SAJC 9758 2023 Prelim P1 Solution

2023 H2 Math Prelim Paper 1 Solutions 1 No Solutions 1 32 2 22 32 00 2 432 0 f( ) f ( 1) 15 (1) f (2) 8 4 2 3 (2) f '( ) 3 2 f '(1) 3 2 0 (3) f ( ) d 6 ( ) d 5 111 5432 84 2 2 5 (4)3 Using GC t x ax bx cx d abcd abc d x ax bx c a bc x x ax bx cx d x ax bx cx dx a bcd = + ++ − = −+−+ = − − = + + += − = ++ = + += − = ⇒ + ++ =  +++= +++= − ∫∫ o solve (1), (2), (3), (4) 1.5, 6, 7.5, 0 abc d= = −= = 2(a) ( ) ( ) ( ) ( ) ( ) ( ) 1 22 22 11 ( 1) ( 1) ( 1) ( ( 1)) 1 2 11 2 n nnuSS nn n n n n nn nn nn n n −= − = +− −+−  = −− +−− =+− −−+ = −+ = The general term 2nun= ( ) ( ) 1 2 2( 1) 2 constant nnuu n n−−= −− = 1Since is a constant independent of , he nce { } forms a GP.nn nuu n u−− 2(b) Let a denote the first term of the geometric progression. Let b and d denote the first term and common difference of the arithmetic progression. ∴ ar2 = b + 6d …(1) ar4 = b + 12d …(2) ar6 = b + 24d …(3) (2) – (1): ar4 – ar2 = 6d …(4) (3) – (2): ar6 – ar4 = 12d …(5) (4)/(5): ( ) ( ) 22 42 1 6 121 ar r d dar r − = −

2023 H2 Math Prelim Paper 1 Solutions 2 2 4 2 1 2 11 2 2 r r r r = = =± Since 0, 2rr>= Since |r| > 1, the geometric progression is not convergent. 3(i) Let the height of the isosceles triangle be a cm. 222 222 4 4 xak xak += = − Therefore, height of the pyramid 222 22 44 2 xxk xk = −− = − Volume of pyramid, V 2 2 22 2 1 base area height3 1= ()32 32 xxx k xx k = ×× − = − Hence 4222 92 xxVk  = −  . (ii) 4 2 24 622 9 2 9 18 x x kx xVk  = −= − Differentiating with respect to x, ( ) 23 5 23 5 32 2d4 6 4 3 12 43d 9 18 9 9 9 V kx x kx xV xk xx = −= −= − dWhen 0, d V x =

2023 H2 Math Prelim Paper 1 Solutions 3 ( ) 32 2 22 1 43 09 Since 0, 430 23 23 or (rejected 0)33 xk x x kx x kx k x −= ≠ −= = = −>  23 5d4 32 d 99 V kx xV x = − Differentiating with respect to x, ( ) 22 22 4 22 2 2 d d 12 15 12 45d 9 93d V V kx xV xk xxx   + = −= −  When 23 3xk= , d 0d V x = , 2 22 2 2 2 2 22 2 d 14 4 4 82 4533 3 9 3d d 1 16 27d VV kk k k k x Vk Vx      = −= −           =− Since 0V > , then 2 2 d 0 d V x < when 23 3xk= . Therefore 23 3xk= will maximise the volume of the pyramid. 4 (i) (ii) y = 4 (2,1) (1, 0) 10, 2   1y=−

2023 H2 Math Prelim Paper 1 Solutions 4 maximum a =1 (iii) ( )1g( ) 2 1 21 xx x ⇒= − −  <  ( )1g( ) 1 2 2 xx = − ( )1Let 1 2 2 xy= − 2(1 )2x y= − ( )2log 2(1 )xy= − 21 log (1 )xy= +− Since 1()xg y −= , 1 2( ) 1 lo (g g1 )yy− = +−∴ 1 2( ) 1 lo (g g1 )xx− = +−∴ 1 (0,1)ggDR− = = (iv) 342 35 1 2 3 ) 6 f( x x x ⇒− = − ⇒= = - O

2023 H2 Math Prelim Paper 1 Solutions 5 ( ) ( ) 1 2 21 342 35 3 35 33 3 22 5 , 2, 13,, 6 13,. 6 , 2, 1 1 2 1 for2 1f ( ) 2 1 for 22 34 for34 5 2 1 for 2 1 for 2 9 154 for21 3,, 6 44 13,. 6 x x x x x x x xx xx x x x x x x x xx − − − − − − − ∈< ∈≤ ∈≥ ∈< ∈   −   = −<    − −−  − = −<  − − − ≤ ∈≥       5 ( ) ( ) 2 2 2 2 2 2 49 49 49 49 xy xy xy xy −+= −= − −= ± − = ±− Since 4x< , 249xy= −− 5 35 5 35 35 5 yx xy yx = −+ −= − −= − ( ) ( ) ( ) ( ) 2 25 2 0 2 5 22 2 5 22 0 0 35 49 d 5 35 16 8 9 (9 ) d5 35 25 8 9 d5 yV yy y y yy y y yy π π π  −= −− − − − = − − − +− − = − −− − ∫ ∫ ∫ Using GC: V = 31.899 units3 (correct to 3 d.p.) 6(i) Consider yk= , k is a constant

2023 H2 Math Prelim Paper 1 Solutions 6 2 2 () () 0 xx a kxa x ax xk ak x a k x ak − =+ −=+ −+ −= For the range of y can take, the line yk= and the curve C should have point(s) of intersection. 2 22 22 ( )4 0 2 40 60 a k ak a ak k ak a ak k ++ ≥ + ++ ≥ + +≥ Consider ( ) 22 22 2 60 6 36 4 6 32 3 2222 k ak a aa a aaka + += −± − −±= = =−± ( ) ( )3 22 o r 3 22k ak a∴≥−+ ≤−− Hence, ( ) ( )3 22 o r 3 22y ay a≥−+ ≤−− (ii) ( 1) 1 xxy x −= + and 5 10 23y x= −+ + 2 ( 1) 5 10 123 2 ( 1)( 3) ( 5( 3) 20)( 1) 2 ( 1)( 3) (5 5 )( 1) 2 ( 1)( 3) 5( 1)( 1) ( 1)(2 ( 3) 5( 1)) 0 ( 1)(2 11 5) 0 ( 1)(2 1)( 5) 0 11 or or 52 xx xx xx x x x xx x x x x xx xx x xx x x xx x xx xx x − = −+++ − += − ++ + − +=− + −+ = − −+ − ++ + = − + += − + += = = −= − (iii) ( 1) 1 xxy x −= + Sketch the curve 5 10 23y x= −+ + Coordinates of intersection 13,22 − and ( )1, 0 and (-5, -15/2)

2023 H2 Math Prelim Paper 1 Solutions 7 Note: the minimum and maximum y values can be found from (i): (0.414, 22 3− ) and ( 2.41− , 223−− ) (iii) Area ( ) 1 1 2 1 1 2 1 1 2 12 1 2 2 2 5 10 ( 1) d23 1 5 10 22d23 1 1 2 10 d2 13 1 2ln 1 10ln 322 1 11 1 1 522ln 2 10ln 4 2ln 10ln22 4 2 2 2 9 8 xxAx xx xxxx xx xx xx xx − − − −  −= −+ − ++ = −+ − −− ++ = −−− + ++ =− − − ++ +  − = −− − + − − − +    = −− ∫ ∫ ∫ 2ln2 20ln2 2ln2 10ln5 10ln2 926ln2 10ln5 8 926, 10, 8ab c + −− + =−− = = −= − 3x=− 1x=− 2yx= − ( 1) 1 xxy x −= + 5 2y=−

2023 H2 Math Prelim Paper 1 Solutions 8 7(i) Since Q lies on the line passing through OB, OQ is parallel to OB. Hence Q has position vector in the form λb where λ is a real constant. [OR] Equation of line OB: ,λλ λ ∈= + ⇒= r b rb 0  Since Q lies on the line, it has position vector in the form λb where λ is a real constant. (ii) By Ratio Theorem, ( )1 34OP → = + ab OQ λ → = b AQ λ → = −ba 22 1 (3 ) ( ) 04 33 0 3 cos 3 cos 0 3 cos 3 cos 0 (3cos 1) 3 cos 3 cos 3cos 1 AQ OP λ λλ λθλ θ λθ λ θ θλ θ θλ θ +− ⊥ ⇒ +⋅ −= ⋅− ⋅+ ⋅−⋅= −= −= += + += + +− ab ba ab aa bb ab ab a b ab (iii) Analytical method 8 3 cos 1 3 3cos 1 3 3cos 1 θλ θθ += = + ++ A P B 1 3

2023 H2 Math Prelim Paper 1 Solutions 9 0 2 0 cos 1 0 3cos 3 1 3cos 1 4 11 14 3cos 1 8 28 3 3 3cos 1 3 8 1 313 3 3cos 1 13 πθ θ θ θ θ θ θ λ << << << < +< << + << + <+ < + << From GC. Graphical Method 13 λ∴< < Q lies on OB produced. Hence, the point Q does not lie between O and B. 8 (i) 22 4 i1 i1 1 1 ii i 2i 2 2 111 22 2 arg 4 1 2 i w w w we π −−= = = ++  = +=   π= =

2023 H2 Math Prelim Paper 1 Solutions 10 14 14 14 4 7 2 7 2 7 1 2 1 2 1 2 i 128 i i i we e e π π π− =   =   =   =− (ii) 2 22 2 24 24 24 4 24 cos isin 1 cos isin i 1 2 sin 2 2cos 24 sin 2 i ii ii i ii i i i w e ee ee e ee e i e e θ π θ θθ θ θπ θπ θπ π ππ θθ θθ θ θπ θ      −      + − −−   −   − +−= ++ −= +  −=  + =  − = 4 4 4 cos cos sin sin24 24 sin 2 1 cos sin222 2 cot 12 2 i i i e e ke π π π θπ θπ θ θθ θ            +       =   +    =  + =