2024 DHS Y5 H2 Math Promo Practice Paper 2 Suggested Solutions

1 2024 Year 5 Math Practice Paper 2 Suggested Solutions Qn Suggested Solution 1 Let the price of a desktop monitor, keyboard and mouse be $x, $y and $z respectively. 317.90 0.9 0.85 0.9 282.66 0.95 0.75 0.8 283.72 x y z x y z x y z + + = + + = + + = x = 219 , y = 69 , z = 29.9 Employees of Company B will pay $[0.9 5(219)+0.75(69)+0.78(29.9)] = $ 283.12 > $282.66. No, it will not be more attractive for employees from Company B to purchase all the three items from their own company since they will still have to pay more than the sale by Company A . Qn Suggested Solution 2 For 0x , solve 2 xax =− − to determine the x-coordinate of the point of intersection. 2 2 2 22 20 2024 822 4 2 4 x ax aax a a a ax + + =  + + − = −=−  − =−  Since 2 9a  , then 2 80a − . From the graph, the soln is 2 8 024 aa x−− +   or 2 8 24 aax −− − . y x O

2 Qn Suggested Solution 3(a) (b) Qn Suggested Solutions 4(a) ( )2 2 2 2 2 2 2 2 2 2 d 3 6 2d (1 ) (1 ) d 9d d 6 1 2 d (1 ) 9 3 (1 ) yt tt tt x tt yt x t t t t −=− = ++ = −−=  = ++ There are no real solutions for t when d 0.d y x = Curve C has no stationary points. Alternatively, 22 2 0 2 0 (Inconsistent) 3 (1 )tt − = − = + . Thus C has no stationary points. y O x y = 1 x = 3 (2, 5)  y O x y = 0 x = –2 –1

3 (b) Qn Suggested Solution 5(a) 2 2 22 2 22 2 e 2 e Differentiate with respect to dee d Differentiate with respect to d d de e edd d d d e dde d d d (shown)ddd yx yx y y x x y x y x x y y y xx x yy xx y y y xxx =+ =    +=     += += When 2 2 d 1 d 20, ln 3, , d 3 9 d yyxy x x = = = = 2 2 12ln 3 ... 3 2! 9 ln 3 ...39 xyx xx    = + + +       = + + + (b) ( ) 2 2 2 ln 2 e ln 2 1 ... 2! 1ln 3 1 ...3 2! 1ln 3 ln 1 ... 3 2! xy xx xx xx =+ = + + + +  = + + +    = + + + +   y x y = 0 (0, 3) O

4 ( )   2 2 222 22 1ln 2 e ln 3 ln 1 let ... 3 2! 1ln 3 ... ( apply std series of ln(1 ) )2 1 1 1ln 3 ... ... ...3 2! 2 3 2! ln 3 . 3 6 18 x xy X X x X X X xxxx x x x  = + = + + = + +    = + − + +     = + + + − + + +              = + + − + 2 .. ln 3 ... (verified)39 xx    = + + + Qn Suggested Solutions 6(a) n Loan balance at beginning of month (after interest) Loan balance at end of month (after repayment) 1 847500 847500 M− 2 (847500 )(1.0025) 847500(1.0025) 1.0025 M M − =− 847500(1.0025) 1.0025 MM−− 3 2 2 847500(1.0025) 1.0025 1.0025MM−− 2 2 847500(1.0025) (1.0025 1.0025 1)M− + + n 1 1 847500(1.0025) (1.0025 1.0025 1) n nM − −− + + +  Loan at the end of second month 847500(1.0025) 1.0025 MM= − − Loan at the end of nth month ( ) 11 1 1 847500(1.0025) (1.0025 1.0025 1) 1 1.0025847500(1.0025) 1 1.0025 847500(1.0025) 400 1 1.0025 n nn n n n M M M −− − − = − + + +  −=−  − =+  − (b) For loan to be repaid in 25 years, ( ) 300 1 300 Monthly re 847500( payment 1.0025) 400 1 1.0025 0 4008.9185 $4008.92 M M − + − = === 4008.92 100 54.9167300 55=  Thus the bank will approve Jane’s loan. (c) Assume that the housing loan is the only debt that Jane has.

5 Qn Suggested Solution 7(a)(i) 22 2 2 22 1/ 2 41 Standard eqn of hyperbola: 1 To find equations of oblique asymptotes: as , , 4 2 ie. Equation of asymptotes: 2 yx y x xy yx xy xy −=  −= → → → = (ii) ( ) ( ) by22 by ( 1)2 2 2 2 22 1 1 1 1 yy yyy x y x y x −− = ⎯⎯⎯ → − = ⎯⎯⎯⎯ → − + = 1) Scaling parallel to y axis by factor of 2 2) Translation by 1 unit in the positive y- axis direction (b) y x O (0,0.5) (0,-0.5) C 2x=− 1 2y= (3,1)A (2,0) x ( )0, 2B − y 0y= 1 f (2 ) y x =

6 Qn Suggested Solution 8a(i) ( ) ( ) ( ) ( ) ( ) ( ) 12 1 2 1 1 1 23 2 2 3 2 2 1 2 3 12 1 2 1 ( )( 1)(2 1)6 ( 1)4 2 1 9 (2 1) 6 14 2 1 ( 1)(4 )3 n r r n n n r r r r n n n rr rr n n n n n nn n n n n + = = = = +− = + −  − = + + −  − + + += − + − + = − + + −     a(ii) Replace r with 1r+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 1 12 14 11 12 33 12 1 2 1 2 11 1 2 3 ( 1) 2 3( 1) ( 1 1) 2 3 3 2 3 2 3 3( 1 3 1) 14 2 1 ( 1)( 1 1)(4 ( 1)) (16) (3 9)3 12 2 ( 1)(5 ) 3 293 where 2, N r r rN r r NN r rr N rr rr N N rr rr rr r r r r N N N N N N N N N BC = += + += −− + == − ++ == − + − − = + + − + − = + − + = + − − + − + − − + = − + − − + − − − + − = + − − + − =     1 , 3, 293 DE= = =− 8b (i, ii) p = 5 Method 1(GC) (preferred) Method 2 (algebraic) 21 32 43 3 2 3(5) 2 13 3 2 3(13) 2 37 3 2 3(37) 2 109 vv vv vv = − = − = = − = − = = − = − = … & so on The sequence increases & diverges.

7 p = 1 Method 1(GC) (preferred) Method 2 (algebraic) 21 32 43 3 2 3(1) 2 1 3 2 3(1) 2 1 3 2 3(1) 2 1 vv vv vv = − = − = = − = − = = − = − = … & so on It is a constant sequence which converges to 1. Qn Suggested Solution 9(a) 5 d(2 3)( 1) xxx−+ 21( ) d2 3 1 xxx=− −+ ln 2 3 ln 1x x C= − − + + where C is an arbitrary constant (b) ( )   22 2 π 2 3 π 4 π 3 2 π 4 π 3 2 π 4 π 3 π 4 1 d sec 1 sec tan dsec tan d (sec 1) d tan π π π πtan tan3 3 4 4 π31 12 x xx        − −= = =− =−    = − − −       = − −     (c) ( ) ( ) 1 2 2 2 2 222 1 122 2 12 12 22 2 1 d 2 1 d21 11 2 1 1 x x m x m x xmmx mx cm m x cm − −+ −+ =− − − − − =− + =− − +  ( )d sec tan dx   = When 2, sec 2 1cos 23 x   == =  = ------------------------------- When 2, sec 2 1cos 42 x   == =  =

8 ( ) 1 22 1 22 2 1 22 2 sin d 1 sin 1 1d sin 1 xmx x mx mx m x xmm mx x m x cmm − − − − = − − − − = − − + +   Qn Suggested Solution 10(a) (i) Every horizontal line y = k , k cuts the graph of f at most once hence f is one-one and 1f − exist. Let 1y xa= + 1 1 11 xy ay xy ay ayxa yy  + =  = − − = = − Therefore 1 1f : , , 0.x a x xx − −   (a)(ii) Since fR \{0}= and gD = fgRD , thus gf exists. 1 22 22 222 dsin d d 1 1 1d 1 vu mx x u x mx m mxmmx vx − − − = −== − = y x = –a x O y = 0 y = k ( )fyx= y x O 2 y x = –a x O y = 0 ( )fyx=

9 Using the graphs of f and g f f gfD RR→→ \{ } \{0} (2, )a− → →  ( )gfR 2,= b(i) b(ii) ( ) 6 2 h d 0xx − = Qn Suggested Solution 11(a) ( ) 2 22 42 3 3 15 4 2 15 4 ( 2) 15 4 4 4 4 15 0 4 15 0 0 (does not satisfy original e qn) xx xx x x x x x x x x x + = − + = − + = − + − − = − − =  2( 3)( 3 5) 0x x x− + + = Since 2 2 3 113 5 0 24x x x + + = + +  for all x, 3, 7xy = = (3,7) is the only point of intersection between the 2 curves. (b) Area of the region R R (3,7) 2 -2

10 3 2 0 3 3 32 0 33 22 2 15 4 ( 2) d (15 4) 23 3(15)2 2 27 2(45 4) 6 (4)45 3 45 686 16345 45 551 16 45 535 45 107 unit9 x x x xx x = + − −   += − +   = + − + −  = − − −= = =  11(c) Volume of revolution ( ) 77 22 22 3 1π 2d 4 d 225 91.7 units y y y y − = + − − =  Alternative (markers’ reference only) Volume of revolution ( ) 77 22 22 72 7 42 2 2 753 2 3 1π 2d 4 d 225 1π 2 8 +16d2 225 81 1 8π 162 225 5 3 81 1 38381 256π 2 225 15 15 81 305π 2 27 1577 π units54 y y y y y y y y y yy y − − = + − − = + − −  = − − +    = − −   =−  =  