EJC 9758 2025 Prelim P2 Solutions

EJC 9758 2025 Prelim P2 Solutions Solution 1(a) 22 22 22 3 1 3 1 3 1 1 1 3 3 3 3 2 Differentiating w.r.t. , dd3 3 3 3 0 dd d 3 3 d 3 3 When 0, , d d Cx yyx y x y xx y y x y x x y x y x y xk xk y xkx y k x k k x k + − − = −−== −− = = = == = − = −  (b) ( )( ) 2 6 3 3 63 33 2 33 1 3 dFor tangent to be parallel to -axis, mu st be undefined.d Hence, 0 Substituting into and 3, 30 2 3 0 Factorising, 3 1 0 3 or 1 3 or 1 yy x yx xy Ck y y y k yy yy yy yy −= = = + − − = − − = − + = = =− = =−

Solution 2(a) Let L be the limit of the sequence ( )1 2 1n n n uuu+ = − . ( )2 1L L L= − 22 0LL− = ( ) 021LL − = 0L= or 1 2L= (b) When 1 0u = , 2 0u = , 3 0u = , sequence is convergent. When 1 0.01u = , 2 0.0198u = (exact), 3 0.0388u = (3 s.f.), sequence is convergent. When 1 0.01u =− , 2 0.0202u =− (exact), 3 0.0412u =− (3 s.f.), sequence is not convergent.

Solution 3(a) (b) x-coordinate of points of intersection are 4.6391, 7.7172 to 5 sf Equation of ellipse: ( ) 2 2 625 1 16 xy  −− =  Required volume ( ) 7.7172 2 4.6391 2 26π 25 1 d 16 2 8 28xx xx x −= − −   −  −  +    358.7 units= O y x (–2, 12) (6, –4) x = 2 y = 6 – x (6, 0) (2, 0) (10, 0) (6, 5) (6, –5) C2 C1

Solution 4(a) Normal vector of p 0 5 5 5 0 10 5 2 2 25 5 tt t                =  = =              −−        plane p: 20 5 t  =− r Since l1 and p are parallel, 2 22 2 3 2 0 15 6 2 5 0 1 1 or 1 tt t tt t tt        −=       −    + − − = = = =− Since point A is not on p, 1 2 2 0 15 4 5 0 1 t t t               −    + −   1t =− (Shown) (b) Let  be the acute angle between l1 and l2 1 1 11 11cos 2 2 66 11 4cos 6 0.841 or 48.2  − − −       =           = = (c) an is the perpendicular distance from A to p 11 122 3015 2 30 2 30 −       =        −    −= = an

(d) Let N be the foot of perpendicular from A to p 1 1 1 2 2 2 2 1 5 1 5 s ON s s s −+           = − = −          −+      Since N is on p, 11 2 2 2 0 1 5 5 1 4 4 5 25 0 1 15 s s s s s s s +−       −=      +−    − − + − − − = =− 14 1 3215 10 ON  =  Let B be the point of reflection of A in p 2 14 1 2 32 215 10 1 13 1 3415 5 OB ON OA=−        =−            =  line of reflection of l2 in p: 13 34 , 5 kk  = r

Solution 5(a) Number of ways to choose the 6 glasses = total number of ways – number of ways with 1 missing colour = 84 61    −       = 24 Required number of ways = ( )24 6 1 ! − = 2880 (b) Number of ways = 4 4 16= (c) Let A be the event that the blue glasses are adjacent. Let B be the event that the yellow glasses are adjacent. Then ( ) 5n ' 4! 2! 480 2A ==   [Explanation: arrange the four non-blue glasses (4!), then slot in the blue glasses ( 5 2 2!C  )] Also ( ) 4 2n ' 3! 2! 2! 144A B C = =   [Explanation: group the two yellow glasses (2!), arrange this group with the red and green glass (3!), then slot in the blue glasses ( 4 2 2!C  )] So the required number of ways is ( ) ( )n ' n ' 336A A B−  = .

Solution 6(a) Stage 1 Stage 2 Stage 3 (b) ( )( ) 2 3 0.5 0.5 0.1 0.8 0.928 p p p p p    (c) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) P wins exactly 2 stages out of 3 | proceeds to Stage 3 P wins exactly 2 stages out of 3 proceeds to Stage 3 P proceeds to Stage 3 0.9 0.45 1 0.225 0.9 0.55 0.45 0.1 0.9 0.45 1 0.1 0.1 0.577125 0.5830.99 = − + += − == 1 – p 1 – p 1 – 0.5p 1 – 0.25p 1 – 0.5p p 0.5p 0.25p 0.5p 0.5p p 1 – 0.5p

Solution 7(a) 0.00208r = (b) There is a non-linear (curved) relationship between x and y. (c) 0.997r=− . This means there is a (very) strong negative linear correlation between y and ( ) 2 10x− . (d) Equation is ( ) 2 0.987 10 9.92yx=− − + When 17x= , 38.4y=− This estimate is not valid because 17x= lies outside the valid range of 7 13x

Solution 8(a) ( ) ( )E 3 E 3 4 3 1XX− = − = − = ( ) ( )( ) ( ) 22 Var 3 E 3 E 3 40 1 39X X X− = − − −  = − = ( ) ( )Var Var 3 39XX= − = Alternatively, ( )( ) ( ) 2 2E 3 E 6 9 40X X X− = − + = ( ) ( ) 2E 6E 9 40XX − + = ( ) 2 5E6 40 54 9X = =+ − ( ) ( ) ( ) 22Var 39 E E 55 16XX X−  = −==  (b) y 0 1 2 3 f ( )y 1 1 2 1 3 1 4 ( )P Yy= 3 2 3   2 3 1 21 33C            2 3 2 21 33C         3 1 3   ( ) ( )f Py Yy= 8 27 2 9 2 27 1 108 ( ) ( ) ( )Ef Pf y Yy Yy =  =    8 2 2 1 27 9 27 108= + + + 65 108= (or 0.602 to 3 s.f.) (c) Since sum of probabilities is 1, 0.15 1p q r+ + + = 0.85pqr+ + = (1) From ( )E 0.2W =− , 3 0.15 2 0.2p q r− − + + =− 3 2 0.35p q r− − + =− (2) From ( ) 3E 3.2W =− , 27 0.15 8 3.2p q r− − + + =− 27 8 3.35p q r− − + =− (3) From GC, solving simultaneous linear equations, 0.2, 0.35, 0.3p q r= = =

Solution 9(a) The probability that any student is selected is the same, and the selection of any student is independent of the selection of other students. (b) Let X be the queue time at the canteen during peak hours of a randomly chosen EJC student and let  be the population mean. 01: 20, H : 20H = Assume that the queue times are normally distributed. Under 0H , 2 15 3.8~ N 20,X   From GC, Given that H0 is not rejected at 5% level of significance, 18.1 21.9x (c) ( ) 320 50 20 19.tt +=−= ( ) ( )( ) 2 22 2 20 201 ( 35)48949 50 929 or 9.4796 9.48 (3 s.f.)8 1 9 1 t st nn  − −−−   −=−   = =   (d) Let  be the population mean (of T). 01: 20, H : 20H = Under 0H , since sample size n=50 is large, by Central Limit Theorem, 9.4796~ N 20, 50T   approximately. At the 10% level of significance, the critical region is 19.442t  . From data, 19.3 19.442t = Hence, we reject H 0 and conclude at 10% level of significance that there is sufficient evidence that EuOrder has reduced the queue time at the canteen. 21.9 0.025 20 18.1 0.025