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. S t S = 6

Qn Suggested Solution 6(a) Number of ways = 5 24! 3! 2! 2880C   = (b) Required probability 36 23 P(DAY grouped and start end vowel) P(start and end with vowel) no. of ways ( _ DAY _) no. of ways (start and end with vowel) 2 2! 3! 1 360 CC = = =    = Qn Suggested Solution 7(a)(i) P( ) P(fall, rise, rise) P(fall, fall, rise) (0.4 0.15 0.6) (0.4 0.85 0.15) 0.087 AB =+ =   +   = (ii) P( ) P( ' ) 0.087 P(rise, rise, rise) P(rise, fall, ri se) 0.087 (0.6 0.6 0.6) ( P( 0.6 0.4 0.15) 0. ) 339 A B A B B  +  = + + = +  = =  +   (iii) P( | ) P( P( ) 0.087 0. ) 0. 175 4 2 BA B A A = =  = (b) Since P( | ) 0.2175 0.339 P( ),BA B== A and B are not independent. (c) Let W be the number of Tuesdays in which the unit price of X rises, out of 12 Tuesdays. ( ) ~ B(12,0.6) P( 5) 0.101 3 s.f. W W == Qn Suggested Solution 8(a) • Set B will have a smaller r.

The data points for Set B lie relatively closer to a straight line with negative gradient whereas Set A’s r value will be closer to 0 since the data points are more scattered with weak linear correlation between x and y. (b) (i) (ii) r = 0.83161 = 0.832 ( 3s.f.) Since r value is close to 1, it indicates a strong positive linear correlation between x and y which is seen in the scatter diagram where the data points lie close to a straight line with positive gradient. x y

(iii) equation of regression line of y on x : 3.4787 0.87005 3.48 0.870 yx yx =+ =+ equation of regression line of x on y: 12.897 0.79487 12.897 0.79487 1.2581 16.225 1.26 16.2 xy xy yx yx =+ −= =− =− ( , )xy = (50.78, 47.66) (iv) Using equation of regression line of y on x : For 50, 3.4787 0.87005(50) 46.9812 x y = = + = The mean household expenditure is estimated to be $46 981. The estimate is reliable since it is an interpolation where x = 50 ( 45.5 55.5x ) and r is close to 1. (v) It is not valid because a strong positive linear correlation between income and expenditure does not imply causation. Qn Suggested Solution 9(a) ~ B(30, ) P( 3) 0.0188 Using GC graph, 0.0199894 or 0.2644435 (rej since 0 0.2) 0.01999 (5 d.p.) Xp X p p p p == = =   =

(b) ~ B(30,0.1) P( 2) 1 P( 1) 1 0.183695 0.81630 0.816 (3 s.f.) X XX = −  =− = = (c) Let Y be the number of boxes with at least 2 defective phones, out of 10 boxes. ~ B(10,0.81630) P( 6) P( 5) 0.023192 0.0232 (3 s.f.) Y XX =  = = (d) E(X) = 30(0.1) = 3 Var(X) = 30(0.1)(0.9) = 2.7 Since n is large, by Central limit theorem X ~ N(3, 2.7 n ) approximately P( 3.5) 0.998 Using GC table, 88, P( 3.5) 0.9978 0.998 89, P( 3.5) 0.998 0.998 90, P( 3.5) 0.9981 0.998 X nX nX nX  =  =  =  = = =  =  Hence, greatest n is 89. (e) The boxes are picked without replacement, hence the trials are not independent.

Qn Suggested Solution 10 Let X be the mass of a randomly chosen mooncake. H0 :  = 150 H1 :  < 150 where  is the population mean mass of mooncakes. Since sample size of 9 is small, assume X follows a normal distribution. Under H0, 26.73~ N 150, 9X   Using GC, the test statistics 148x= gives 0.89153calcz =− and p-value = 0.186322 0.186 (3 sf) Since the p-value = 0.186 > 0.1, we do not reject H0 and conclude that there is insufficient evidence at the 10% significance level that the mean mass of the mooncake is less than 150 g, i.e. insufficient evidence to reject owner’s claim. (b) (i) Assign each teacher in the country a number in consecutive order. Among these numbers assigned, use a calculator to generate n different numbers randomly and choose the corresponding numbered teacher. (ii) Let Y be the working hours of a randomly chosen teacher in the school. H0 :  = 60 H1 :  ≠ 60 Under 0H, 26.5N 60, nY   In order to reject 0H, p-value = 2 )P 05( 62 0.Y  Using G.C n 2P( 62 ) 0.05Y  − 40 0.00165 41 0.0012− ≤ 0 42 0.0039− ≤ 0 Least n is 41. (iii) There is a probability of 0.05 that we reject the null hypothesis that the mean working hours of teachers in the school is 60 hours when it is actually true.

11(a) (b) 2~ N(580,22 )X Expected number 300 P( 600) 300 0.18165 54.495 54.5 (3 s.f.) X=   = = = (c) No. By combining the masses, it would give a distribution with 2 peaks instead of a single peak. (d) Let K and L be the selling price of a randomly chosen rock melon and watermelon respectively. 22 12 22 12 12 0.003 , 0.0028 ~ N(0.003 580, 0.003 22 ) ~ N(1.74, 0.004356) ~ N(3.48, 0.008712) ~ N(0.0028 870, 0.0028 30 ) ~ N(2.436, 0.007056) ~ N(5.916,0.015768) P( 5.9) 0.44930 0.449 (3s.f.) K X L Y K K KK L L K K L K K L ==  +  ++ + +  = = (e) 2 2 P( 1.70) P( 2.50) 0.27224 0.77694 0.057583 0.0576 (3s.f.) KL   = = = (f) Although the probabilities for both events are for at most $5.90 payment, part (e) is a subset of (d) as part (e) is a special cas