AM4 4049 BSSS 2023 P2 MS

Marking Scheme BSSS AM4 4049 P2 2023 1 By Factor Theorem, ( ) ( 2)(2 1)( ) By Remainder Theorem, (1) 6 (1 2)(2 1)(1 ) 6 ( 1)(3)(1 ) 6 12 1 fx x x x k f k k k k = − +− =− ⇒− + −= − ⇒ − −= − ⇒ −= ⇒= − M1 M1 A1 2a 3 2cos 2Period = 5min/10 = = 0.5min 4 rad/min d kt k k π π = − = M1 A1 2b >2, 3 2cos 4 2 1 cos 42 1Let cos 4 = , 2 , 4 lie in quad 1&43 4 ,2 ,2 ,4 ,4 ,. . . 5 7 15 17,,,,, . . .3(4 ) 3(4 ) 3(4 ) 3(4 ) 3(4 ) 1 5 7 15,,,, . . .12 12 12 12 from the sketch, for >2, 51 12 12 d t t t t t t t d t π π π παπ π απαπαπαπα πππππ πππππ −> > = =−+−+ = = = − 1 = 3 10In 5 min, for 10 cycles, time = min.3 M1 M1 M1 A1 3a 2ln ( 1) when 2, 2 2 ln( 1) 1 yx y xa a ae = − = = = − = + M1 A1 3b 2 22 2 00 2 2 0 1 2 2 0 2ln ( 1) 1 d 1 d 2 22 2 2 2 ln( 1) d = 2( +1) d = 2( +1) 2 = 2 (Show y y y e yx xe xy e y ye e e xx e xy ee + = − = + = + = +  = +− = −= − − ∫∫ ∫ ∫ n) M1 - x M1 - intg M1-intg A1 - ans M1 A1 4a 2 22 2 sec 1 1 sec cos 1 2 2cos 2 2 cos 2 1 2 2 (Shown)cos 2 3 x xx x x x =++ = + = ++ = + M1 M1 A1 4b 2 2 sec 2 4 1 sec 2 7 24 cos 4 3 7 14 cos 4 34 2cos 4 4 5 7 114 , , , ,...33 3 3 5 7 11, , , ,...12 12 12 12 for 2 2 , there will be 16 solutions. x x x x x x x x πππ π πππ π ππ =+ ⇒= + ⇒= + ⇒= ⇒= ⇒= − ≤≤ M1 M1 A1 A1

5a G1- Shape of Graph + y- Intercept G1-Labels 5b 22 22 22 d d Since 0 and 0, d 0, for all values of .d the function is an increasing function. xx xx xx y eex ee y ee xx − − − = + >> = +> ∴ M1 A1 5c 22 22 22 For equation of function, 11 22 when 0, 3, 3 11 322 xx xx xx yee ye e c xy c ye e − − − = + =−+ = = = ∴= − + ∫ M1 M1 A1 6a (Angle in alternate segment) = = (Given) DAG DCA ECF ECB ∠= ∠ ∠ ∠ M1 A1 6b is midpoint of AC (Given) F is midpoint of DG (Given) ACD is similar to ECF. (Midpoint Theore m) BCA= BDA (Angle in the same segment ADCB) BEC= AED (Angle in the same segment AB) BCE is similar t E ∴ ∠∠ ∠∠ ∴   o DCA. (AA) BCE is similar to FCE.∴   M1 M1 M1 A1 6c Since ACD is similar to ECF. 1 2 Since BCE is similar to ECF 1 2 1 = (Shown)2 AC CD DA EC CF FE BC CE EB EC CF FE EC BE BC FE BC DA AD BC = = = = = ∴×=× = × ×   M1 A1 Alternatively Since BCE is similar to ECF 1 (Midpoint Theorem)2 1 = (Shown)2 BC CE EB EC CF FE EC BE BC FE BC DA AD BC = = ∴×=× = × ×  M1 A1 7 ( ) ( ) ( ) 13 2 32 32 32 32 32 2 2 74 52 2(2 ) 4 7(2 ) 5(2 ) Let y = 2 , 2 47 5 2 7 5 40 Let ( ) 2 7 5 4, ( 1) 2( 1) 7( 1) 5