VJC 2023 H2 MATH CT SOLUTIONS

1 2 + – + 2023 J2 Common Test SOLUTION (For Students) No. Solution Comments 1 2 22 2 2 7 7 32132 32 49 32 x x xx xx xx xx xx − −− + −−=−+ −+ −+= − −+ 2 2 2 2 2 2 7 132 49 032 49 032 ( 2) 5 0( 2)( 1) x xx xx xx xx xx x xx − <−+ −+−< −+ −+ −+ + −− > − > Since 2( 2) 5 0x − +> for all real values of x, ( 2)( 1) 0xx− −> 1 or 2xx<> Multiplying -1 on both sides of an inequality will cause the inequality sign to change. Since no GC is allowed, you need to show that 2 49xx −+ is always positive. In this case, we complete the square to show the expression is always positive. Showing 2 4 90xx − += has no roots does not allow you to conclude ( 2)( 1) 0xx− −> . You need to also mention that coefficient of x2 > 0 and hence, 2 4 90xx − +> for all real x. 2(a) Differentiating e 13y xa −= + w.r.t. x, 0 () () 0 () de (3 ) ln 3d d3 ln 3de 3 ln 313 3 ln 3 33 ln 3 , 3 (shown)3 3 3 1 y xa xa y xa xa xa x a x x a a y x y x k −− − − − − −− − − − = × =   =  + = + = = ×× + Alternatively, ( ) ( ) 0 ln 1 3 ln 3 3d d 13 ln 3 1 ln 313 3 ln 3 33 ln 3 , 3 (shown)13 xa xa xa xa ax ax ax y y x k − − − −− − − = + = +   =   +   = + = =+ To remember: For c > 0, ( )d lnd xxc ccx = To remember: You can derive it from scratch. Let xuc= . ln lnuxc= Differentiate w.r.t. x 1d lnd d lnd lnx u cux u ucx cc  = = = This is a “show” question, so provide clear working on how 3 13 xa xa − −+ can be simplified to 1 13 ax−+

(b) Tangent to C at x = 0 makes an angle of 45° with the positive x-axis. 0 d ln 3 tan 45 1d 13 a y x −= = =+  ln 3 1 3 3 (ln 3) 1 ln 3 ln ((ln 3) 1) 2.1086 2.11 a a a a = + = − = − = −= − At x = 0, 0 ( 2.1086)e 13y −−= + 2.1086ln(1 3 )y = + Equation of T: 2.1086ln(1 3 ) 1( 0) 2.41 yx yx −+ = − = + Alternatively, using GC Equation of T: 2.41yx= + To remember: Grad of line = tanθ where θ is the angle which the line makes with the positive x-axis. Working should be in 5 s.f. and final answer can be given in 3 s.f. Alternatively, Use GC to find equation of tangent is allowed. Sketch ( ) 2.1086ln 1 3 xy += + , then on the graphing screen, press y¼ and select 5: Tangent. Press 0. Notice that the gradient of the tangent given by GC is 1.00. But from the question, you know it should be 1. Hence, equation of T: 2.41yx= + 3 ( ) 24 24 224 24 24 4 ln sec 1ln cos ln(cos ) ln 1 ... (from MF26)2 24 = ln 1 ...2 24 1 ... (from MF26)2 24 2 2 24 1 ... ...2 24 2 4 x x x xx xx xx xx xx x =  = − = − −++   − +− + +    = − −+ −−+ +    = −− + − + +   = − 244 24 ...2 24 8 ... (shown)2 12 xxx xx −+ −+ =++ Follow the given instruction: Use standard series from MF26. Repeated differentiation is NOT allowed. Standard series which are found in MF26 are: • ( )1 n x+ • ex • sin x and cos x • ln(1+x) We do not have standard series of sec x so use trigo identity and law of logarithm to change 1ln cos x   to ln(cos )x− . When expanding 24 ln 1 ...2 24 xx − +− + +   , treat 24 2 24 xx−+  as a single term and apply MF26 formula.

Putting ,4x π= 24 11ln sec 4 24 1 24 ππ π   ≈+       24 24 24 24 1 2 1ln 2(16) 12(256) ln 2 2(16) 12(256) 1 ln 22 2(16) 12(256) ln 2 , 16 and 1536 (Shown)16 1536 mn ππ ππ ππ ππ   ≈+ ≈+ ≈+ ≈+ = = Follow the given instruction: You are asked to substitute 4x π= . ( ) 24 ln sec ...2 12 xxx =++ From MF26, ( )tan d ln secxx x C= +∫ . Differentiating w.r.t. x, 3 tan ...3 xxx= ++ Make full use of MF26 You need to “Deduce” the series expansion for tan x. You need to use the result 24 ln(sec ) ...2 12 xxx =++ 4(a) (i) ( )f3yx= − Method 1: f( ) f( 3) f( 3)yx yx y x= →= + →=−+ Step 1: Replace x with x + 3 Translate graph 3 units in the negative x-direction Step 2: Replace x with -x Reflect the graph in the y-axis Method 2: ( )f( ) f( ) f( 3 ) f(3 ) y xy xy x yx = →=− →=− − = − Step 1: Replace x with -x Reflect the graph in the y-axis Step 2: Replace x with x – 3 Translate graph 3 units in the positive x-direction (a) (ii) ( ) 1 fy x= You need to sketch the tail-end behaviour correctly. As x → ±∞, f( )x → ±∞ and 1 0f( )x → . Label the features of your graph as stated in the question. Note that ( ) 1 22,− should be positioned lower than ( ) 1 107,−− . ( )2,0 ( )1, 0

4(b) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 12 f 2 21 ax a x ax a y yx a −= −  −= − −+= According to the sketch on the diagram in part(a)(ii), 2 11 24a >= for there to be 2 distinct real roots. In (a)(ii), 1 f( )y x= . So replace ( ) 2 1 f x as y2 ( ) 2 2 21 yx a−+= is an ellipse with centre (2, 0) 5(a) 21 , 3 3 zxy −+= = can be written as 11 3 0 , 23 r λλ −     = +∈        Normal to 1π : 1 31 31 0 20 3  −−         −×             21 20 23 63 4 22 21  = ×    = −= −  −−  Hence, 1π is perpendicular to 3 2 1  −− 33 2 1 9 2 11 10 −     − ⋅ = −− = −    −   Equation of 1π : 3 2 11x yz− −= − Note the difference between “Show” and “Verify”. To show, we are not allowed to use the given result. You need to show the given result. Hence, you should not be verifying 13 13 0 2 and 1 2 31 11          ⋅− ⋅−        −−     equal to 0. This is a “show” question. You should write 63 4 22 21    −= −  −−  to illustrate that the normal is // to 3 2 1  −− and provide a conclusion. (b) Method 1 Eq of perpendicular which pass through B: 53 6 2 , 10 1 ββ    = −+ − ∈   −  r  5 33 6 2 2 11 10 1 1 15 9 12 4 10 11 14 28 2 β β ββ β β      − + − ⋅− = −     −−   + + + − += − = − = − Method 1: See N as the intersection of line and plane Method 2: See BN   as projection of BA  on normal of plane and ON OB BN= +      (or NB   as projection of AB    on normal of plane) A(-3, 1, 0) (-1, 3, 2)

5 31 6 22 2 10 1 12 ON −     = −− −= −     −     Method 2 35 8 1 67 0 10 10 33 228 117 941 94110 33 24 14 10 2 2214 11 BA BN ON OB BN −−       = −− =       −         −−  −   −−  = ⋅ ++ ++−    −−+  = −= −−  −−  = +          5 61 64 2 10 2 12 −−     = −+ = −        Sketch a diagram with crucial information to help you visualize what is going on. (c) Method 1 Let A’ be the reflection of A in line BN ' 2 1 31 '2 2 1 5 12 0 24 OA OAON OA += −−       = −− = −               15 4 '5 6 1 24 10 14 BA −     =− −− =         Equation of line BA’: 54 6 1 , 10 14 δδ −   = −+ ∈    r  Method 2 Let A’ be the reflection of A in line BN ' 2 684 '24 7 1 2 10 14 BA BABN BA += −−−       =−=      −          Equation of line BA’: 54 6 1 , 10 14 δδ −   = −+ ∈    r  N B (5, -6,10) A (-3, 1,0) N B (5, -6,10) A (-3, 1,0) A’

6(a) 2 19 25 322 xx xxx −+ =−+++ You should do long division or compare coefficient to make 2 19 2 xx x −+