger PARENT'S SIGNATURE 1 90 Vetter: Ms Lim Fen Niu This document consists o f 2 4 printed pages.

Mantematical Fomia 1 . ALGEBRA Quadratic Equation F o r the e q u a t i o n ax + b x +c=0, 2a Binomial Expansion w h e r e i r a posie i n t e g e r ma () a -_п(п-1)... (п -r+1) r!(n-r)! Identities 2 . TRIGONOMETRY sin' A+cos A = 1 sec? A = 1 + tan' A cosec' A = 1 + c o t * A s i n ( A + B ) = s i n Acos B ‡ cos A s i n B cos(A B ) = cos Acos B F s i n A s i n B tan(A‡ B ) = - tan A tan B 1 7 t a n A t a n B sin 2A = 2 sin Acos A c o s 2 A = cos' A-sin' A=2cos' A-1=1-2sin' A . 2 tan A t a n 2 A = - 1 - t a n A Formulae f o r ABC a C ɾ ＝ ʔ sin A sin B sin C a ' = b 2 + * -2bccos A 1 . A==bcsinA

S w A p (presentation) § F o r all questions m i s s i n g essential brackets table without borders x3-4x2-24x+41 1 Express i n partial fractions. x2-4x-21 [ 5 ] X : 4 4 0 6 2 1 ) 2 3 - 4 2 2 - 2 4 2 + 4 1 - ( 2 3 - 4 2 2 - 2 1 2 ) - 3 5 + 4 1 X - 4 2 8 - 2 4 8 + 4 1 2 2 - 2 2 - 2 1 - 3 + 4 - = x + - 1 ʔ 4 2 0 - 2 1 - M I let -301+41 ( 1 - 7 ) ( X + 3 ) x-7 + - 0 1 + 3 - 3 X + 41 = A ( x + 3 ) + B ( X - 7 ) l e t 2 （ ： - 3 - 3 （ - 3 ） + 4 1 = 3 1 - 3 - 3 ） 5 0 = - 1 0 B - 5 = B let d : 7 , - 3 8 7 ) + 4 1 = A ( 7 + 3 ) 2 0 = 10A 2 = A ʔ M I - M I - M I 2 2 4 2 - 2 1 S W A p (statement) : original expression missing from h a l statement o R fixt) used without clefining if i m p r o p e r f r a c t i o n is net split u p , maximum 2/5 marks fet A R B v a l u e s 5m [Turn over

4 2 ( a ) I n a quiz q u e s t i o n , the quadratic e q u a t i o n k (x*+1) - 8x- 6 x ' = 0 is g i v e n . Find the range o f values o f the constant k , f o r w h i c h the equation has n o real roots. [4] x ( - 1 ) kx' + 4 - 8 2 - 6 7 7 = 0 4x7-67-826+1=0 a=K-6, b=-8, C= K 5 3 - 4 a c < 0 ( - 8 ) - 4 1 k - 6 1 6 k ) 5 0 6 4 - 4 k * + 2 4 K < 0 4 k * - 2 4 k - 6 4 > 0 - M I 2X(-1) - M I (simplify b - 4 a c , a n d change t o positive coefficient at ka and change inequality sign) k ' - 6 K - 16 > 0 ( 1 - 8 ) 1 1 + 2 ) > 0 F o r (K-8) (K+2) = 0 k=8 ar k = - 2 i: K < - 2 a k > 8 - M I (factorsation) coptional w o r k i n g ) - Al SWAP (presentation): "and" used in fral answer 4m

5 (b) Another question i n the quiz is shown below . F i n d the range o f values o f the constant k , f o r w h i c h the curve y = ( k - 6 ) x * - 2 x lies c o m p l e t e l y above the l i n e y = 6x-k. - 0 - 2 Jeremy stated that the range o f values o f k found i n part (b) is t h e same as that i n part ( a ) . Is J e r e m y correct? Explain y o u r a n s w e r clearly . S u b . Dinto 2 (1-6)27-206=67-K (476) 27-226-67+k=0 （ K - 6 ） 8 - 8 x + k = 0 b'-цас< 0 ( - 8 ) - 4 ( K - 6 7 K ) < 0 fran (9), K8-2 o r k> 8 H o w e v e r , the curve s h o u l d be a U-shaped curve o R the curve s h o u l d o p e n u p w a r d s . o R coefficient o f d'should b e p o s i t i v e ) YM 5 , K = 6 > 0 k > 6 [ 2 ] i t h e solution fursb) is o n l y k > 8 con: the solutionder c b ) cannot incude kr-2) (as a n d o b ) have different ranges u f values o f k :: N o , J e r e m y is not c o r r e c t . Al 2m [Turn over

6 3 Solve the simultaneous equations. 14] 2x+y=10 - 0 3 + 2 • = 2 x From D , y=10-226 - 3 3x+2y=2xy - 4 sub.@ i n t o D 3X+2(10-22) =2x610-2X) -MI (Substitution) 3 2 + 2 0 - 4 2 0 = 2 0 8 - 4 2 6 3 2 + 2 0 - 2 0 2 4 4 2 2 = 0 4 2 2 - 2 1 2 6 + 2 0 = 0 - MI ( simplify c o r r e c t l y ) （ ϝ ʔ 4 ） （ 4 X - 5 ） = 0 x = 4 d x = 5 S u b . a l 4 i n t o ᶅ y=10-264) = 2 Sub.X=E i n t o B , y=10-2(4) = 1 6 :. x = 4 , y = 2 o r x=5, y = 1 5 M e t h o d a S u b . 3 intoX. 1 2 2 x + ܳ = 2 3x+2(10-2x) - = 2 X ( 1 0 - 2 7 ) 3 X + 2 0 - 4 2 0 - = 2 1 0 2 - 2 x 2 - 8 + 2 0 = 2 0 1 - 4 8 2 4 2 2 - 2 1 2 0 4 2 0 = 0 - MI (subsertution) - M I (simply correctly) 4 m

7 4 I n triangle A B C , B C = ( V 7 - V 2 ) cm a n d angle A B C = 3 0 ° . The area o f triangle A B C i s (47 - V 2 ) c m ? . W i t h o u t u s i n g a c a l c u l a t o r , e x p r e s s the l e n g t h o f A B i n t h e form ( p + 9 V 1 4 ) c m , where p a n d q are constants, a n d state the value o f p a n d o f q . [ 4 } ? 30° （ 厚 - 五 ） C area=≤* A B X B C * S i n L A B C ⼀ x A B x （ 有 ⼀ 下 ） x s i n 3 o ° = 4 万 - 五 占 x A B x （ 阿 ⼀ 五 ） x ⼠ ⼆ 4 万 ⼀ 下 A B * （ 再 - 五 ） = 1 6 ⽉ - 4 万 A B ≤ _ 1 6 万 - 4 五 万 + ʔ ϝ N 7 - N 2 N 9 + J I - M I - M I cratioalise denominator ) (allow e r f ) = 1 6 ( 3 ) + 6 1 : 4 J - W 2 - M ( e r p a n s i o n ) 7 - 2 callow ecfuf similar diffralty) = _ 1 0 4 + 1 2 , J 1 4 = ( 1 0 4 + 1 I ) cm p = 1 0 4 , 9 = 1 _ ^ Al 4m [Turn over

8 5 A c i r c l e C 1 , w i t h centre C , has e q u a t i o n x*-8x+y*+2y-17=0. (a) F i n d t h e coordinates o f C a n d the radius o f t h e circle. [ 3 ] ( + - 4 3 - 1 6 + 1 9 + 1 3 - 1 - 1 7 = 0 - M 1 （ x - 4 3 + 1 9 + 1 ） 3 = 3 4 centre C ( 4 , - 1 ) - Al radius = 1 3 4 units - Al ( 0 2 5 . 8 3 u n i t s ( 3 5 . 8 . ) s w A p caccuralys: m i s s i n g unit / degree ut accuraly . Method 2: compare with x+ y2+291+2fy+C=0 2 g 2 - 8 2 4 = 2 C = - 1 7 g=-4 ⼨ ⼆ 1 M I (correct 9 , f,c) Centre C ( 4 , - 1 ) - Al r a d u s = 1 1 + 1 4 9 - 0 - 1 ） = N 5 4 units ( b ) Determine whether the point S (-1, - 5 ) lies inside, outside o r o n the circle. Show y o u r working clearly. [ 2 ] - M I (allow e r f ) = 7 4 =6.40 units (35.P .) > 5.83 units The distance u f s from centre o f circle is l a r g e r than the radius or circle. . . S lies outs