YI 2026 H2 Math JC2 Prelim Paper 1 (Solutions with comments)

1 ©YIJC 9758/01/JC2PE/26 1 A function f is defined by ( ) 32f. x ax bx cx d= + + + The graph of f ( )yx= has turning points at ( )1, 4− and 1 11,24 − . Find the values of a, b, c, and d. [4] 1 f(x) = ax3 + bx2 + cx + d passes through the points ( )1, 4− and 1 11,24 − . At ( )1, 4 , 4 (1)a b c d− − + − + =  At 1 11 11, , (2)2 4 8 4 2 4 a b c d − + + + =−  At the turning points, 2f ( ) 3 2 f ( 1) 3 2 0 (3) 13f 0 (4)24 x ax bx c a b c a bc  = + +  − = − + =  = + + =   From GC, 4, 3, 6, 1a b c d= = =− =− Generally well done. 1. Wrong approach: 1a. Some mistakes were in writing 21f ( ) ( 1)( ) 3 2 2x x x ax bx c = + − = + + where the roots are correct but the coeff of a would be wrong. 1b. Those who wrote 21f ( ) ( 1)( ) 3 2 2x k x x ax bx c = + − = + + obtained something which ended up being not very useful anyway. Either way, not a good approach. 2. Did not form equations/use GC 2a. Few students couldn’t form 4 equations, getting 2 only. 2b. After forming 4 equations, some did not recognise this is a system of 4 linear equations (SLE) that can be solved using a GC (Polysmlt2). 2 A geometric series has first term a and common ratio r. The sum to infinity of the series is 16 and its fourth term is 3 16 . (a) Show that r satisfies the equation 43256 256 3 0rr − + = . Given that 0.5r  , solve this equation and hence show that 12a= . [5] 2(a) ( ) ( ) 3 3 3 43 43 16 --- (1)1 3 ---- (2)16 3 316(2) (1) : 1 16 256 256 1 3 256 256 3 0 256 256 3 0 (shown) Using GC, 0.25 or 0.988 (rejected as 0 .5 ) a r ar rr rr rr rr rr =− =  − = = −= − + − = − + = = Some students wrote (1 )S = 16 1 nra r − =− , without making sense that n, number of terms is infinite. Did not know that they could write S = 161 a r =− directly. Students need to show all answers for r and reject with reason. This is not a complex number question. So, complex solutions can be omitted. But if written, need to write/reject correctly:

2 ©YIJC 9758/01/JC2PE/26 Sub 0.25 into equation (1) or (2) e.g. 161 0.25 12 (shown) r a a = =− = 0.118 0.183 (rejected, complex root) ri= −  Some students wrongly wrote 0.118 0.183r =− + or 0.118r =− , did not read complex number answers in GC correctly. (b) Find the exact value of the sum to infinity of the even -numbered terms of the geometric series. [2] 2(b) ( ) ( ) 2 Sum to infinity of the even numbered terms 12 0.25 1 0.25 3.2 = − = Students who listed the even numbered terms appear to have a better chance of getting this right. 3 It is given that 2 for 1 1,f () for 1 3, 10 94 2 x xx x  −  =  −  and that ( ) ( )f f 4xx=+ for all real values of x. (a) Evaluate f ( 49)− [1] 3(a) 2 f ( 49) f ( 49 4 12) f ( 1) 10 9 4( 1) 2 − = − +  =− = −− = • Evaluating ( )f 49− means the answer should be a value rather than left in terms of the function f. • Note that x = –49 is outside the domain of the expressions given for ( )f x by the question. Hence you cannot just substitute and find ( ) 2 10 9 4 49−− . • ( ) ( )f f 4xx =+ means that the function is periodic with period 4. Use this to obtain ( ) ( ) ( ) ( )f 49 f 49 4 f 49 4 12 f 1− = − + = = − +  = −  where –1 is within the domain of the given expression for ( )f x . Hence evaluate ( )f1− to obtain the answer.

3 ©YIJC 9758/01/JC2PE/26 (b) Sketch the graph of ( )fyx= for 27 x−   . [3] 3(b) Given: 2 for 1 1,f () for 1 3, 10 94 2 x xx x  −  =  −  ( ) ( )f f 4xx=+ • Note that the graph of ( )fyx= is required for –2 ≤ x ≤ 7 and not just for –1 ≤ x < 3. In this case, use the periodicity of the function to sketch the graph as required. • Since the function is periodic with period 4, ‘copy and paste’ the graph for –1 ≤ x < 3 (which has ‘width’ 4) to 3 ≤ x < 7. Realise that the ( )f2 x = for –2 ≤ x < –1 such that the graph would be a horizontal line here. • Note that the requirement for –2 ≤ x ≤ 7 means that x = 7 must be included for the sketch graph. Consider that ( ) ( ) ( )f 7 f 7 4 2 f 1 2= −  = − = . Hence this endpoint must be closed and not open. • There are no open endpoints in the sketch graph. For instance, consider x = 1; although the expression 2 10 94 y x = − should not be used, realise that as x → 1, y → 2, and at x = 1 itself, y = 2 exactly since x = 1 is in the domain such that ( )f2 x = . This means that there are no gaps or ‘overlaps’ of closed and open circles representing this and other such points where the graphs are ‘joined’. • Be alert to the number of marks available. It would not be reasonable for 3 marks to be awarded where the coordinates of only the two i.e. left- and right-most endpoints are given. The coordinates of all 6 endpoints must be given for full credit. In particular, the y- coordinate(s) must not be excluded. (c) Find the exact value of ( ) 6 4 fd xx . [3] 3(c) ( ) ( ) ( ) ( ) ( ) 16 24 0 1 22 0 1 0 10 d 94 102 d 32 1 1 3 22 10 ln 2(3) 2 3 2 102 ln 5 ln112 52 ln 5 f1 6 d2 xx x x x x x x − = −  += +   −  =+  = + − + + =     • You should use the part (b) sketch graph to visualise the definite integral as the area under the graph between x = 4 and x = 6. Realise that this area will be composite, including an area under the curve and the area of a rectangle. • Note that the limits for the area under the curve, specifically x = 4 and x = 5, are outside the domain of the expression for the curve by the question. This means that you cannot just evaluate 2 5 4 d10 94 x x −  . The periodicity of the function (with period 4) leads to x y O

4 ©YIJC 9758/01/JC2PE/26 Alternative ( ) ( ) ( ) 2 16 4 0 2 1 1 0 0 10 94 1 94 31 2ln 332 22 102 ln 5 ln112 52 ln 5 f d 1 2 d 102d 4 102 6 4 x x x x x x x x − −  +  − +  =+ + = = − + = + =      ( ) ( ) 0 2 151 4 0 fd 1f 0d 9 d 4 x x x x x x == −  to be evaluated, using integration techniques since the question specified for an exact value. • Realise that this can be integrated using the integral for 22 1 ax − provided in MF27. However, note that ( ) 2 2x rather than just 2 x is involved. This means there must be another factor of 1 2 to the integral and not just ( ) 11 2 2 3a = . In addition, the expression that logarithm is taken of must be in modulus, since there is no a priori reason for it to be always non-negative. (NB: The integral provided in MF27 does not involve modulus since the domain is explicitly restricted therein.) Modification: Change f(x) to the following and solve (d) and (e). 2 for 1 1,f () for 1 3, 10 94 3 x xx xx  −  =  − −  (d) Sketch the graph of ( )fyx= for 27 x−   . 3(d) • • • • • Label 8 pairs of coordinates (can also use dotted lines) with open circle at appropriate points x y O

5 ©YIJC 9758/01/JC2PE/26 (e) Find the exact value of ( ) 6 4 fd xx . 3(e) ( ) ( ) ( ) ( ) ( ) 16 24 0 1 22 0 1 0 10 d 94 3 10 d2 32 3 1 1 3 210 ln2 2(3) 2 3 2 3 10 ln 5 ln12 12 35 ln 5 1 2 1f d (2 1) 6 2 xx x x x xx x − = −  += +   − = + − = = + + + +     Area of trapezium: ( )( ) 2 1 1 1 1 2 or 3 d2 xx+−  4 (a) Find the exact solutions of the equation 2 6 2 7xx− = + . [4] 4(a) ( ) ( ) ( )( ) ( ) 2 2 2 6 2 7 2 13 0 2 2 4 1 13 21 xx xx x − = + − − = − −  − − −= or ( ) 2 1 14 0x− − = 1 14x= ( ) 2 6 2 7xx− =− + ( ) 2 2 2 1 0 10 1 xx x x + + = += =− • Recall the definition of the modulus function, where ( ) 2 2 2 6 for some values of 6 6 for another set of values of xx x xx − −= −−    For the purpose of solving this equation, these respective values of x are not essential, since the roots of the equation would naturally lie within