HCI 9758 2025 Prelim Paper 2 Solutions

2025 C2 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Section A: Pure Mathematics (40 marks) 1 The region A is bounded by the curves 1yx=+ , 72yx=− , the x–axis and the y–axis. (a) Find the exact area of A. [4] (b) Find the volume of the solid obtained when A is rotated through 2 radians about the y–axis. [3] Qn Suggested Solutions (a) Coordinates of point of intersection: ( )2, 3 Method 1: Using x-axis. ( ) ( ) 2 3.5 02 2 3.5 02 2 3.533 22 02 2 Area d d 1 d 7 2 d 2 1 7 2 33 2 27 2 27 3 3 3 22 3 3 3 23 3 units3 y x y x x x x x xx =+ = + + −    +−   =−             = − − −          = − + =−   y (2, ) (3.5,0 ) x A (0,1) O

Method 2: Using y-axis. 33 01 233 2 01 3333 01 2 Area d dy 7 d 1 d2 7 2 6 3 7 3 3 3 3 3 1 312 6 3 3 23 3 units3 x y x y y y y y y y y =− −= − −    = − − −           = − − − − −              =−   (b) 2 72 7 2 yx yx =− −= ( ) 2233 22 01 3 7Vol d 1 d 2 47.38326 47.4 units (3 s.f.) y y y y  −= − − =  

2 It is given that ln sin 4yx  =+   , where 3 44 x−   . (a) Show that 22 2 dd 10dd yy xx + + = . Hence find the first four non-zero terms of the Maclaurin expansion of y, leaving your answer in exact form. [6] (b) Verify the result obtained in part (a) is obtained using standard series from the List of Formulae (MF27). [5] Qn Suggested Solutions (a) e sin 4 y x =+  de cosd4 y y xx =+  2 2 2 dde e sind d 4 yy yy xxx    + =− +       2 2 2 dde e edd y y y yy xx  + =− 2 2 2 dd 10dd yy xx  + + = Alternative Solution cosd 4 cotd4 sin 4 xy xx x     + = = +  + 2 2 2 d cosecd4 y xx =− +  22 22 2 dd 1 cot cosec 1 0d d 4 4 yy xxxx      + + = + − + + =           Applying further implicit differentiation 23 23 d d d20 d d d y y y x x x  +=  When x = 0, 23 23 1 d d dln , 1, 2, 4d d d2 y y yy x x x = = =− =

Maclaurin expansion of y is 2312ln ... 32 y x x x= + − + + (b) ln sin 4yx  =+   ln sin cos sin cos44xx =+  ( ) 11ln sin cos 22 1ln sin cos 2 xx xx =+  =+  231ln ln 1 262 xxx + + − − 22 3 2 3 323 11ln 2 6 2 2 62 1 ... 3 2 6 x x x xxx xxx    = + − − − − −       + − − + ( ) 22 3 2 3 23 2 3 3 1 1 1ln ... 2 6 2 2 32 1 1 1ln ... 2 6 2 32 x x xx x x xxx x x x    = + − − − − + +       = + − − − − + + 2312ln ... 32 x x x= + − + + (verified)

3 The parametric equations of the curve C are 1 3cosecx=−  and 2cot 3y=−  , where 0 . (a) Show d2 secd3 y x =− . Hence find the equation of the normal to C at the point where 4  = . Give the equation in the form ,=+y Ax B where A and B are exact constants to be found. [4] (b) Show that the normal found in part (a) will cut C again. [2] (c) Find the Cartesian equation of C. [2] (d) Sketch C, indicating clearly its key features. [3] (e) Find the range of values of m such that there is no intersection between the line ( )13y m x= − − and C. [2] Qn Suggested Solutions (a) ( ) 2 2 dd 3 cosec cot 2cosecdd d 2cosec 2cosec 2 secd 3cosec cot 3cot 3 xy y x        =− − =− =− =− =− When 4=  , 1 3 2 1 x y =− =− ( ) d 2 1 2 2 d 3 3 cos 4 y x   =− =− Equation of normal: ( ) ( )( ) ( ) 31 1 3 2 22 3 1 3 2 1 22 3 3 9 122 2 2 2 yx yx x − − = − − = − + − = − + − 3 3 7 22 2 2 2 yx = − +

(b) Substitute 1 3cosecx=−  and 2cot 3y=−  into ( )3 3 72cot 3 1 3cosec 22 2 2 2 − = − − + By GC graph, Enter 1 2cot 3Y =−  and ( )2 3 3 71 3cosec 22 2 2 2 Y = − − +  , and find points of intersection. 2.95319= or 0.7853982 = 4  Since there are 2 values of  , the normal line will cut the curve again. (c) 1 3cosec and 2cot 3 13cosec and cot32 xy xy   = − = − −+== For 0 , 0 sin 1 cosec 1 3cosec 3 1 3cosec 2 2x   − − − −  −     22cot 1 cosec+= 22 22 31 123 13 1, 232 yx xy x +−    +=       −+   − = −      

(d) (e) When x = 1, ( )1 1 3 3ym= − − =− The line passes through the centre ( )1, 3− , of the hyperbola . For no intersection between the line and C, The range of values for m is 22 or 33mm−  . x y ( )1, 3− O

4 Let A, B and C be the points on the same plane with position vectors a, b and c respectively. It is given that a, b and c are unit vectors such that + + =a b c 0 . (a) (i) By considering cc , find the value of ab . [3] (ii) Find the angle AOB . [2] (iii) Draw the position vectors a, b and c on a single diagram. Using your diagram, identify the type of triangle ABC. [2] (b) The point D has position vector a + b. Find the area of the quadrilateral ACBD. [2] Qn Suggested Solutions (ai) =− −c a b ( ) ( ) = +  +c c a b a b 2 22 | |= +  +a a b b 1 1 2 1= +  +ab 1 2 =−ab (aii) 1 cos2 AOB− = ab 1cos 120 2AOB AOB =−  =  (aiii) Since 120AOB =  , 30OAB =  with OA OB OC== . It can be shown that the triangles OAB, OAC and OBC are congruent triangles and 60ACB ABC BCA = = =  . Hence triangle ABC is an equilateral triangle. Triangle ABC is an equilateral triangle. (b) Since OD= + =−a b c and OD and AB are diagonals of the rhombus OADB, OD and AB are perpendicular to each other. Hence ADBC is a kite. A B O C 120o 60o

Method 1 Area of ACBD = 12 2 AC DC ( ) 2= − c a c ( )2= − − b a a ( )2 sin120 3=  = Method 2 Area of ACBD = ( )( ) 11 2 3 322 DC AB  = = A B O C D 120o 60o

Section B: Probability and Statistics (60 marks) 5 The eleven letters in the word INSPIRATION are each printed on separate, identical cards. (a) Find the number of ways in which the cards can be arranged in a row if, (i) there are no restrictions, [1] (ii) the letters N are together or the letters I must all be separated, but not both. [3] (b) Three of the eleven cards are removed at random. Find the probability that the letters on the eight cards left behind are all distinct. [2] Qn Suggested Solutions (a)(i) Number of ways 11! 33264002!3!== (a)(ii) Number of ways for N,N together 10! 3! 604800 = = Number of ways for I,I,I to be separated 9 3 8! 2! 1693440 C= = Number of ways for both NN together and III separated 8 37! 282240 = = C Number of ways the letters N must be together or the letters I must be separated, but not both 604800 1693440 2(282240) 1733760 = + − = A S B A: ‘N’s Together B: ‘I’s separated