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