TMJC H2 Mathematics Prelims Paper 1 (A)

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 1 of 21 2024 H2 MATH (9758/01) JC 2 PRELIMINARY EXAMINATION – SUGGESTED SOLUTIONS Qn Solution 1 Transformation of Curves (i) ( )2 4 92 1 9 24 4 4 xxy x x x −++= = = +− − − where 2a= and 9.b= (ii) Note: For sequence of transformation questions, you MUST describe the transformations (using the keywords) and not just write the replacements Template {delete as appropriate}: Translate ___ units in the {positive / negative} {x-direction / y-direction} x-direction [Replace x by x – (k)] y-direction [Replace y by y – (k)] • Sign of (k) determines positive / negative • Magnitude of (k) determines no. of units of translation Stretch by a factor of k parallel to the {x-axis / y-axis} Parallel to the x-axis Replace by xx k   Parallel to the y-axis Replace by yy k   • k is the stretch factor Reflection in the {x-axis / y-axis} In the y-axis [Replace x by –x] In the x-axis [Replace y by –y] Method 1: Stretch parallel to y-axis ( ) ( ) ( ) ( )1 2 31 1 1 9 9 24 9 4 4 4 yy y y yx x x x x = ⎯⎯ → = ⎯⎯ → =  = ⎯⎯ → = +− − − −  (1) Translate 4 units in the positive x-direction (2) Stretch by a factor of 9 parallel to the y-axis (3) Translate 2 units in the positive y-direction Alternative: In the sequence (2), (3), (1) or (2), (1), (3) Method 2: Stretch parallel to x-axis ( ) ( ) ( ) ( ) ( ) 1 2 31 1 9 9 9 9 224 4 4 9 y y y y y xx x x x x= ⎯⎯ → = = ⎯⎯ → = ⎯⎯ → − =  = +− − − (1) Stretch by a factor of 9 parallel to the x-axis (2) Translate 4 units in the positive x-direction (3) Translate 2 units in the positive y-direction Alternative: In the sequence (1), (3), (2) or (3), (1), (2)

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 2 of 21 Qn Solution 2 Application of Differentiation (Maxima/Minima) 2 Let V be the volume of the cylinder (in 3cm ). By Pythagoras Theorem, 2 22 2 22 2 --- (1)4 h rk hrk  += =− 2 2 2 23 π π ,from (1)4 ππ 4 V r h hkh k h h = =−  =− Differentiate w.r.t h, 22d3 ππd4 V khh =− For maximum V, d 0d V h = ( ) 22 22 2 2 d3 ππ0d4 3 04 4 3 2 0 3 V khh kh kh khh  = − = −= = = Testing V is maximum Method 1: 1st derivative test h 2 3 k −   2 3 k 2 3 k +   d d V h + 0 − Slope  V is maximum when 2 3 kh= . r k ℎ 2 Question asks for ‘exact value of h’ You should keep h and get rid of r so that you can directly solve and answer the question Note: k is a constant (radius of sphere) Reminder: When you square root both sides of the equation, you will have ±, i.e. 2 2 42 3 3 kkhh=  = So you should state why you only want 2 3 kh= from context of the question Question asks for ‘volume of the cylinder is maximum’ Hence the formula you need is the volume of cylinder Steps to solve Maxima/Minima Problems 1. Draw a clear diagram and