VJC 9758 2025 Prelim P1 Solutions

2025 JC2 H2 Math Prelim P1 Solutions 2025/VJC/Math Dept 1 Using an algebraic method, solve the inequality 2 43 14 3 1 x xx − +− . [3] Hence, find the set of values of x that satisfy 2 4ln 3 14(ln ) 3ln 1 x xx − +− . [2] No. Solution 1 2 2 2 2 2 43 14 3 1 43 104 3 1 4 3 4 3 1 0( 1)(4 1) 42 0( 1)(4 1) 42 0( 1)(4 1) x xx x xx x x x xx xx xx xx xx − +− − −+− − − − + +− − + − +− −+ +− 22 2 114 2 4 2 88x x x    − + = − − +       2 1 314 8 16x= − + Since 2 1 08x − for all real values of x, 2 1 3140 8 16x − +  OR Since coefficient of 2x is 40 and the discriminant of 242xx−+ is 2( 1) 4(4)(2) 31 0,− − =−  24 2 0xx− +  for all x . We have ( 1)(4 1) 0xx+ −  . 1x− or 1 4x Replace x with ln x : ln 1x− or 1ln 4x 10e x − or 1 4ex –1 1 4

2025/VJC/Math Dept 2 The function f is defined by 1f : , xx x − x , 0,x 1x . (a) Show that ( ) ( ) 21ff xx −= . [3] (b) Find ( ) 3f x in simplified form. [1] (c) Find ( ) 2030f5 . [2] Functions g and h are defined by 1g : , xx x − x , 1x , h : sin ,x ax − x , where a is a positive constant. (d) Find the value of a given that the range of hg is ( 1,0− . [2] 2a 11Let 1 xy xx −= = − 1 1 1 1 yx x y =− = − Hence ( ) 1 1f 1x x − = − ( ) 2 1ff 1 1 1 1 1 1 1 xx x x x x x xx x x −=  − − = − −−= − = − ( ) ( ) 21ff xx −= 2b ( ) ( ) 21ff xx −= ( ) ( ) ( ) 32 1 f f f ff xx x x − = = = 2c ( ) ( ) ( ) 2030 3(676) 2 2 11f 5 f 5 f 5 1 5 4 += = = =− −

2025/VJC/Math Dept 2d Range of g =  )0,1 = new domain of h Given: range of hg is ( 1,0− ππ 122 aa =  = 3 Referred to the origin O, points A and B have position vectors a and b respectively. Point C lies on OA, such that : 2:1OC CA = . Point D lies on OB, such that ::OD DB = . It is given that the area of triangle ABD is half the area of triangle ABC. (a) Show the area of triangle ABD is given by ( )2   + ab . Hence find the ratio : . [4] (b) The point E has position vector 15 48 +ab . Show that A, E and D are collinear. [3] It is further given that the angle AOB is π 4 and O lies on the perpendicular bisector of the line segment AB. (c) Find the length of projection of a on b , giving your answer in terms of b . Hence find the position vector of the point F, the foot of perpendicular from A to OB. [3] No. Solution 3a ( ) ( ) ( ) ( ) 1Area of triangle 2 1 2 2 02 2 ABD DB AB b b a b b b a ab ab         = =  −+ =  − + = + + = + ( ) 1Area of triangle 2 11 23 1 6 1 6 ABC CA AB a b a a b a a ab = =  − =  −  = π 2a

2025/VJC/Math Dept Area of triangle ABD = 1/2 (area of triangle ABC) ( ) ( ) 11 2 2 6 1 2 12 6 5 : 5:1 a b a b          =  +  =+ =+ = = 3b 15 48 35 48 AE OE OA=− =+− =− + a b a ab 5 6 4 3 5 3 4 8 AD OD OA=− =− + = − + ab ab Since 4 3AD AE= and A is a common point, A, E and D are collinear. 3c Length of projection of a on b = . ab cos 4 cos 4 1 2   = = = ab a b 1 2 1 2 OF = = bb b

2025/VJC/Math Dept 4 (a) A sequence is such that 1up= , where p is a constant and 1 5 81 n n n uu u + = + , for 1n… . (i) Describe how the sequence behaves when 1p= . [2] (ii) Find the value of p for which 6 3125 6253u = . [2] (b) Another sequence 1v , 2v , 3v , … is such that for all 1n… , 21 2n n nv v v k++− + = , where k is a constant. Let 1n n nw v v +=− for 1n… . Explain why the sequence  nw is an arithmetic progression. [2] No. Solution 4ai When 1p= , 1 1u = , 2 5 ,9u = 3 25 49u = , 4 125 249u = , 5 625 ,1245u = The sequence is a decreasing sequence, and it converges to 1 2 . 4aii 1 1 5 1 8 1 8 1 8 1 5 5 5 nn n n n n n uuu u u u u + + +=  = = ++ 1 11 1 1 1 1 1 1 8 5 8 5 5 5 55 58 58 n n n n n n n n n n u u u u uu u uu u + ++ + + + + −= − = = − = − Using the GC, 1 1 5u = 1 5p= 4b ( ) ( ) 1 2 1 1 21 2 which is a constant n n n n n n n n n w w v v v v v v v k + + + + ++ − = − − − = − + = Hence  nw is an arithmetic progression.

2025/VJC/Math Dept 5 The function f is given by 22 22 2 2 ( 2) , 0 < 4,f ( ) 2 2 ( 6) , 4 < 8. xxx xx  + − − =  − − −  It is given that ( ) ( )f f 8xx=+ for all real values of x. (a) On the diagram in the Printed Answer Book, sketch the graph of ( )fyx= for 6 7,x−   indicating clearly the coordinates of the end points and the points where the graph cuts the axes. [3] (b) Without integrating, write down the exact area of the region bounded by f( )yx= , the line 4x= , the x-axis and the y-axis. [1] The curve C has equation ( ) ( ) 2 2 2 3 21y xa − − + = , where a is a positive real constant. (c) State the equations of the asymptotes of C in terms of a. [1] (d) Determine the range of values of a if there is at most one intersection between C and the graph of ( )fyx= . [2] No. Solution 5a 5b Area = 2124 π(2 ) 8 2π2 + = + 5c Equations of asymptotes: 2 2 2 ( 3) ( 2)y xa − =+ 3 ( 2)y a x− = + 3 ( 2) and 3 ( 2)y a x y a x= + + = − + 5d C is a hyperbola centred at ( 2, 3)− with vertices at ( 2, 3 ) and ( 2, 3 )aa− + − − . When 3a= , there is exactly one intersection at ( 2, 0)− . Hence there is at most one intersection between C and the graph of f( )yx= when 3a .

2025/VJC/Math Dept 6 It is given that ( ) ( )2 2 1 1 1 4 1 2 1 21 n r nnrr= = − +− + . (a) Show that ( ) 2 2 1 1 n r rr= − is less than 1 4 . [2] (b) Give a reason why the series ( ) 2 2 1 1r rr  = − converges, and write down its value. [2] (c) Find the smallest value of n for which ( ) 2 2 1 1r n rr= − differs from ( ) 2 2 1 1r rr  = − by less than 0.0007. [2] (d) Find ( )( )( )1 1 12r N m r m r m r m=+ − − + − + , where m and N are integers with 0Nm . (There is no need to express your answer as a single algebraic fraction.) [2] No. Solution 6a ( ) ( ) ( ) ( ) 11 1 1 0 for all 22 1 2 2 1 2 1 nn nn n n n n n −+ −− = =  + + + ( ) ( )2 2 1 1 1 1 4 2 1 2 4 1 1 n r rr nn= =− − + + Alternative ( ) ( ) ( ) 2 1 2 for all 2 11 2 1 2 11 02 1 2 n n n nn nn +   + −+ ( ) ( )2 2 1 1 1 1 4 2 1 2 4 1 1 n r rr nn= =− − + + 6b As n→ , ( ) 11 0 and 0 2 1 2nn →→+ , ( ) 2 2 1 4 1 1 n r rr=  − → , a constant. Hence ( ) 2 2 1 1r rr  = − converges and ( ) 2 2 11 41r rr=  = − .

2025/VJC/Math Dept 6c ( ) ( ) ( ) ( ) 22 22 0.0007 1 1 1 1 0.00074 2 1 2 4 11 0.0007 1 21 1 11 2 rr n nn n r r r r n ==  − + − − + − −− +  Using GC: Alternative n ( ) 11 2 1 2nn −+ 26 0.000712 > 0.0007 27 0.000661 < 0.0007 smallest value of 27n= 26.231 smallest value of 27nn  = 6d ( )( )( ) ( )( ) ( )( ) ( )( )( ) ( )( )( ) ( ) ( ) ( ) 2 1 1 2 2 1 1 12 1 1 1 ...1 2 3 2 3 4 1 2 1 1 11 1 1 1 1 1 4 222 r N Nm Nm m r r r m r m r m N m N m N m rr N r r N r mm =+ = = −+ −+ − − + − + = + + + − − + − + = −+ = − = + − − + − +    Alternative ( )( )( ) ( )( )( ) ( )( ) ( ) ( ) 1 1 11 1 2 2 1 12 1 11 1 1 1 1 1 4 2 2 2 1 r m m rm r N rm N m N r m r m r m r r r rr N m N m =+ + − = + + − = −+ = − − + − + = −+ = − = + − − + − +   

2025/VJC/Math Dept 7 Do not use a calculator in answering this question. (a) Find the complex number z which satisfies the equation 4 5i15 * z z =− . [3] (b) The complex number w is such that ( ) 3 iiw− =− . (i) Given that one possible value of w is 2i , find the two other possible values of w. Give your answers in cartesian form iab+ . [4] The points 1W , 2W and 3W on the Argand diagram represent the three roots of the equation ( ) 3 iiw− =− , and the point A represents the complex number ik , where k is a positive real number. (ii) Show that