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 − − − − − −