2024 TJC JC2 Prelim Exam H2 Maths Paper 2 (Solutions) For Studens (1)

1 2024 TJC Preliminary Exam H2 Mathematics Paper 2 (Suggested solutions) Section A: Pure Mathematics [40 marks] 1 A curve C has equation 3 1 xy x −= − . (a) Sketch C, stating the equations of the asymptotes. [2] (b) Find the exact volume of the solid generated when the region bounded by C, the x-axis and the line x = 9 is rotated through 2 radians about the x-axis. Leave your answer in the form ( )ln2ab + , where a and b are constants to be determined. [5] [Solutions] Remarks 1(a) 32 111 xy xx −= =− +−− Use dashed lines for asymptotes. When sketching additional lines to solve other parts of the problem, it is recommended that a side sketch be redrawn to avoid damaging the solution. 1(b) Required volume ( ) ( ) ( )( ) ( )( ) ( ) 2 2 9 3 9 3 9 23 91 3 11 3 3 d1 21d 1 441 + d 1 1 4ln 1 4 1 9 4ln 9 1 4 9 1 3 4ln 2 4 2 15 4ln 2 4ln 22 15 8ln 22 x xx xx xx x x x x        − −− −=  − = − + − =− − − = − − − − = − − − − − − − = − + =−     An additional diagram drawn by the side is useful in helping identify the solid of revolution. It is more efficient to perform a long division first then squaring. Squaring first will result in a more complicated algebraic fraction that would still have to be dealt with by long division (or further splitting), which would increase the complexity of the working. x y 3 −3

2 2 (a) Find the series expansion of ( ) 1 38 cos 2 x x + in ascending powers of x up to and including the term in 2x . [5] (b) Find the range of validity of x for the expansion to be valid. [2] [Solutions] Remarks (a) ( ) ( ) ( ) ( ) ( ) 11 1 33 3 2 2 2 12 2 2 2 2 2 12 33 818 8 cos 2 cos 2 121 3 2! 1 2 2 1 1 224 576 2 1 1 224 576 2 1 2 24 576 1 11512 1 8 8 2 2 2 8 8 x x xx x xx x xx x xx x xx x x − ++ =  −   + + +       = −+ = + − + − + = + − + + + = + + − + = + + + • Do not differentiate and use Maclaurin’s theorem. • Use binomial expansion/standard series in MF26: ( ) ( ) 211 1 , 1 2! n x x x n xnn −+ = + + +  21cos 1 2x x= − + , all x • Expand until x2 term. Do not waste time expanding more than what is required. • Rewrite ( ) 2 1 12 x− as ( ) 1212 x − − so that we can further expand it using binomial expansion again (b) For ( ) 1212 x − − : 2 2 1x  1 2 x  11 22 x−   For 1 3 1 8 x− : 18 x   88 x−   Taking intersection, the range of validity is 11 22 x−   Note that there are two main binomial expansions, 1 3 1 8 x+ and ( ) 1212 x − − . We need to find the validity range for each expansion and then find the range of values which satisfy both validity range by taking intersection.

3 3 A sequence  na is defined by 0 2a = and 2 1 21 where 1.33 n nna a n − − = −   (a) By considering ( )1 1 N nn n aa − = − , find an expression for Na . [4] (b) Hence explain whether the sequence is convergent. [1] [Solutions] Remarks 3(a) 2 1 21 where 133 n nna a n − − − =−   ( ) 2 1 11 21 33 nNN nn nn aa − − ==    − = −        ( )1 1 10 21 32 21 12 1 0 LHS 2 N nn n N NN N NN N N aa aa aa aa aa aa aa aa a − = −− −− − =− =− +− +− +− +− +− =− =−  2 2 1 1 0 1 1 21RHS 33 2 1 1 1 ... 3 3 3 3 11 1332 13 1 3 1 1 3 1 3 3 nN n N N N − = − − −    =−        = − + + + +       −       =− −  =− −   1Thus 2 3 3 3 1 3 1 3 N N N N a a − = −  = −  - This is an equation. So when we take sum, we take sum ON BOTH SIDES of the equation. - For summation, if you cannot observe what sum is that, it is always good to write out few terms to see. - LHS: observe is sum of DIFFRENCE of 2 SIMILAR TERMS, so use MOD. - In MOD presentation, it is a MUST to show the cancellation pattern. - Pay attention to the SUBSCRIPT use. - RHS: Can write out a few terms to see as well to observe is sum of GP to FINITE term. - Sum of GP, learn to count the number of terms correctly. For summation number of terms can be obtained by “upper − lower limit + 1” if you did not remove any of the terms. Pay attention to the use of capital letter vs small letter again. - Simplify all answers.

4 3(b) 1As , 0 3 N N → → . Hence 1Na →− , a finite value. Thus the sequence is converges to −1. - Check the concept of converge and watch necessary presentation. - Note the difference of SEQUENCE converge vs SERIES converge.

5 4 The line 1l passes through the point A with coordinates ( )1,0,4 and is perpendicular to the plane 1 with equation 2 4 3x y z− + =− . (a) Find the coordinates of the point B where 1l meets 1 . [4] (b) Verify that the point C with coordinates ( )5,9, 1− lies on 1 . [1] (c) Find a vector equation of the line 2l which is a reflection of the line AC in 1 . [3] (d) Find a vector equation, in scalar product form, of the plane 2 which contains 1l and 2l . [3] [Solutions] Remarks 4(a) B is the foot of perpendicular from A to 1 . B lies on 1l : 12 01 44 OB         = + −          for some  B lies on 1 : 2 13 4 OB  − =−  Thus 1 2 2 0 1 1 3 4 4 4             + − − =−                  ( ) ( )2 16 4 1 16 3+ + + + =−  21 21 =−  1 =− Thus 1 2 1 0 1 1 4 4 0 OB −           = − − =                i.e. coordinates of B are ( )1,1,0− Take note the the direction of l1 is indirectly given in the normal to 1 . Notations: Remember to mention that  , and write “ r= ” when writing vector equations. Pay attention to details: Coordinates are required. 4(b) 2 5 2 1 9 1 10 9 4 3 4 1 4 OC            − = − = − − =−          −       Thus C lies on 1 . (Verified) Do not just write the calculation; include some statements. A(1,0,4) B

6 4(c) Let A be the point of reflection of A in 1 . Using ratio theorem, 2 OA OAOB +=  2OA OB OA=− 1 1 3 2 1 0 2 0 4 4 −−           = − =           −      The line of reflection 2l passes through C and A . 3 5 8 2 9 7 4 1 3 AC −−            = − = −          − − −      Thus equation of 2l is 58 97 13 r         =+      −    ,  . Draw a good diagram. Think carefully about the information provided; B is the midpoint of A and A, not C. Other variations include using 2 CA CACB += and so on, but these require more calculation. 4(d) 2 contains 1l and 2l A normal to 2 is 8 2 31 7 1 26 3 4 22             − = −           −      Equation of 2 is 31 31 1 31 26 26 0 26 22 22 4 22 r OA                − = − = −              − − −           i.e., 31 26 57 22 r  − =−−  Alternative solution 2 contains 1l and 2l  2 also contains line AC A normal to 2 is 2 4 2 31 1 9 1 26 4 5 4 22 AC                 − =  − = −              −−        Equation of 2 is 31 31 5 31 26 26 9 26 22 22 1 22 r OC                − = − = −              − − − −           i.e., 31 26 57 22 r  − =−−  Refer to your diagram. There are multiple ways to obtain this answer, even without solving part (