HCI 9758 2025 Prelim Paper 1 Solutions

HCI 2025 C2 H2 Mathematics Preliminary Examination Paper 1 Suggested Solutions 1 A curve has equation 31 1yx x= + + − . Using differentiation, find the set of values of x where the curve is strictly decreasing. Give your answers in exact form. [4] Suggested Solutions ( ) 1 31 1 1 3 1 yx x xx − = + + − = + + − Hence ( ) 2 2 d 1 0 3( 1)( 1)d 31 1 y xx x −= + + − − =− − For strictly decreasing, ( ) 2 d 0d 310 1 y x x   −  − Method 1 ( ) ( ) 2 2 2 2 2 1 3 0 1 22 0 1 xx x xx x − + −  − −−  − Critical values: 1x = and 2 2 2 0xx− − = ( )2 4 4 2 132x  − − = =  + - - + 13− 1 13+  |1 3 1 or 1 1 3x x x  −     +

2 Method 2 ( ) 2 31 1x  − Since ( ) 2 10x− , ( ) 2 13x− for 1x  (IMPT!!) 2 2 2 1 3 0 2 2 0 xx xx  − + −  − −  Consider 2 2 2 0xx− − = ( )2 4 4 2 132x  − − = =   |1 3 1 3, 1x x x  −   + 

3 2 A sequence of real numbers, nx , satisfies the recurrence relation ( ) 2 1 3 n nnx a bn cx+ = − + + , for , , ,n a b c and 1n . Given that 1 2 31, 8, 70,x x x= =− = and 4 377x =− , find the values of a, b and c. [3] Hence, find the values of 10x and 11x . [2] Suggested Solutions ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 3 32 4 3 8 3 (1) 3 2 8 70 9 4 8 (2) 3 3 70 377 27 9 70 (3) x a b c abc x a b c a b c x a b c a b c = − + + − =− + + −−−−−−−−− = − + + − = + − −−−−−−−−− = − + + − =− + + −−−−−−−− By GC, a = 2, b = 3, c = –5 From GC, 10 11 7210868 36172738 x x =− =

4 3 It is given that 2 2 d12 cosd yxy y xx xy  += , where 0 2x  and 0y . (a) Using the substitution 2u xy= , show that the differential equation can be reduced to d sec d ux xu= . [3] (b) Hence find the general solution to the differential equation. [3] Suggested Solutions (a) 2 2dd 2dd u xy uy y x yxx = =+  2 2 d1cos 2 d d1cos d d sec d yx y xy x xy ux xu ux xu  +=  = = (b) ( ) ( ) 2 24 d sec d 1 ln sec tan 2 sec , tan 0 for 0 2 2ln sec tan u u x x u x x C x x x x y x x C  = = + +     = + + 

5 4 The diagram shows a sketch of the function 3f ( ) e 1, xx =+ for 01 x The region bounded by the curve and the lines 0y= , 0x= and 1x = is A The region A is split into 5 vertical strips of equal width h, as shown in the diagram (a) State the value of h and using a suitable sketch, explain whether ( )( ) 5 1 f k h kh =  is less or more than the area of A. [3] (b) A is now split into n vertical strips of equal width . Using calculus, f ind the exact value of 3 6 9 3 3 31lim e e e ... e e n n n n n n nn − →  + + + + + +  . [3] (a) Suggested Solutions (a) 1 5h= ( ) 5 1 f k h kh =  = 1 1 2 3 4 5f f f f f5 5 5 5 5 5          + + + +                   = Area of 5 rectangles > Area of A O y x h y x

6 (b) ( ) 3 6 9 3 3 3 lim 1 e e e ... e e n n n n n n nn − → + + + + + + ( ) ( ) ( ) ( ) ( ) 3 6 9 3 3 3 lim 1 e 1 + e 1 + e 1 ... e 1 e 1 n n n n n n n − → = + + + + + + + + 1 3 0 13 0 30 3 e 1 d e 3 ee 1033 e2 33 x x x x =+ =+ = + − − =+ 

7 5 The tourism board plans to construct a tram track around a tourist attraction. The diagram above shows part of the blueprint of the track, where the tram will run along a circular path with centre O and radius 450 m, from a fixed point A to a variable point B, and then straight across to a fixed point C. The angle AOB is denoted as  , where 0 . The speed of the tram along the circular path will be maintained at 8 m/s and the speed along the straight path BC will be maintained at 6 m/s. (a) Show that the time taken for the tram to travel from point A to C is 225 150 6 cos42 +  . [3] (b) Use calculus to find the maximum time taken for the tram to travel from point A to point C. (You need not show that your answer gives a maximum.) [3] Suggested Solutions (a) Arc AB 450r== Time taken along arc AB 450 225 84 == 1 2OBC = (  at centre = 2  at circumference / ext.  of a  = 2 int.  in  ) BC = 2 450cos 900cos22 == O  C A B O  450m C B

8 Time taken along BC 900cos 2 6 150 6 cos 2   = = Total time 225 150 6 cos42 =+ (Shown) Method 2 (To find BC) 2 2 2 22 2 22 22 450 450 2(450)(450)cos( )] 2(450 ) 2(450 )( cos ) 2(450 )(1 cos ) 2(450 )(2cos ) 2 900 cos 2 BC = + − − = − − =+ = =      900cos 2BC = (b) Let 225 150 6 cos42T =+ . d 225 75 6 sind 4 2 T   =− For longest time, d 225 75 6 sin 0d 4 2 3sin 2 46 0.622368 T     = − = = = ( )225 0.6223680.622368 150 6 cos42 384.784942 seconds 6.41308 minutes 6.41 minutes (to 3.f.) T = + = = 

9 6 Given that b is a real constant such that 04 b , describe fully a sequence of transformations that transforms the curve 2yx= to the curve 241y x bx= + + [4] Sketch the curve 2 1 41y x bx= ++ Give the equation of any asymptotes and the coordinates of any axial intercepts and turning points, in terms of b where appropriate [3] Suggested Solutions 2 2 2 2 2 2 41 14 44 14 8 4 64 41 8 16 y x bx bxx bbx bbx = + + = + + = + + − = + + − Method 1 (1) Translate the graph 8 b units in the negative x-direction. (2) Scale the graph parallel to the y-axis by a scale factor of 4. (3) Translate the graph 2 1 16 b− units in the positive y-direction Method 2 (I) Translate the graph 4 b units in the negative x-direction. (II) Scale the graph parallel to the x-axis by a scale factor of 1 2 . (III) Translate the graph 2 1 16 b− units in the positive y-direction y x

10 7 (a) Use the formula listed in the List of Formulae (MF27) to explain why 0 1 e!r r  = = . [1] (b) Show that 0 1 2e!r r r  = + = . [3] (c) Find the value, in terms of e, of 6 1 !r r r  = +  . [3] Suggested Solutions (a) Given in MF27 23 0 e1 2! 3! ! ! r x r r x x xx r x r  = = + + + + + + = Letting 1x = , 1 0 0 1e ! 1 e! r r r r r  =  = = =   (b) ( ) 0 0 0 1 1 0 11 ! ! ! e! 1 e1! 1 e! e e 2e r r r r r r rr r r r r r r r    = = =  =  =  = + =+ =+ =+ − =+ = + =       (c) 5 6 0 0 1 1 1 ! ! ! 2172e 40 r r r r r r r r r  = = = + + +=− =−   