TJC 9758 2023 Prelim P1 Solution

Solutions to 2023 TJC Prelim Paper 1 Solution to Q1 (a) 1tan dx xx − ∫ 22 1 2 2 1 2 2 11 1tan d21 2 11tan 1 d22 1 1tan tan22 xxxx x xxx x xx x xc − − −−  = −  +  = −−  +  = −− +  ∫ ∫ (b) sin2 cos5 dx xx∫ 1 sin7 sin3 d2 11 1 cos7 cos327 3 11cos7 cos314 6 x xx x xC x xC = − = −+ + = − ++ ∫

2 Solution to Q2 (a) 4x2 − 24x + 9y2 =0. ( ) ( ) ( ) 22 2 2 2 2 22 4 690 4 3 9 36 3 132 xxy xy x y −+= −+ = − += (b) The required sequence of transformations is A: A translation of magnitude 3 unit in the negative direction of the x-axis. B: A translation of magnitude 1 unit in the positive direction of the y-axis. C: A scaling parallel to the y-axis by a scale factor of 1 α . OR A: A translation of magnitude 3 unit in the negative direction of the x-axis. B: A scaling parallel to the y-axis by a scale factor of 1 α . C: A translation of magnitude 1 α unit in the positive direction of the y-axis. ( ) ( ) ( ) 22 22 22 1(c) 194 1 13 2 yx yx α α α −+= −+= For shape of a circle, 2 3α = x y O (0,0) (6,0) (3,2) (3,−2) × ×

3 Solution to Q3 ( ) ( ) 1 2 1 2 1 2 1 2 (a) g 4 4 1 4 1 1 4 4 2 xx x xx xx xx − − − − = − = − = −  = −  2 244 13 11 22122 ! xx x   −−      = +− + +      −−  2311 3 2 16 256xx x= ++ +  Therefore 11 3,,2 16 256ab c= = = (b) Percentage error = () () 100% 4%() f x gx gx − ×< 2311 3 2 16 256 4 0.04 4 xxx x x x x  ++ − −⇒< − Using GC, 0 1.87x⇒ << (corr. to 3 s.f.)

4 Solution to Q4 23 8 51 8(a),(b) 3 1(3 ) 3 xxyx kx k x − +− = = +−−−  When 0x= , 5 3y k=− When 0y= , 23 85 ( 35 ) (1 ) 0xx x x− + −= − + −= ⇒ 51 or 3x= Equation of asymptotes: 1(3 1)yx k= + and 3x= 1 8 37(c) 3 1 3 8 31 37 3 8 8 3 31 2 xxk xkk xx x x xx  +− = − − +− = −− −= −− −=⇒= Therefore, from the graph, 2 or >3xx≤ x = 3 7 ,03   x y (1,0)

5 Solution to Q5 (a) First term of AP: 500a= Common difference of AP: 10d = Formulation of problem: [ ]2(500) 10( 1) 100002 n n+ −> Using GC (table of values to be shown), 18n= Date of 18th month: 1 June 2024 (b) Formulation of problem: Month n Start of month End of month 1 X 1.005x 2 1.005xx+ 21.005 1.005xx+ 3 21.005 1.005x xx++ 321.005 1.005 1.005x xx++    At the end of Nth month, account has 11.005 1.005 ... 1.005NNxx x −+ ++ 1(1.005 1.005 ... 1.005) 1.005(1.005 1) 1.005 1 201 (1.005 1) NN N N x x x −= + ++  −=  − = − Thus we have 60N = at the end of 31 December 2027 60201 (1.005 1) 50,000 $713.0747xx −≥ ⇒ ≥ Least $714x⇒=

6 Solution to Q6 (a) 2 2 1 ( )( )sin2 1 ( 1)(4 )4 1 ( 1)(16 8 )4 A PQ PR QPR xx x xx = ∠ =+− = + −+ 23 21(16 8 16 8 )4 x x x xx= − ++−+ 321( 7 8 16)4 xxx= − ++ (Shown) Or let N be the foot of perpendicular from Q to PR. ( )1sin 30 1 2QN PQ x= = +  2 1 ( )( )2 1 ( 1)(4 )4 A QN PR xx = =+− 21 ( 1)(16 8 )4 x xx= + −+ 23 21(16 8 16 8 )4 x x x xx= − ++−+ 321( 7 8 16)4 xxx= − ++ (Shown) (b) 2d1 (3 14 8)d4 A xxx = −+ At stationary values, d 0d A x = 21(3 14 8) 04 2 or 4 (rejected since it is given that 4)3 xx xx x ⇒ − += ⇒= = < 22 33 2 2 d1 (6 14) 2.5 0d4 xx A xx == = − = −< (maximum) To find QR: When 2 3x= , 5 3PQ= and 100 9PR= Using cosine rule, 22 2 5 100 5 100 2 cos303 9 39 9.70 (3 s.f.) QR QR      = +−           ∴≈ 

7 Or 12 5 123 6QN = += 35 53cos30 23 6PN PQ = = =   100 5 3 96RN PR PN=−=− 22 22 5 100 5 3 9.706 96QR QN RN = += +− =   

8 Solution to Q7 (a) 23 d 1d 8 ,1dd xy tt tt= − = −− ddd ddd y yx xtt= ÷ ( ) 2 3 8 1 tt = −− − 28 tt= + For stationary point, 2d8 00d y txt=⇒+= 3 8t⇒= − 2t⇒= − When 2t =− , ( ) 2 13 42 , 2322 2 xy= += = +=− − Coordinates of A is 3 ,32   Equation of tangent is 3y= (b) When 3y= , 2 4 3tt −= 3243 tt⇒ −= 32 3 40tt⇒ + −= Using GC, 1 or 2 tt= =− (Reject 2t =− is pt A) [If GC is not allowed, note that t = -2 must be one of the solution since one of the intersection point between the tangent and C is at A ( )( ) 22 20t tt⇒ + +− = ( ) ( ) 2 2 10tt⇒ + −= 2 or 1tt⇒= − = ] ∴coordinates of B is (3, 3). Gradient of normal at B 28 1 11 19= −= −+ Equation of normal is ( )133 9yx−= − − , i.e. 1 10 93yx= −+

9 (c) The point F has coordinates 100, 3   . Height of triangle = 10 1333−= AB has length 333 22−= Area of Triangle 11 3 1 23 2 4   = =     unit2

10 Solution to Q8 (a) Let izxy= + *( i i) 2i( i) 4 i( 1) 2i 2 4 xy xy xy xy ++= + + − += − + Comparing real and imaginary part ( ) 4 2 (1) 1 2 (2) solving: 1 2(4 2 ) 39 3 xy yx yy y y = − −−− − + = −−− −−= − = = Solving, 2, 3xy= −= (b)(i) ( ) 2 1 3 12 1arg tan 63 w w π− = += = −= − 6 * 2 3 6 2 3 6 62 36 2 3 1 3 3i2 1 16 1 16 1 216 4 4 114 cos sin33 134i 22 2 2 3i i i i ii ii i we z ew z ew ze e e e i π π π ππ ππ π ππ − − −− −− = −=  = = =   = =   = +    = + = +