TJC 9758 2023 Prelim P2 Solution

1 (a) 44 11 22nn  + −−     2 34 43 2 1 2 34 43 2 1 44 4 1 1 11 12 3 2 2 22 44 4 1 1 11 12 3 2 2 22 nn n n nn n n       = + + + +−                  −+ − +            3 3144 112213 22nn   = +      34nn= + (b) 44 3 11 11 (4 )22 NN nn n n nn = =   + −− = +     ∑∑ 44 44 3 1 44 44 31 22 53 = 4 ( 1)22 2 75 22 11 22 N n NnN NN =  −+    −+ ++    −+     + −−     ∑  44 3 1 11 = 4 ( 1)22 2 N n NN nN =   +− + +     ∑ ( ) 44 3 1 1141 22 2 N n NnN N =   = + − +−    ∑ 34 3 2 2 1 311 11142 2 2 16 2 2 16 N n nN N N N N N = =+ + + +− − −∑ 34 32 1 42 N n nN NN = = ++∑ 3 4 32 1 1(2 )4 N n n N NN = = ++∑ 4 321(2 )4 NS N NN∴= + +

(c) 4 2 1 2114 NNS NN − = ++ As N →∞ , 2 21,0NN → Thus 4 1 4 NNS− → . Limit = 1 4

2 (a) ( ) 222 28 8 6xx x−−= −− Therefore the largest value of a = 6 (b) (c) For 04 x<< Let 2yx= + ( ) 2 2xy= − For 45 x≤≤ 28 ( 6)yx= −− 2( 6) 8xy−= − 68xy= ±− Since 5x≤ 68xy= −− , 47 y≤≤ 2 1 ( 2) , 24f: 4768, x xx xx −  − <<∴  ≤≤−−  (d) Since fgR (2, 7] D += ⊂=  , ∴gf exists. (0,5] (2,7] [1,5]→→ gfR [1,5]= x y (0,2) (4,4) (5, 7) f g x y (0,4) (3,1) (7,5)

3 (a) ( ) ( ) ( ) ( )( )( ) ( )( ) ( ) 2 1 22 22 22 2 2cos 2 1 sin 2 221 1 2 2! 122 12 1( 1 ) 2 2 2! 2 12 12 4 2 1 24 2 2 44 xy x x x xx x x xx xx x xx − = +  ≈− + −−≈ − +− +   = − −+ ≈ −+ − = −+ (bi) ( ) 22 2 22 2 ln 1 sin2 e 1 sin2 Differentiatewrt de 2cos2 (1)d Differentiatewrt dde e 4sin2dd dde 4sin2 (2)dd 2, 4 y y yy y yx x x y xx x yy xxx yy xxx hk = + = + = −−− += −  + =− −−−   = =−

Method 2 (not recommended, most students’ working) ( ) ( )( ) ( )( ) ( ) 2 22 2 22 2 2 2 ln 1 sin 2 d 2cos 2 d 1 sin 2 Differentiate wrt 1 sin 2 4sin 2 2cos 2 2cos 2d d 1 sin 2 4sin 2 2cos 2 1 sin 2 1 sin 2 d 4sin 2 d d 2cos 2 (since and e 1 sin 2 )d e d d 1 sin 2 dee d y y y yx yx xx x x x xxy x x xx xx y xy y x xx x xx y x = + = + +−−= + − = − ++  − = −= = + + + 2 22 2 d 4sin 2d dde 4sin 2 (2)dd 2, 4 y y y xx yy xxx hk  =−  + =− −−−   = =− (ii) 232 2 32 2 Differentiate wrt d dd d d de 2 e 8 cos 2 (3)d dd d d d yy x y yy y y y xx xx x x x    + + + =− −−−        When x = 0, 0y= From (1) : 0 dde 2cos 0 2dd yy xx= ⇒= From (2) : ( ) 22 20 22 dde 20 4dd yy xx  + = ⇒= − From (3) : ( )( ) ( ) ( )( ) 33 200 33 dde 22 4 e 2 4 2 8 8dd yy xx  + − + −+ = −⇒ = ( ) ( )2 3 2348 42 ... 2 2 ...2 3! 3y x x x xx x−= + + += − + +

Method 2 (not recommend) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 2333 3 323 23 sin 2 sin 2ln 1 sin 2 sin 2 ...23 222 ... 2 ...3! 3!22 ...3! 2 3 2222 ...3! 2 3 42 2 ... 3 xxxx xxxx xx xxxx xx x += − + +   −+ −+   = −− + + = −−++ = −++ (c) ( ) ( ) 23 23 2 4ln 1 sin 2 2 2 ... 3 differentiating both sides, d d4ln 1 sin 2 2 2 ...d d3 2cos 2 24 41 sin 2 x xx x x xx xxx x xxx + ≈− + + + = −++  = −++

4 (ai) ×ab denotes the perpendicular distance of the point B to the line OA. Or The area of parallelogram with adjacent sides OA and OB Or Twice the area of triangle OAB (ii) Since c is perpendicular to ×ab , c which denotes OC   (1) lies on the plane containing O and (2) parallel to a and b. Hence c can be written as µλ= +c ab . Or Since c is perpendicular to ×ab , it denotes c is (1) parallel to a and b and (2) since a, b and c are position vector containing same point O, (3) hence c will lie on the same plane as a and b (OR point O, A, B and C lies on the same plane). Therefore c can be written as µλ= +c ab . (iii) Since angle 2AOC π= , we have ( ) 0 0µλ ⋅= ∴ + ⋅= ca a ba 0aa baµλ⇒ ⋅+ ⋅= 2 cos 03 πµλ⇒+ = a ba ( ) ( )( ) 2 11 21 0 2µλ ⇒+ =  µλ⇒= − Hence c abλλ⇒ = −+ 2 33c abλλ= ⇒−+ = 22 3abλ⇒ −+ = ( ) ( ) 2 3ab abλ⇒ −+ ⋅−+ = ( ) 2 23aa ab bbλ⇒ ⋅− ⋅+⋅ =

222 2 cos 3 (*)3 πλ ⇒ − + = −−− a ab b ( ) 22 122 3λ⇒ −+ = 2 1 1 λ λ ⇒= ⇒= ± Method 2 (to find λ ) 3 3 3 3AB λλ λ λ = −+ = −= = c ab ba   Using cosine rule, 222 2 2 cos 1 2 2(2) cos 3 3 3 AB OA OB OA OB AOB AB π =+− ∠ = +− = =             3 33 1 1 ABλ λ λ λ ∴= = = =±   (bi) Let D (0, 2, 0) be a point on plane p1 (OR let POSITION VECTOR of a point on plane p1 be 0 2 0 OD  =   ) Distance of A from p1 ( ) ( ) 2 22 11 2 1 11 2 1 AD  ⋅ −= + +−  

( ) ( ) ( ) 2 22 2 11 0 11 42 2 10 0 1 11 2 1 22 4 10 44   −  −⋅     −−  = + +− +−= = (ii) Let 11 2 1  = − n . Therefore  11 2 1 4    −=n . The equations of p1 and p2 are therefore  1⋅=rn and  3⋅=rn respectively. Therefore, distance between p1 and p2 = 3 -1 = 2 Since distance of A from p1 is larger than the distance between the 2 planes, therefore the point A will not lie between the 2 planes. Method 2 (for finding distance between 2 planes) Let E (0, 6, 0) be a point on plane p2 00 0 62 4 00 0 DE  =−=   Distance between 2 planes ( ) ( ) 2 22 11 2 1 11 2 1 DE  ⋅ −= + +−   = ( ) ( ) 2 22 0 11 42 01 8 2411 2 1   ⋅ − = = = + +− (iii) For the line BC to be perpendicular to both p1 and p2, C will be the foot of perpendicular of B onto p2. Since BC is parallel to normal of p2,

11 2 1 BC k  ∴=  −   0 11 42 12 1 OC k  ⇒ = −+  −−    Since C is a point on p2, we have 11 11 2 4 2 12 12 1 k k k      −⋅ =     −− −   11 4 8 12 12kk k⇒ + −++ = 1 2k⇒= 11 2 3 25 2 OC    ∴= −   −  