NYJC 2026 FM Prelim P1 (Ans)

1 NYJC 2025 JC2 Preliminary Examination 9649/01 2026 Preliminary Examination Further Mathematics Paper 1 Solutions 1 A sequence {}nu is given by 1uM= and 3 1 6 for 1,7nn uun+ += where M is a constant. It is given that the sequence converges to the limit L. (a) Find the possible values of L. [1] (b) Show that if 12 nu , then 112 nu + and 1nnuu+  . [3] (c) Given that 1.5M = , deduce the value of L. [1] 1(a) As ,n→ 1, .nnu L u L +→→ 3 36 7 6 07LL L L +=  − + = Using G.C., 3, 2 or 1L=− (b) 333 1 61 6 2 61 2 7 7 7 1 2 n n n uu u + +++        ( )( )( )33 1 1 2 36 7 6 7 7 7 n n nn n n n n n u u uu u uu u u+ − − ++ − +− = − = = 1 2 nu   1 0, 2 0 and 3 0 n n nu u u−  −  +  Thus 11 0 n n n nu u u u++−    . (c) By (b), since 1 2 31 2 1 , ,... 2u u u     and since the sequence is decreasing, 1L= .

2 The complex variables w and z, represented by the points W and Z respectively in an Argand diagram, are related by the equation 4i 2 zw z+= − , 2z . (a) Given that the locus of W is the imaginary axis, find the cartesian equation of the locus of Z and give a geometrical description of this locus. [4] (b) Given instead that the locus of W is the real axis, show that the locus of Z is a straight line, and write the equation of the line in the form pz qz r+= where z denote the complex conjugate of z and p, q and r are complex numbers to be determined. [3] 2(a) Since W lies on the imaginary axis, ( )Re 0w = . Let iz x y=+ where ( ) ( ), 2,0xy  . ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 4i i 4i 2 i 2 4 i 2 i 2 i 2 i 2 4 2 4 i 2 z x yw z x y x y x y x y x y x x y y x y xy xy + + +== − + −  + +  − −   =  − +  − −    − + + + − + − = −+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 Re 0 2 4 0 1 2 5, , 2,0 w x x y y x y x y =  − + + =  − + + =  which is a circle of radius 5 and centre at ( )1, 2− with the point ( )2,0 removed. (b) Since W lies on the real axis, ( )Im 0w = . ( ) ( )( )Im 0 2 4 0w x y xy=  − + − = from (a) ( ) ( )2 4, , 2,0x y x y − =  which is the cartesian equation of a straight line with the point ( )2,0 removed. ( ) ( ) 2Re 2 and 2 2iIm 2i 2i zzz z z x x zzz z z y y     ++ = =  = −− = =  = Substitute above into 24xy−= gives 112 4 1 i 1 i 42 2i 2 2z z z z zz    +−    − =  + + − =              .

3 (a) It is given that for non-negative integer n, 6 0 sin dnnI  = . Show that ( ) 2 3 1 2 nn nnI n I −=− + − , for 2n . [3] (b) The region R is bounded by the curve 22 1y x x=− , the x-axis and the line 1 2x= where 10, 2x  . Using the shell method and the substitution sinx = , show that the volume of solid generated when R is rotated through 2π radians about the y-axis is given by ( ) π 3mnk + , where k, m and n are integers to be determined. [5] (a) ( ) ( ) ( ) ( ) ( ) ( )  ( ) ( ) 6 66 6 1 0 12 0 0 1 22 0 22 2 2 2 sin sin d cos sin cos 1 sin cos d 31 1 cos sin d22 3 1 1 sin sin d 2 3 1 sin d sin d 2 3 1 2 3 11 2 n n nn n n n n nn n nnn nn I n n n n n I I n I n I                       − −− − − − − − − = = −  − −  − =− + −  =− + − −  =− + − −  =− + − − =− + − − −      ( ) 2 3 1 2 n nn nnI n I −=− + − (b) V olume of solid generated ( ) ( ) ( ) 1 2 1 2 6 6 6 66 5 22 0 32 0 32 0 32 0 32 0 35 00 3 2 1 d 2 1 d 2 sin 1 sin cos d 2 sin cos d 2 sin 1 sin d 2 sin d sin d 2 x x x x x x x II                          − =− =− = =− =−  =−       From (a), ( ) 2 31 2 nn n nII nn −−=− + .

  6 6 1 00 3sin d cos 1 2I     = =− = − ( ) 31 3 3 2 3 2 3 2 3 133 24 3 2 3 832II =− + =− + − = −  ( ) 5 3 3 5 3 4 3 4 5 160 552I I I=− + =− + ( ) ( ) 53 3 3 3 3422 160 5 312 160 5 3 1 2 323 160 5 3 8 2 1123 15 160 64 33 3240 I I I I I       − = − − +   =+    = + −  =−  =− 240, 64, 33k m n= = =− 4 Ethan uses the expression ( ) ( )1f 0 f f 1 2E a b c = + +  to estimate the integral ( ) ( ) 1 0 1 f d ,I x x x=+ where a, b and c are constants. It is given that Ethan’s method gives the exact value of I for ( ) 2f 1, and x x x= . (a) Find the values of a, b and c. [4] (b) The Simpson’s rule with two strips is applied directly to estimate I. Show that Simpson’s approximation is identical to Ethan’s method. [2] (c) Tang applies the trapezium rule with two strips to estimate I. Show that his approximation can be written in the form ( ) ( )1f 0 f f 1 2T    = + +  where , and    are values to be determined. [2] (d) If ( )f x is a quadratic function, determine whose estimate is more accurate. [2] 4(a) ( )f 1:x = ( ) 1 0 1 d 1.5I x x= + = 1.5abc++= ----- (1) ( )f: xx= ( ) 1 0 51 d 6I x x x= + =

15 26bc+= ----- (2) ( ) 2f: xx= ( ) 1 2 0 71 d 12I x x x= + = 17 4 12bc+= ----- (3) Solving (1), (2) and (3) using GC: 1 6a= , 1b= , 1 3c= . (b) Using Simpson’s rule, I  ( ) ( ) ( ) ( ) ( )1 1 11 0 1 0 f 0 4 1 f 1 1 f 16 2 2     − + + + + +         ( ) ( )11f 0 6f 2f 162 = + +   ( ) ( )1 1 1f 0 f f 16 2 3 = + +  = E . (c) Using Trapezium rule, T = ( ) ( ) ( ) ( ) ( )101 1 1 1 0 f 0 2 1 f 1 1 f 12 2 2 2−     + + + + +        ( ) ( )11 f 0 3f 2f 142 = + +   ( ) ( )1 3 1 1f 0 f f 14 4 2 2 = + +  1 3 1,,4 4 2  = = = . (d) Since f(x) is quadratic, the integrand (1+ x)f(x) is a cubic polynomial. Ethan’s method which is the same as Simpson's Rule would provide exact value for the integrand for cubic polynomial whereas Tang’s method which uses the trapezium rule uses straight, linear segments to approximate the curve results in larger gaps under the curve . Hence, Ethan’s estimate is more accurate. 5 The variables x, y and z are related by the equation ( )f,z x y= where f is a function of the independent variables x and y. We define the independent differentials dx and dy to be the small increments of x and y respectively. The total differential dz , which is the small increment of z corresponding to small increments of x and y, is given by d d d zzz x y xy=+ . A right circular cone has a circular base radius of 9 cm and a height of 20 cm. Denote • the radius of the circular base of the cone by r,

• the height of the cone by h, • the volume of the cone by V. (a) Find expressions in terms of r and h where appropriate, for V r   and V h   . [2] In one instance, in measuring r and h, a man made a maximum error of 0.01 cm in each of them. (b) Use the above result to estimate the maximum possible error in calculating V, leaving your answer in the form πk where k is to be determined to two decimal places. [3] In another instance, a man used the same faulty instrument to measure r and h and gave an overestimation of each by an amount  cm but maintained that the error in calculating V is no more than π cm3. (c) Determine in fractional form, the largest possible value of  . [3] 5 (a) ( ) 21, π3V V r h r h== . 2 π3V rhr = and 21 π3V rh = . (b) d d d VVV r h rh=+ 221π d π d33 rh r r h=+ from (a) ( )( ) ( ) 221π 9 20 d π 9 d33 120πd 27πd rh rh =+ =+ Since d 0.01 and d 0.01rh , ( ) ( )maxd 120 π 0.01 27π 0.01 1.47πV = + = cm3. (c) We want largest  ( 0  ) such that d πV  . That is