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