DHS_8865_2022_Prelim_Solutions_vetted

1 DHS 2022 Year 6 H1 Mathematics Preliminary Paper Qn Suggested Solution 1 2 1 log 8 4 (1)3 x y+= 2 3 (2)x y y+ = − ( ) 2 3 2 2 2 22 22 From (1), 1 log 8 43 1 log 2 43 1 (3 ) log 2 43 4 (3) Substituting (3) into (2), ( 4) ( 4) 3 41 4 2 1 3, 7 x x y y xy xy x x x x x x x x x x xy += += += += + + = + − + + = + + + = + + == Qn Suggested Solution 2 ( ) 2 720 4kx k x k+ − + +  0 and discriminant 0k ( ) ( )( )   2 2 72 4 0 4 3 11 4 0 0 and 1 3 4 0 14 or 3 4 :4 k k k kk k k k kk k kk − − +  − − +   − +  −   −  −

Qn Suggested Solution 3(a) 2 π3 2 π2 2 π2 d (5 3e )d 3(5 3e ) (10 ) 30 (5 3e ) xx xx xx − =− =− (b) ( ) 2 11 22 13 22 3 dx 96= dx 9= 6 dx 9 6 dx 2=18 6 3 218 6 3 x x xx x x xx xx x x x C x x x x C − −+ +− +− = + − + − + = + − +     (c) 2 d lnd 32 x x x     − ( ) ( ) 2 2 2 2 d1 ln ln 3 2d2 1 1 4 2 3 2 12 32 3 32 xxx x xx x xx xx = − − −=−  − =+ − = − ( ) ( ) 11 11 22 22 1 2 1 2 1 2 63 d 2 d 3 2 3 2 2 ln 32 1 1 22 ln ln 15 2 12ln 10 2ln10 ln10 1, 10 xx x x x x x x ab − = −−  =   −      =−     =− =− =  = = 

Qn Suggested Solution 4(a) HA: 72 3 2 VA: 3 y y x x = =+ − = (b) Axial intercepts: 1(0, ) 3− & 1( ,0)2− (c) 21 3 2 3 6 7 3 2 2 x x x x  − −  − −  32 72 3x x−+ From GC, 0.240x− or 3 6.24x (d) Using GC, area of bounded region = 6.2404 4 7 3 2(2+ ) d 3 xxx− − = 5.06 units2 OR 6.2404 4 18(2+ ) d ( 4.1602)(6.2404 4)2 7 3 3x x−+− − (e) 2 22 2 22 7( 3) 3 7( 3) 2 2 3 xm x xm x − + = − − + + − = − For the equation 2 22 7( 3) 3xm x − + = − to have no real roots, there is no intersection between 72 3y x=+ − and ( ) 222( 3) 2x y m− + − = which is a circle centred at (3,2) radius m . Using GC, possible values of m are 1,2,3.m= x = 3 y = 2 x y

Qn Suggested Solution 5(a) 22 2rxy p q x s x += + + + ( ) ( ) ( ) 2 2 2 (0,9) : 2 9 (9,1787) : 81 512 2 3 1787 (4, 201) : 16 16 2 2 201 (1, 29) : 2 2 29 r r r r p p q s p q s p q s += + + + = + + + = + + + = 2From GC, 7 9 3, 2 2 1, 9 rp q q r s= =  = =  = = The key is (7,3,1,9) (b) From G.C, maximum S occur at t = 4 Maximum S = 95.4 million (3 s.f.) (c) 20.5 40.03e 6 ttS −+=+ ( ) 20.5 4d 0.03 4d ttS tet −+= − + ( ) ( ) ( ) ( ) ( ) 2 2 15 0.5 5 4 5 2 15 0.5 5 4 5 2 when 5, d 0.03 5 4 e 0.03ed 0.03 6 0.03e 6 t S t Se −+ −+ = = − + =− = + = + ( ) 15 15 22 15 15 22 Equation of tangent: 0.03e 6 0.03e 5 0.03e 0.18e 6 St St − + =− −  =− + + (d) (e) For t ≥ 7, as t increases from 7 as t→ , d d S t increases from 42.980− (gradient less negative) until it approaches 0. Hence the manager expects the sales to decrease at a slower/decreasing rate until it stabilises at 6 million.