2021 NTSS Prelim 4E AM P2 MS v20210805a

This document consists of 16 printed pages. Setter: Mr Alan Cheng NEW TOWN SECONDARY SCHOOL Preliminary Examination Secondary 4 Express NAME Mark Scheme CLASS INDEX NUMBER Additional Mathematics Paper 2 4049/02 2 August 2021 09:05 – 11:20 2 hours 15 min READ THESE INSTRUCTIONS FIRST Write your name, register number and class in the spaces provided above and on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. For Examiner’s Use

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax , 2( 4 ) 2 b b acx a −  −= . Binomial Expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 221 21)( , where n is a positive integer and ( ) ! ( 1) ( 1) ! ! ! n n n n n r r r n r r  − − +== − 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2tan1 tan22tan −= Formulae for ABC A a sin = B b sin = C c sin a2 = b2 + c2 − 2bc cos A  = 2 1 bc sin A

3 1 Express ( )( ) 2 2 2 19 13 xx xx +− −+ as a sum of three partial fractions. [5] ( )( ) ( ) ( ) 2 22 2 19 131 3 3 x x A B C xxx x x +− = + +−+− + + [M1 ( ) ( )( ) ( ) 22 2 19 3 1 3 1x x A x B x x C x+ − = + + − + + − [M1] Sub 1x= , 16 16 A−= 1A=− [A1] Sub 3x=− , 16 4 C− =− 4C = [A1] Sub 0x= , 19 9 3A B C− = − − Sub 1A=− , 4C = , 19 9 3 4 B− =− − − 2B= [A1] ( )( ) ( ) ( ) 2 22 2 19 1 2 4 131 3 3 xx xxx x x +− =− + +−+− + +

4 2 The table below shows experimental values of two variables x and y. x 2 4 6 8 y 8.48 5.99 4.90 4.24 It is known that x and y are connected by the equation nyx k= , where k and n are constants. (a) Plot ln y against ln x , using a scale of 4 cm for 1 unit on both axes, for the given data and draw a straight line graph on the grid on page 5. [2] (b) Use your graph to estimate the value of n and of k. [3] ln ln nyx k= ln ln lny n x k+= ln ln lny n x k=− + [M1] 2.5 0.95 03n −−= − 31 60n= Accept 0.517 0.1 [B1] ln 2.5k = 12.2k = Accept 12.2 1.2k = [B1] (c) Use your graph to estimate the value of x when 2ye= . [2] ( ) 2ln 2e = Draw 2Y = [M1] From graph, ln 1x= xe= [A1]

5 (d) On the same diagram, draw the straight line representing the equation 3yx= and hence find the value of x for which 3 nxk+ = [3] 3yx= Draw ln 3lnyx= [B1 – line] 3 nxk+ = ( )3 ln lnn x k+= 3ln ln lnx n x k+= 3ln ln lnx n x k=− + [M1] 3ln lnxy= 0.7ln 0.7x x e =  = Accept 2.01 [A1] 0 1 2 3 1 2 3     B1 – Straight line through correct points plotted ln x 0.69 1.39 1.79 2.08 ln y 2.14 1.79 1.59 1.44 B1 - Correct table (c) (d)

6 3 The diagram shows part of the curve ( ) 2 9 1 3 y x =− − which intersects the x-axis at the origin and at the point P. The normal at the point Q on the curve cuts the x-axis at R. The gradient of the curve at Q is 18. (a) Find the coordinates of Q. [3] ( ) 2 9 1 3 y x =− − --- (1) ( ) 3 18 3 dy dx x = − [M1] ( ) 3 18 18 3 x = − [M1] ( ) 3 31 x−= 2x= Sub 2x= to (1) ( ) 2 9 18 32 y= − = − ( )2,8Q [A1] y x Q P R

7 (b) Find the coordinates of R. [3] Gradient of QR 1 18=− [B1] Sub ( )2,8 , ( )182 18yx− =− − [M1] 1 73 18 9yx=− + Sub 0y= 1 730 18 9x=− + 146x= ( )146,0R [A1] (c) Find the area of the shaded region. [4] Sub 0y= , ( ) 2 901 3 x =− − ( ) 2 39 x−= 33x= 0,6x= ( )6,0P [B1] ( ) 2 20 299 14 033 dx x xx − = − = −− [B1] ( )( )1 8 146 2 5762 −= ( ) 146 26 1469 9 196001 63 1433 dx x xx − = − =− −− [B1] Shaded area 196004 576 717 143= + + = unit2 (3 s.f) [A1]

8 4 (a) The equation of a curve is ( ) ( ) 22 2 4 8y b ac x a c x= − + + − . (i) Show that the roots of the equation are real if a, b and c are real. [3] Discriminant ( ) ( )( ) 2 24 4 2 8a c b ac= + − − − [M1] 2 2 216 32 16 32 64a ac c b ac= + + + − 2 2 216 32 16 32a ac c b= − + + ( ) 2 2 216 2 32a ac c b= − + + ( ) 2 216 32a c b= − + [A1] For all real values of a, b and c, ( ) 2 0ac− , 2 0b  Since Discriminant ( ) 2 216 32 0a c b= − +  , therefore roots are real. [A1] (ii) State the conditions that the roots are equal. [2] Discriminant ( ) 2 216 32 0a c b= − + = 0ac−= ac= [B1] 232 0b = 0b= [B1] (b) The equation of a curve is 2 43mx x m− + − , where m is a constant. Find the range of values of m for which 2 43mx x m− + − is not negative for all real values of x. [5] 2 40b ac− ( ) ( )( ) 2 4 4 3 0 mm− − −  [M1] 216 4 12 0mm− +  24 12 16 0mm− −  2 3 4 0mm− −  ( )( )4 1 0mm− +  [M1] [M1] 1m− or 4m [M1] Since curve is not negative, it must be U-shaped, hence m must be positive. Therefore, 4m [A1] -1 4

9 5 (a) Show that 31 11 3 1 3 x xx=−−− and hence find 3 13 x dxx− . [4] 31 11 3 1 3 x xx=−−− [B1] 31 11 3 1 3 x dx dxxx =−−− ( )1 ln 1 33 x x c=− − − + [B3 – minus 1 for each error, needs +c] (b) Given that ( )ln 1 3y x x=− , find an expression for dy dx . [2] ( )3 ln 1 313 dy xxdx x −= + − − [M1 for correct dvu dx or duv dx ] ( )3 ln 1 313 x xx −= + −− [A1] (c) Using the results from (a) and (b), find ( ) 0 1 ln 1 3 x dx − − . [4] ( ) ( )3ln 1 3 ln 1 3 13 dx x x xdx x −− = + − − ( ) ( )3ln 1 3 ln 1 3 13 xx dx dx x x x− = + − − [M1] ( ) ( ) ( )1ln 1 3 ln 1 3 ln 1 33x dx x x x x− =− − − + − [M1] ( ) ( ) ( ) 0 1 01ln 1 3 ln 1 3 ln 1 3 13x dx x x x x − − = − − − + − − ( )10 ln 4 1 1 ln 43  = − − + + −  [M1] 4 ln 4 13=− [A1]

10 6 In Figure 1, 3AE= cm and angle BAE x= . T