2023 JC1 Promo Practice Paper C (solutions) VJC

2023 PROMO PRACTICE PAPER C Solutions Qn Suggested solution Comments 1 Method 1 As 2 2 1 11 1130 2 44xx x ++= ≥ > ++  for all real values of x . OR Method 2 As (coefficient of 2 1 ) 0x = > and discriminant 2 4(1)(3) 11 0(1) − = −<= , we have that 2 30x x++> all real values of x . 2 2 2 2 2 26 1 12 4 26 1 0, 4 and 3( 4)( 3) 4 2 6 ( 3) 0( 4)( 3) 3 0( 4)( 3) xx xx x xx xxxx x xx x xx xx xx ++ ≥−− − ++ − ≥ ≠ ≠−−+ − + +− + ≥−+ ++ ≥−+ This implies 1 0( 4)( 3)xx ≥−+ since 2 30x x++> all real values of x . ( 4)( 3) 0xx− +> 3 or 4xx<− > 2 ( ) ( ) ( ) 2 2 2 2 4i 1 i 1 i1 i i 4i 4 1 5 4i 1 z λλ λλ λλ λ λ λλ λ −+= ×−+ + −+ = + +− = + arg( )z π= ⇒ , 0 and Im( ) 0zz z∈< = 2 40 2λλ−= ⇒ = ± When 102, 2 5zλ = = = (Rejected since 0z< ) When 102, 25zλ −= −= = −

2023 PROMO PRACTICE PAPER C Solutions Alternatively, Sketch indicates that λ< 0. 4// 1OA OB λ λ −−⇒= 2 4 2 (since < 0) λ λλ −= − =− Hence, ( )2 1 2i2 4i 21 2i 1 2iz −+−−= = =−++ 3i ( ) ( ) ( ) 222 22 2 cos 5 6 2(5)(6)cos 25 36 60 cos cos sin sin 3461 60 cos sin55 61 36cos 48sin PQ PR QR PR QR PRQ PRS PRS PR S θ θθ θθ θθ =+−⋅ ∠ =+− ∠ + =+− ∠ − ∠ = −−  = −+ ( ) 1 2 61 36cos 48sin (shown)PQ θθ∴=− + A(1, ) B( , –4) RE IM O

2023 PROMO PRACTICE PAPER C Solutions 3ii ( ) ( ) ( )( ) 1 2 1 2 1 2 1 2 2 2 2 211 222 22 2 61 36cos 48sin 61 36 1 48 2 25 48 18 48 1851 25 25 1 48 18 4851 2 25 25 2! 25 24 9 28851 25 25 625 24 6351 25 625 245 5 PQ θθ θ θ θθ θθ θθ θ θθ θ θθ θ = −+  ≈− −+  =++ = ++  −  ≈+ + +       = ++ − = +− = + 263 24 63, , 125 5 125pqθ−= = − 4i d ed xy xyx −+= 2 2 dd edd xyy xyxx −+ += − 32 32 d d dd ed d dd xy yyyxx x xx −+ ++= When 0x= , 1y= , d 1d y x = , 2 2 d 2d y x =− , 3 3 d 1d y x =− . Let y = f(x). The Maclaurin series for y is: 23 f (0) f (0) f (0) f (0) ...2! 3! xxyx ′ ′′ ′′′= ++ + + 23 11 ...6y xx x=+− − + f(0) = 1 ( )1 when 0yx= = 4ii Since 23 11 ...6y xx x=+− − + , differentiating y w.r.t. x, 2d1 1 2 ...d2 y xxx = −− + 5ai 2 1 d25 xxx++ ⌠⌡ ( ) 2 2 1 d 12 x x = ++∫ 111tan22 x C− += +

2023 PROMO PRACTICE PAPER C Solutions 5aii ( ) 3 ln dx xx ⌠ ⌡ ( ) 4 ln 4 x C= + f(x) = ln x, 1f '( )x x= ( ) ( ) 3 4 f ' ()f () d f( ) 4 xxx x C= + ∫ Note that ( ) 3 ln 3lnxx ≠ 5b [ ] [ ] ( ) π 2 ππ 22 π 2 π 2 0 0 0 0 0 cos d (sin ) sin d π sin d2 π cos2 π 012 π 12 x xx x x xx xx x = − =+− = + = +− = − ∫ ∫ ∫ dd d d dd uvv x uv u xxx = −∫∫ Take note the correct choices of function to be d d u x (integrated) and v (to be differentiated) choose d sind u xx = and v = x 6a 22 0x ay bx cy+ ++= Since C passes through ( )5, 3− and 13,22  − , ( ) ( ) ( ) 225 35 3 0 9 5 3 25 - 1 a bc abc + − + +− = ⇒+−= − ( ) 22 1 31 3 02 22 2 19 1 3 044 2 2 9 2 6 1 - 2 a bc abc abc    + − + +− =       ++