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