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 glas