2024 DHS Y5 H2 Math Promo Practice Paper 1 Suggested Solutions

1 2024 Year 5 H2 Math Practice Paper 1 Suggested Solution 1 chocolate and walnut cakes respectively. Using GC, 2 2 44 xy=+ C : Replace y by –y 2 44 xy− = + B : Replace x by 2x ( ) 2 2 2 44 4 xy yx − = + − = + A : Replace y by y – 4 ( ) 2 2 2 44 44 yx yx yx − − = + − + = + =− Equation of the original curve: 2yx=− 3(a) For 1 f ( )y x= to be increasing, f ( )yx= will be decreasing. Range of values: 2x . (b) Commented [OMFJ1]: SY - vertical asymptote, origin Commented [KSY2R1]: Is this ok?

2 4(a) Since f(1) = f(4) = 5, f is not one-one. Hence, f does not have an inverse. (b) Let 2 4f( ) , 0 2xy x x x += =   2 2 2 2 2 2 2 2 2 2 2 4 40 4024 16 024 16 24 16 24 16 24 16 2 xy x x yx yyx yyx yyx yyx yyx yyx =+ − + =  − − + = − − − = − −= −− = −= −= Since 02 x , 2 16 2 yyx −−= Hence, 2 1 16f ( ) , 4. 2 xxxx− −−= 1f [4, )D − =  Alternative (to check plus or minus): When 211, f ( ) 4 51xx += == When 5,y=

3 22 (rej +ve)5 5 16 5 5 16 42 , 12xx + − − −= = = = Hence 2 1 16f ( ) , 4. 2 xxxx− −−= 5 ( ) 2 2 2 2 2 ( 2 ) ( 1) 1 ( 1) bx bx b x x bx b x b − =− = − − = − − Maximum value at b− when x = 1. Hence turning (minimum) point is at 11, b   − 2 1 2 y bx bx = − 2 20 ( 2) 0 0 or 2 bx bx bx x xx −= −= == Line of symmetry: x = 1. 6(i) ,35 nsec tayxa a == 22 22 tan ,53 1 sec ( ) 1 5() += = +yx xaaa (ii) x = 0 x = 2 x y 11, b   −

4 Qn Suggested Solution 7(a) (b) 2 2 2() 2 2 b x axa xa b xa b xa b =−− −= − = = Hence, for 2 .b x axa −− 22 , a x a x abb−   +  (5a,0) x O y ( )5 ,0a x = a x y ab a Commented [CMYE3]: With graph, to accept answer with symmetry deduction, to award full credit

5 Qn Suggested Solution 8(i) ( ) ( ) 24 2433 9 27cos3 1 ... 1 ...2! 4! 2 8 xxx x x= − + + = − + + 24 24 24 2 24 24 2 4 4 24 9 27ln(1 cos3 ) ln 1 1 ... 28 9 27ln 2 1 ...4 16 9 27ln 2 ln 1 ... 4 16 9 27 ...9 27 4 16ln 2 ... ...4 16 2 9 27 81ln 2 ...4 16 32 9 27ln 2 4 32 x x x xx xx xx xx x x x xx  + = + − + +   = − + + = + − + + − + + = + − + + − + = − + − + = − − ...+ (ii) 0.5 0.5 24 00 0.5 35 0 0.52 46 0 9 27ln(1 cos3 )d ln 2 ... d 4 32 9 27(ln 2) ... d4 32 99(ln 2) +...2 16 64 0.04929 (5 d.p.) x x x x x x x x x x x x xx + = − − +  = − − + = − −     (iii) Using GC, 0.5 0 ln(1 cos3 )d 0.04900 (5 d.p.)x x x += (iv) From the diagram, it can be seen that the graphs of ( )ln 1 cos3y x x=+ and 2 49 27ln 2 4 32y x x x= − − are close to each other mostly from x = 0 to x = 0.5. Hence, the approximated value of 0.5 24 0 9 27ln 2 ... d4 32x x x x − − + from (ii) is approximately equal to the actual value of 0.5 0 ln(1 cos3 )d 0.04900 (5 d.p.)x x x += in (iii).

6 Suggested Solution 9a (i) ( ) ( ) ( ) ( ) 1 00 01 0 1 1 1 3 4 3 4 3 4 1 4 3 4 3 13 4 4 4 1 3 3 3 ...4 4 4 4 11 41 11 4 1 11 r rnn r rr nn r n n n x x x x x x x x x x x + == = + + + + + + − + + +=   + + +     = + + +           −= − −=  =− −   (ii) Common ratio, r of G.P. = 3 4 x+ When x = −5, 5 3 1 42r −+= =− . Since 1 12r = , the G.P. converges. Hence, the series ( ) 1 0 3 4 rn r r x + = +  converges. ( ) ( ) ( ) ( ) 1 1 0 11 2 53 4 53 1lim lim 14 1 5 1lim 16 1 6 rn n rnn r n n + +→ →= + → −+−+ =− −− = − −  =  (b)(i) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) 22 2 66 2 5 2 22 1 1 6 3 2 3 2 32 253 2 1 4 1 6 1166 2 6 12 6 2 2 2 1 4 1 165 2 2 5 3 2 1 4 1 2 2 5 3 165 kk rr kk r r r r r r r r r r k kk k k k k k k k k k k k k == = = = − = − = − − = + + − −+−+ = + + − − − + = + + − − + −    

7 Qn Suggested Solution 10(a) ( ) ( ) ( ) 2 22 2221 1 3 2 1 6 11tan 36 1 13 d19 13 dd1 9 1 9 1 18 dd 19 3 ln 9 9 1 x xx xxxxx xxx xx x x c− − + = ++ = − − ++ = − + +     (b) d2sin 2cos d xx  =  = 22 22 2 2 2 12 ( 1) (2sin 1)d (2cos ) d 4 4 4sin (2sin 1) (2cos ) d2cos 4sin 4sin 1 d 1 cos 24 4sin 1 d2 3 2cos 2 4sin d 3 sin 2 4cos 43sin 2 422 x x x c x x x xc                 − −− = −− − = − + −= − + = − − = − + + −= − + − +       (c) ( ) ( ) 12 22 4 d4 sin (2 )d 12 4 14 xxx x x x − = − = − ( ) ( ) ( ) 1 2 1 2 3 1 2 2 1 2 4 3 2 1 2 4 4 2 1 2 1 2 2 1 2 4 42 sin (2 ) d sin (2 ) d 14 1 16sin (2 ) d 4 14 141sin (2 ) 4 1sin (2 ) 1 4 2 xx x x x x x x xx x x x x x x c x x x c −− − − − =− − −=+ − − = + + = + − +  

8 Qn Suggested Solution 11 (i) The amount of money in the investment account after 1 month is 111% of the investment at the beginning of the month, i.e. GP with a = 1500 and r = 1.11 Hence the projected profit at the end of 1 full year = 1500(1.11)12 – 1500 = $3747.68 11 (ii) End of rth month Amt in account 1 200(1.11) 2 [200 + 200(1.11)](1.11) = 200(1.11) + 200(1.11)2 3 [200 + 200(1.11) + 200(1.11)2](1.11) = 200(1.11) + 200(1.11)2 + 200(1.11)3  the projected amount of money = 200(1.11) + 100(1.11)2 + … + 100(1.11)n = ( )200(1.11) 1.11 1 1.11 1 n− − = 22200 111 111 100 n − 11 (iii ) Projected amount of money in Carl’s investment account at the end of one year = 12 22200 111 1 $5042.33 (2 d.p.)11 100  −= 11 (iv ) Let n be the number of months after Carl started his investment.  22200 111 1 1500(1.11)11 100 n n −  Let Tn = 22200 111 1 1500(1.11)11 100 n n −−  From GC, T13 = –5.937 < 0 T14 = 215.41 > 0 T15 = 461.11 > 0 On 31 Jan 2023, Betty has $5.94 more in her account than Carl. So, Carl would have more money in his account on the following day, ie. 1 Feb 2023. On 31 Jan 2023, Carl’s projected amount in his account = 13 22200 111 111 100  −  = $5818.98 Betty’s projected amount in her account = 1500(1.1113) = $5824.92 Carl still has less money in his account than what Betty on 28 Feb 2023 On 1 Feb 2023, Carl’s projected amount will increase to $6018.98 > $5824.92. Hence, Carl will have more money in his account than Betty has in hers on 1 Feb 2023. (v) The amount of money lost due to scam cases each year forms a GP with first term $633.3 million and common ratio 0.95.

9 The estimated total amount of money lost due to scam cases = 633.3 1 0.95− million = $12666 million Qn Suggested Solutions 12(a) (i) (A) 1 2 , nnu u k+ =− 1 4, ua== 1k = : The population is always on the uptrend. Population will grow to infinity (B) 1 4, 5uk== 2 3 4 The population 2(4) 5 3 2(3 will bec ) 5 1 2( ome by 1) 5 3 0 4th year u u u = − = = − = = − =−  extinct (ii) 2 2(4) 8u k k= − = − ( )32 2 2 8 16 3 52 12 u u k k k k k = − = − − = − =  =− (b) nnv uk=− 11 2 2 (a constant) nn nn n n uk uk uk uk v v k ++ − −= − −== − Thus nv is a geometric progression 1 1 1 11 1 1 )2(2) ( ( ( ( 2) 2) )2 n n n n n nn n n v v u a k u a k k ua k v k k k − − − − + = = = − −+ − − = + = += (c) 2n → as n→ For population to stay stabilized as ,n→ 0,ak−= i.e. ak= (d) Rewriting ( ) ( ) ( ) 1 4 9 79 5 80 4 5 16 2 ( ) 2 2 ... 2 16 2 (2 1621 16 2 1 13 ) 61 1 n nu a k k S a k k a k k a k k −= − + = − + + + + = − + − = − − + − Commented [OMFJ4]: amended parts

10 13(a) ( ) ( ) ( ) 1211 2 2 2 2 00 0 122 12 0 0 112 2 2 0 0 122 1122 0 0 0 112 2 2 00 2e 2 e d 2e 2 e d 2 eee 2 2 2 d 11 e 2 e 4 e d eee 2 e 4 1 d 11 e 2 e 4 e xx xx xx xx x xx xx x x x x x x x x x x x x x xx −− −− −− −− − −− − = −  = − − −− = + −  = + − −   −−    = + + −        12 0 2 e4 1 11e 4e x− − =− (b)(i) 22 22 ( 6) 36 36 ( 6) xy xy + − = = − − When ( ) 2 2