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 define all variables, where necessary. 2. Form equation(s) relating the variables. 3. Express quantity to be maximized/minimized in terms of a single variable, say x (if there are 2 variables, express one in terms of another). 4. Differentiate w.r.t. x and equate to 0 to find the stationary point(s). 5. Use 1st or 2nd derivative test to determine/prove nature of the stationary point. Remember to verify using the 1st or 2nd derivative test that the value of h gives the maximum volume For 1st derivative test, you need to state clearly the value of h For 1st derivative test, you need to state the signs of d d V h and corresponding slope

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 3 of 21 Method 2: 2nd derivative test Differentiate w.r.t h, 2 2 d3 π d2 V hh =− When 2 3 kh= , ( ) 2 2 d3 π2 3π 0 0d2 3 Vk kkh =− =−    V is maximum when 2 3 kh= . For 2nd derivative test, you need to (1) find the second derivative, 2 2 d d V h (2) explain clearly why it is < 0 for the value of h (3) conclude that the volume is maximum

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 4 of 21 Qn Solution 3 Techniques of Integration (i) Using ( ) ( )f f 3xx=+ , we have ( ) ( ) ( ) ( )πf 7 f 4 f 1 2sin 2 or 1.41 3s.f.4 = = = =  (ii) Step 1: Use of GC to sketch the graph from 0 to 3xx== using given equation. Step 2: Since ( ) ( )f f 3xx=+ for all real values of x, the curve is a periodic curve with a period of 3, that is, the graph repeats itself every 3 units. Thus, sketch the curve from 4 to 7xx=− = using this property. ( ) ( ) ( ) 5 2 0 2 22 0 2 0 2 0 Volume πd π12π 4sin d π 2 143 π44π 1 cos d π23 2 π44π sin π 23 44π 2 0 0 π3 28 π3 yx x x x x xx  = =+  = − +   = − +   = − − + =    y x 2 O 2 5 3 6 y x 2 O 2 5 3 6 Label the axial intercepts and end points clearly on the diagram. Note: The equation of the curve from 0 to 2xx== is given as π2sin . 4 xy =  The equation of the curve from 3 to 5xx== is NOT π2sin . 4 xy =  Observe that the volume generated between 3 to 5xx== is identical as the volume generated between 0 to 2xx== . Thus, the volume generated between 0 to 2xx== and 3 to 5xx== is given by 2 2 0 π2π 4sin d4 x x  . Rotating the triangle about the x-axis gives a cone. Thus volume is found using 21 π.3 rh r =2 h = 1

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 5 of 21 Qn Solution 4 Maclaurin Series (i) Using cosine rule, 2 2 2 2 π2 2 cos 3r r r = + − −  22 π4 2 2 cos 3rr = − −  ( ) 2 4 π2 1 cos 3 4 ππ2 1 cos cos sin sin33 4 132 1 cos sin22 4 shown 2 3 sin cos r     =  −−   =  −+  =  −+   = −− (ii) 2 4 2 3 sin cos r  = −− 2 4 12 3 1 2  − − −  2 4 113 2 = −+ 11 2222112 1 3 2 1 3 22r     −−     − + = + − +         2 22 13 1 1 1 222 1 3 3 ...2 2 2! 2       −−          = + − − + + − + +          223 1 92 1 ... 2 4 8  = + − + + 2372 1 ... 28= + + + 2723 4 + + 3a= and 7 4b= Use MF26: “ is a sufficiently small angle” means use small angle approximation Use MF26: Use MF26: Answer question Formula NOT given in MF26: 1) Cosine Rule: 2) Sine Rule:

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 6 of 21 Qn Solution 5 AP and GP (i) Method 1 10br a d=+ ----- (1) 2 2br a d=+ ----- (2) 3br a= ------ (3) 2 2 10 a a dr a d a d +== ++ 2( 10 ) ( 2 )a a d a d+ = + 2 2 210 4 4a ad a ad d+ = + + 264ad d= 2 3ad= since 0d  Hence, 2 13 224 23 dar ad dd = = =+ + Since 1| | 1 4r = , hence the series is convergent. Method 2 10br a d=+ ----- (1) 2 2br a d=+ ----- (2) 3br a= ------ (3) Eqn (2) – (1) 2 8br br d− =− -----(4) Eqn (3) – (2) 32 2br br d− =− -----(5) (5) (4) 32 2 2 8 br br d br br d −− =−− 32 2 1 4 rr rr − =− 2(1 ) 1 (1 ) 4 rr rr − =− Since 01dr   , 1 4r = Since 1| | 1 4r = , hence the series is convergent. fourth term of third term of 2 third term of 2 second term of 10 Gar G a d G a dr G a d == + +== + To prove geometric series G is convergent, show | | 1r  . Need to reject 1r = .

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/01)/Math Dept Page 7 of 21 (ii) 2 2 2 2 4: , , ,.....H b b r b r Sum to infinity of H = 22 2 2 9 118 1 9 bb br ==− − Sum to infinity of G = 3 112 1 3 bb br ==− − ( )( ) 2 2 9 3 3 8 2 2 3 4 4 0 3 2 2 0 bb bb bb − − −  + −  2 3