2020 HCI H2 Prelim P1 Soln edited

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 1 of 18 Qn Solution 1 22 d 1d 12, 1dd d d d dd d xy t ttt y y t xx t = += − ∴= d d y x ( ) ( ) ( ) 2 2 2 2 2 2 2 2 11 12 1 21 132122 21 13 2 22 1 13 2 22 1 t t t t t t t t − = + −= + +− = + = − + = − + 22 2 1 2 21 1 1 2 3 2 tt t +− + −−− − (ii) since 0t ≠ ( ) ( ) ( ) 2 2 2 2 2 11 110 222 1 33 02 22 1 1311 22 22 1 t t t t +> << + − <− < + −< − < + ∴ d11 d2 y x−< <

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 2 of 18 Qn Solution 2 ( ) ( ) ( )( ) ( )( )( ) 5 5 5 5 23 1 2 2 21 2 56 56715 1 ...32 2 2! 2 3! 2 ax ax ax ax ax ax − − − + = + = +   −− −−− = −+ + +    Coefficient of 3x = ( )( )( ) 3 35671 35 32 3! 2 256 a a−−− ×= − ( )( ) 1 2 211 22 4 21 4 12 1 ...2 4 2! 4 ax ax ax ax − = −  − = +− + − +    Coefficient of 2x = 21 64 a− 32 32 35 1 5256 64 35 20 7 4,since 0 4 7 aa aa aa a −= −  = = ≠ =

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 3 of 18 Qn Solution 3(i) 2 2 22 32 4 2 26 22 26 8 x x xx xx x x + − −+ − + − ( ) ( ) 2 21 8f 2233 xxx xx −= = ++−− The asymptotes are y = 2x + 2 and x = 3. 3(ii) y x y x

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 4 of 18 Qn Solution To have an odd number of roots, there are 7 points of intersection between the two curves, hence 2 3ak= + 4(a) Algebraic Method: ( )( ) ( )( ) 2 2 2 2 2 2 2 2 1 ( 1) 01 22 01 22 01 ( 1) 3 011 ( 1) 3 ( 1) 3 011 xx x xx x xx x x xx xx xx +− − ≤− −++ ≤− −− ≥− −− ≥−+   −− −+   ≥−+ ( 1) 3 ( 1) 3 ( 1)( 1) 0x x xx  −− −+ − +≥   + – + – + 1 or1 3 1 or 1 3x xx− −≤ ≥< +< 4(b) 23 2ay bx cx+ += Differentiate wrt x, 2 0d23 d yay bx cx+ += … (*) Sub ( )1, 3 into eq C and obtain 3 2 ..... (1)abc++= Sub ( )1,1− into eq C and obtain 2 ..... (2)abc−−= Sub ( )1,1− into eq * and obtain 3 3 0 ..... (3)a bc− + += –1 13− 1 13+

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 5 of 18 Qn Solution Using GC, 1, 2, 3ab c= = =− Eq of C is 23 2 32y xx+ −= 5 Note that 2 22l hr= + . Total surface area of cone, 22 2rh r r kπ ππ ++ = 22 2rh r k r⇒ += − ( ) 2 22 2 24 2r h r k kr r⇒ += − + 2 24 2 24 2r h r k kr r⇒ +=− + ( ) 2 2 2 2 kr hk ⇒= + (Shown) From above equation, 2 2 2 2 kr hk= + ( ) 22 2 2 2 11 332 32 k khV rh hhk hk πππ = = = + + ( ) ( ) ( ) 22 22 22d d3 2 h k hhVk h hk π +− = + ( ) ( ) 22 22 2 32 k kh hk π − = + At stationary point, d 0d V h = ( ) 22 0 2 0kh h k h−=⇒= >  h ( )2k − ( )2k ( )2k + d d V h +ve 0 -ve Alternatively, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2222 42 2 222 32 22 32 2 (2) 2 2 2 2d d3 2 2 222 3 2 26 0 when 23 2 hk h k hhk hVk h hk h k khkh hk k hh k hk hk π π π + −−− + = + −+− − = + −= <= + Volume is maximum when 2hk=

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 6 of 18 Qn Solution ( ) 3 2 2 32 22 = units3 2 2 12 12 k kk k kV kk ππ π= = + 6(a) 3e d5 0.3e x x x−∫ e d5 0.3e 10ln 0 5 0.3 .1 e 30 x x x x C = − = −− −− + ∫ where C is an arbitrary constant. 6(b) Let ( )cos ln dI xx=∫ ( ) ( ) cos ln 1' sin ln ux ux x = =− , '1v vx = = ( ) ( ) ( ) ( ) 1cos ln sin ln d cos nl sin nl d I x xx x x x xx xx = −−  = + ∫ ∫ ( ) ( ) sin ln 1' cos ln ux ux x = = , '1v vx = = ( ) ( ) ( ) ( ) ( ) ( )o n 1 s cs c cos ln cos ln in l ssin ln o n ld d n l I xxxx xx x x xx x x xx − = + =+− ∫ ∫ ( ) ( ) ( ) ( ) 2 cos ln sin ln cos ln sin ln2 Ix x x x xI x xC = + = ++ where C is an arbitrary constant. 6(c) When 2e 5 0, ln 2.5x x−= = ( ) . 2e 5 ln 2 2 2e , e5 5 .5 2, ln 5 x x x x x −   −< −≥ − 3 0 2e 5 dx x−∫

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 7 of 18 Qn Solution ln 2.5 3 0 ln 2.5 2e 5 d 2e 5 dxx xx= − −+ −∫∫ ln 2.5 3 0 ln 2.5 5 2e 2e 5xxxx = − +−  ( ) ( ) ( ) ln 2.5 0 3 ln 2.5 5ln 2.5 2e 2e 2e 15 2e 5ln 2.5 = − ++  −− − ln2.5 310ln2.5 4e 2e 13= − +− ( ) 310ln 2.5 4 2.5 2e 13= − +− 310ln2.5 2e 23= +− 7(i) d 2sin cosd sin 2 0 2 2 1 y xxx x x x y π π =− = −= =− =− =− Minimum point = ,12 π−− 7(ii) 0 2 /2 0 /2 0 /2 2 Area 2 sin d2 1 cos 2 d2 sin 2 24 0 4 3 units4 xx x x xx π π π π π π ππ π − − − = ×+ − −= − = −−  = − −−   = ∫ ∫ 7(iii) Volume ( )( ) 2 20 1 1 3 (2 1) sin d2 7.75156917 2.304987524 10.05655669 10.1 units (to 3 s.f.) yyπππ − − = − + −− = + = = ∫ x y −π 0 π -3 -2 -1 0 1 R x = –π/2 y = –2

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 8 of 18 Qn Solution 8(i) ( )( ) ( ) ( ) * 2 ** 22 2 Re zwzw z w w z ww z wz w −− =−+ + = −+ Since ( ) 2 Re ,ww ∈  , therefore the product of the factors is a quadratic polynomial with real coefficients. 8(ii) ( )( ) ( )( ) 115 cos tan 2 isin tan 2 125i 55 12 w i −−= − +− = −  = − 8(iii) ( )( ) ( ) 2*2 2 2Re 25 zwzw z w z w zz − −= − + =−+ Method 1 ( )( ) 43 2 2 246 125 2 5 25 0z az z bz z z z kz+ + ++ = −+ ++ = Compare coefficient of 2z : 46 25 5 2 8 kk= +− ⇒= − Compare coefficient of 3z : 2 10ak a=−⇒= − Compare coefficient of z: 50 5 90b kb= −+ ⇒= − Method 2 Substitute 1 2iz= − into the equation, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 43 2 1 2i 1 2i 46 1 2i 1 2i 125 0 7 24i 11 2i 46 3 4i 1 2i 125 0 20 11 160 2 2 i 0 ab ab ab a b −+−+−+ − += − ++ − ++− −+−+ = − − ++− + − = Comparing Real and Imaginary part respectively, 11 20 10 2 2 160 90 ab a ab b −= − = − ⇒ −= = − 2 1

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 9 of 18 Qn Solution The remaining roots are ( ) 2 2 8 25 0 4 90 4 3i zz z z ∴−+= − += = ± The roots to the equation are 4 3i, 4 3i, 1 2i, 1 2i+−−+ . 9(i) 2 d ,d xa bxtx= − where a & b are positive constants. dW 0he 5 d 8 n 0 2 ., 4 x t a a x b b = = = = ( ) ( ) 3 22 3 2 18d 8d 18d d axxa axtx x kxx tx − = −= − ∴= 9(ii) Method 1: ( ) ( ) ( ) 3 2 2 3 2 3 2 3 3 d 18 d d d 18 d1d dd 18 1 24 d24 18 1 ln1 824 xx ktx tx x kx txkx x x x xkt x x kt x C  −=   = − = − −=− − = − −+ ∫∫ ∫ 3 3 24 24 3 24 24 ln 1 8 24 24 18 e e 1 8 e where e C kt kt C x kt C x xA A − − −= −+ −= −= = ± When t = 0, x = 3, A = −215 3 241 8 215e ktx −∴− = − When t = 1, x = 2, 24 63e 215 k− =

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 10 of 18 Qn Solution 3 631 8 215 215 t x ∴− = −  When t = 3, 1 3 31 631 215 0.928772 215x  = +=   ∴ Number of fish is 929 (nearest integer) 9(iii) As 63, 0, 0.5215 t tx →∞ → → In the long run, the number of fish decreases and tends to 500. 10(i) 10 (ii) When ,2 ax≤ every horizontal line yk= cuts the graph of f( )yx= at most once. Hence f is one-one and therefore 1 f − exists. The greatest value of k is 2 a . 10(iii) Let 2 3 22 24 aay a x ax a x =−= −−+  2 3 3 3 1 24 1 24 1 24 aaxy a aaxy a aaxy a  −= −   −= ± −  = ±−  x y O ( ,0)a 3 , 24 aa   (0,0)