2021 H2 Math RI Prelim Paper 1 Solns

_____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 1 of 25 2021 Year 6 H2 Math Preliminary Paper 1: Solutions with Comments 1 Given the polynomial 42x ax bx c+ ++ has a factor ( ) 2x− and gives remainders 12 and 26 when divided by ( ) 3x− and ( ) 4x− respectively, find the values of a, b and c. [4] Solution Comments [4] ( ) ( ) ( ) ( ) 42f f 2 0 16 4 2 0 4 2 16 ...(1) f 3 12 81 9 3 12 9 3 69 ...(2) f 4 26 256 16 4 26 16 4 230...(3) x x ax bx c a bc a bc a bc a bc a bc a bc =+ ++ = ⇒ + + += ⇒ + += − = ⇒ + + += ⇒ + += − = ⇒ + + += ⇒ + += − Solving (1), (2) and (3), 54, 217, 234abc= −= = − The majority of the students did well for this question. However, there were some students who inefficiently did long division to arrive at the 3 equations, with some making errors along the way. This should not be an unfamiliar question as it makes use of factor and remainder theorems which students are assumed to have prior knowledge of.

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 2 of 25 2 A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is o60 , as shown in the diagram. At time t = 0, the tank is filled with 394 cmπ of water. At this instant, a tap at C is turned on and water begins to flow out at a constant rate of 312 cm sπ − . Denoting h cm as the depth of water at time t s, find the rate of decrease of h when t = 15, leaving your answer in exact form. [4] [The volume V of a cone of vertical height h and base radius r is given by 21 3V rh π= .] Solution Comments [4] Volume of water in the conical container at time t seconds, 21 3V rh π= tan60 3r rhh=⇒= Therefore 3Vh π= 2dd 3dd Vh htt π⇒= When t =15, ( )( )94 2 15 64V ππ π= −= , 3 64 4hhππ = ⇒= ( ) 2d1 2 34d 24 h t ππ∴ = −÷ = − ∴Rate at which h is decreasing at the instant when t =15 is 11 cms .24 − Most students were able to handle the question with ease as it was something familiar that they had done before with some standard steps outlined in their lecture or revision notes. However, there were students who made the following common errors: 1. 3 64 8hh= ⇒= 2. 2 2 d22 d3 3 h hth π π −−= = 3. d 2d V t π= 4. Differentiating V without realising that both h and r are variables:

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 3 of 25 Eg: 21 3V rh π= ⇒ 2d1 d3 V rh π= or d2 d3 V rhr π= 5. Wrote either “ d1 d 24 h t = ” or “Rate of decrease of h 1 24=− ” 3 (a) Find 1tan d .x xx − ∫ [3] (b) (i) Using the substitution 1u x= , or otherwise, find 2 1sin dx x x ⌠ ⌡ . [2] (ii) Given that n is a positive integer, evaluate the integral ( ) 1 2 1 1 1sin d n n x x x π π π + ⌠ ⌡ , giving your answer in the form aπ, where the possible values of a are to be determined. [3] Solutions Comments (a) [3] 22 11 2 2 1 2 2 11 1tan d tan d2 21 11tan 1 d2 21 1tan tan22 xxx xx x xx x xx x x x x xc −− − −−  = −  + = −− + = −− +  ⌠⌡ ⌠⌡ ∫ 2 11 tan22 xx xc−+= −+ Most students were able to obtain the 1st line, but a number had difficulty proceeding on.

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 4 of 25 (b) (i) [2] Given the substitution 2 1 d1, we have d uu x xx= = − ( )1 22 sin 11dd sin sin d cos 1cos x xx x xx uu uc cx = −−   =− = + = +  ⌠⌠ ⌡ ⌡ ∫ OR ( )1 22 sin 11dd sin 1cos x xx x xx cx  = −−   = +  ⌠⌠ ⌡ ⌡ Generally this part was ok. (b) (ii) [3] ( ) ( ) ( ) ( ) ( ) ( ) 11 1 21 1 1 1 sin 1d cos cos cos 1 nn x n n x xx nn ππ π π ππ ππ π + +  =   = −+ ⌠⌡ ( ) ( ) ( ) ( ) ( ) [ ] ( ) 1 1 21 1 1 1 21 1 sinIf is even, d 1 1 2 2 sinIf is odd, d 1 1 2 2 21 n x n n x n n n xa x n xa x OR a π π π π π ππ π ππ + + = −− = = = −− = − = − = − ⌠⌡ ⌠⌡ Quite a number of students wrote down 2a=± .

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 5 of 25 4 (a) Show that ( ) ( ) 33 22 1 2 1 12r r kr k+−−= + , where k is a constant to be determined. Use this result to find 2 1 n r r = ∑ , giving your answer in the form ( )( )12 1pn qn qn++ where p and q are constants to be determined. [5] (b) Raabe’s test states that a series of positive terms of the form 1 r r a ∞ = ∑ converges when 1 lim 1 1 n n n an a→∞ +  −> , and diverges when 1 lim 1 1 n n n an a→∞ +  −< . When 1 lim 1 1 n n n an a→∞ +  −= , the test is inconclusive. Using the test, explain why the series 3 1 1 r r ∞ = ∑ converges. [3] Solutions Comments (a) [5] ( ) ( ) 33 32 32 2 21 21 8 12 6 1 8 12 6 1 24 2 2 rr r rr r rr r k +−−    = + ++− − +−    = + ∴= OR ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 33 22 2 22 2 2 21 21 21 21 21 21 21 21 2 4 4 14 14 4 1 2 12 1 24 2 2 rr rr r r rr rr r rr r r k +−− = + −− +++ − +−  = + + +− +− + = + = + ∴= Very well done.

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 6 of 25 ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 332 11 23 3 11 33 33 33 33 3 24 2 2 1 2 1 24 2 1 3 1 5 3 7 5 ... 21 23 21 21 21 1 nn rr nn rr r rr r nn nn n = = = = += +− − += − +− +− + + −− − + +− − =+− ∑∑ ∑∑ ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 32 1 32 1 2 24 2 2 1 1 1 21 2124 21 21 124 21 2 2224 21 2 2224 1 12 16 1 ,16 n r n r rnn r nn n n n nn n nn nn n pq = = += +− = +− + + = +−  += + += + =++ ∴= = ∑ ∑ Generally well done. Some common mistakes include having cubes of even numbers in the cancellation and wrongly concluding 1 22 n r= =∑ .

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ _____________________________________________________ 2021 Y6 H2 Math Preliminary Paper 1: Solutions with comments Page 7 of 25 (b) [3] ( ) ( ) 3 3 1 3 3 3 32 3 3 2 2 2 1Let . 1 11 1 1 1 1 3 31 3 31 313 n n n a n a nnn a n nn n nnn nn n nn n nn + =    −= −    +  += −   + + +−=  ++= =++ 2 1 31lim 1 lim 3 3 1n nn n an a n n→∞ →∞+  − = ++ =>  . Therefore 3 1 1 r r ∞ = ∑ converges. Many students attempted to show that the expression 1 11n n an a +  −>  and hence the limit is more than 1. However, this is not true in general. Some solutions had the n and r mixed up