NYJC 2024 FM CT2 - modified (Solutions)

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 1 of 15 SOLUTIONS: Qn Suggested Solution 1(a) Equation of line l is ( )tanyx=  . Unit direction vector of l is cos sin     . Projection of 1 0   onto l = 1 cos cos 0 sin sin                   = coscos sin     Projection of 0 1   onto l = 0 cos cos 1 sin sin                   = cossin sin     12)) ))( ( (=A A e A e 2 2 cos sin sin cos cos sin       =  A .

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 2 of 15 (b) 2 2 2 cos cos sin cos 0 0cos sin 0sin si s ta n c n o xy xy x y               =        += =−  Basis for nullspace for T is tan 1  −   [or 1 cot  − ] Alternatively, Note that all position vectors that are perpendicular l will be mapped to the zero vector. Gradient of l = tan Equation of line perpendicular to l : ( )cotyx =− . Basis for nullspace for T is 1 cot  − . (c) 2 2 2 2 35 1c ) o 3 cos cos 35 4 35sin cos os sin 4 5sin --- (1) 35sin 3 cos 5si sin 5 - cs 4 n -- (2 4  +  =    + + += ++=             Method 1: 1(1) 5 (2) : 5 26sin cos (20 26 3)4 + + = +  3 1313sin 2 ( 3) 2 3sin 2 or 2 6 = =  =   When 2 s 2.92, cos 5 in cos 2.423 3 3 3 ,LH RHSS    = == = + for (1), thus rejected. Thus 6  = Method 2: ( ) ( ) ( ) 3 t2 a: n 35 4sin (cos 5sin ) 3 5 1 1 cos (cos 5si ) 35n 3 33 4 5        +++ = = = =+  +  + 6  =

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 3 of 15 Qn Suggested Solution 2 (a) Since 11 1 0 1 11        ==          A0 , A has eigenvalue 0 and corresponding eigenvector 1 1 1    . (b) Since nullity(T) = 1, rank(T) = 2. Since 5 2 3 2 0 2 4 2 2 −           − =− −          −      , take 2nd and 3rd columns of A as basis of range space of T. Since 21 0 2 0 21        =          and 3 1 1 2 2 0 2 2 1 0            =+                , hence 1 0 1    and 1 2 0    form a basis of the range space of T. Need to link to range space of T. Just saying they are L.I. is insufficient. (c) Since 1 0 1    and 1 2 0    are basis of the range space of T, 11 02 10 km p q r        +           =  A 5 2 3 22 4 2 22 p q r pr pqr km m k  − + +  − + =− + + +      Thus 4 2 2 rk pq−+= + , m p r=− + ( ) ( ) 1 11 11 to both sides of eq 11 02 10 App uly 11 02 10 11 2 0 1 2 1 ao 0 ti nn n n n nnn m km k p qk r p q r p q r m  − −− −−        +                 +                 − + −         =   =   =    A A A A A A  Thus ( )( ) 1 4 2 2 2 n pqr − + −−= + , ( ) 1( 1) npr −−= − +

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 4 of 15 (d) ( ) ( ) ( ) ( ) ( )( ) 1 4 4 4 4 14 4 1 − =− −= −= −= −= = −     B A A I B A I x Ax B Ax Ix Ax B x x x B x x Bx x Thus the eigenvalues of B are 0, 2 and 8 3 . 3 Source: RVHS FM Promo 2025 Qn 8 (a) f ( , ) 3 ln( 1)x y xy y x= − + 3f 1 x yy x=− + f 3ln( 1)y xx= − + 2 3f ( 1) xx y x= + f0yy = 3f f 1 1 xy yx x= = − + (b) For stationary points, f0x = and f0y = . f0x =  3 01 yy x−= +  310 1y x −= +  0y= or 2x= f0y =  ln( 1) 0xx− + =  0x= or 5.71144x=  Stationary points occur at (5.71, 0) and (0, 0) (reject as we want x > 0). At (5.71, 0) , ( ) 2 f f f 0.306 0xx yy xyD= − =−  . Hence, (5.71, 0, 0) is a saddle point. (c) At (2, 1), f ( , ) 2 3ln 3xy =− , f0x = and f 2 3ln 3y =− Equation of tangent plane: 2 3ln 3 (2 3ln 3)( 1)zy= − + − − Therefore, an approximate the value of f (2.05,0.98)

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 5 of 15 Qn Suggested Solution 4 (a) Equation of the perpendicular bisector 12 33 y x − =−+ 212 3yx− =− − 2 13yx=− − The foot of perpendicular from ( )7,1− to 2 13yx=− − will give the minimum 7zi+− Consider 2D , the square of the distance from ( )7,1− to 2 13yx=− − , ( ) ( ) 222 71D x y= + + − But 2 13yx=− − , So ( ) 2 22 272 3K D x x = = + + − −  ( ) 2 2 26 502 7 2 2 3 3 9 3 d d K x x xx   = + + + = +     Set 0d d K x = , we have 50 9 75 26 3 13x =− =− 2 75 37 13 13 13y =− − − = 2 3ln 3 (2 3ln 3)(0.98 1) 1.27 (to 3 sf) = − + − − =−

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 6 of 15 2 2 10 03 d d K x = minimum K, hence the complex number is 75 37 13 13 i−+ Alternative Eqn of perp bisector: 1 2/ 3 1 0 3 0   − − =+    r To find shortest distance from ( )7,1,0− from the line We have 11 2 / 3 2 / 3 0 0 00 37 11 0  −           − − • − =            −       +          0 4 1 0 0 1 2 / 3 2 / 3 00           −    + • − =           34 4 3610 91+ =  =−+  Thus the position vector on the line is ( ) 75131 36 37 2 / 313 13 0 0 3 1 0 −   − − =     −  +       Alternative: Equation of ⊥ bisector: Line with grad= 2 3− passing through ( )3,1− ( ) ( )21 3 1 3yx− =− +  A line passing through ( )7,1− . grad 3 2= ( ) ( )31 7 22yx− = +  Solving (1) & (2), we have ( ) ( )23 3732 xx− + = + 4 12 9 63xx− − = + 75 13x=− , thus 37 13y= Therefore the complex number is 75 37 i13 13−+

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 7 of 15 (b) ( ) ( ) 22 8 3 3 2 26AC = − + + − = Hence angle  1 13sin 426 CAB  − ==  =  Angle  1 1tan 5CAD  − ==  ( ) atant a ta n ta1 nn tn  + −− = ( ) 1 4 1 611 5 1 5==  − + Thus max ( ) 1 2a arg 8 3 tn 3iz  − −=−=  +  Qn Suggested Solution 5(i) Given that x kg be the amount of salt in the tank t minutes. In 1 min, 2 kg of salt is going into the tank. | 2 in d d x t = Since 2 gallons of mixture is flowing out per min, hence amount of salt flowing out per min is 2 100 x , Thus | 1 50out d d x xt =− Thus 12 50 d d x xt =− 50 100 ddxtx =−   50ln 100 x t C− − = +

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 8 of 15 1 50100 e t xA − −= where 1 50e C A − = When 0t = , 0x= , 100A= 1 50100 100 e t x − =− (ii) Given that x kg be the amount of salt in the tank t minutes. In 1 min, 2 kg of salt is going into the tank. | 2 in d d x t = At time t, amount of mixture in the tank is 100 2 100t t t+ − = + gallons. Since 1 gallon of the mixture flows out of the tank per minute,, amount of salt flowing out of the tank is 1 100 xt+ kg. Hence we have the DE: 12 100 d d x xtt=− + 1 2100 d d x xtt+= + IF = . Multiplying the DE by IF gives Since when t = 0, x = 0, 2100C=− . Thus the particular solution is . When amount of mixture is 150, 0100 150 5t t+= = . Amount of salt in the tank at this instant = ( )50 200 50 83.3100 50 kg+ + .

2024 NYJC JC2 Common Test Exam 9649/2 Solution Page 9 of 15 (iii) ( )200 100 0.9100 tt t t + + =+ Using G.C. 216.23t = min Qn Suggested Solution 6 (a)  is the proportion of adults of breeding age in the previous month that become ageing adults.  is the proportion of ageing adults that died in the previous month.  is the proportion o