RI 2020 Y5 H2 Math Promo (Soln)

RAFFLES INSTITUTION 2020 Year 5 H2 Mathematics Promotion Examination Questions and Solutions with comments Page 1 of 25 1 Mary runs a fruit stall and keeps a record of the daily sales revenue. The weights of fruits and total amount s received from the sales of three particular types of imported fruits in the past three days are shown in the following table. Black seedless grapes (kg) Premium red cherries (kg) Premium black plums (kg) Total amount received from sales in $ Day 1 17.5 8.1 15.5 580.94 Day 2 12.1 7.5 13.1 431.93 Day 3 18.5 9.6 12.6 602.75 On Day 1, all the fruits were sold at their original prices. On Day 2, Mary gave a special one-day discount of 6% for all purchases. On Day 3, Mary increased the price per kg of premium red cherries by $0.40 due to a price increase by the supplier. Find the price per kg of each fruit on Day 1. [4] Solution Comments 1 [4] Let , and xy z be the price per kg of Black seedless grapes, Premium red cherries and Premium black plums on Day 1 respectively, in $. For Day 1, 17.5 8.1 15.5 580.94xy z++ = ---(1) For Day 2, 431.9312.1 7.5 13.1 459.50.94xy z++ = = ---(2) ( )For Day 3, 18.5 9.6 0.4 12.6 602.75 18.5 9.6 12.6 598.91 xy z xy z + ++ = ⇒ ++ = ---(3) Using GC, we get 16.30, 18.90, 9.20xyz= = = This question was answered well by most candidates. However a significant number of responses were characterized by the following errors: (1) Increase the price of all three fruits when question states only the red cherries had a price increase; (2) writing “ ( )9.6 0.4 ,y− 9 0.4,× 9.6 0.4+ ” instead of “ ( )9.6 0.4y+ ”.

2020 Year 5 H2 Mathematics Promotion Examination: Questions and Solutions with comments Page 2 of 25 2 Referred to an origin O, points A and B have position vectors given respectively by 48++i jk and 623++i jk . The point P on AB is such that : 1:AP PB λλ= − . (i) Show that the exact area of triangle OAB is 5 1012 . [1] (ii) Find the exact value of λ for which OP is perpendicular to AB. [3] (iii) Find the exact value of λ for which angles AOP and POB are equal. [3] Solution Comments 2(i) [1] Area of triangle OAB 16 1 422 83= × 4 1 452 22 1 25252 5 101 (shown)2 − = − = = Most candidates used the correct method by finding the vector product of two vectors, but a significant minority made multiplication or sign errors. (ii) [3] ( ) ( ) 1 16 41 2 83 65 22 53 OP OA OBλλ λλ λ λ λ = +−    = +−    − = + +          61 5 24 2 38 5 AB OB OA    =−=−= −    −          For OP to be perpendicular to AB, 0 65 5 22 20 53 5 OP AB λ λ λ ⋅= −     + ⋅− =    +−         5(6 5 ) 2(2 2) 5(5 3) 0λλ λ− − +− += 1154 11 54λλ=⇒= While many candidates wrote down the correct expression for OP   in terms of the parameter λ using Ratio Theorem, some candidates gave unnecessarily long answers by alternatively setting up a vector equation of a line. Those who chose to use this method were often unsuccessful in obtaining the correct answer due to not realizing that the parameter λ should not be used unless it adheres to : 1:AP PB λλ= − . For those candidates who introduced another parameter e.g.µ , many attempts to eventually find λ generated errors.

2020 Year 5 H2 Mathematics Promotion Examination: Questions and Solutions with comments Page 3 of 25 (iii) [3] 1 65 6 65 11 422 22297 853 353 301 266 99 441 400 175 7 16 AOP BOP OA OP OB OP OA OP OB OP λλ λλ λλ λλ λ λ ∠= ∠ ⋅⋅ = −−           ⋅+ = ⋅+          ++      += −+ = =                         This part proved challenging, with a significant number simply concluding that 12λ = with the wrong concept that for the angles to be equal, AP = PB. This is only true if OA = OB. There were also many attempts who put modulus signs across the equation, albeit unnecessarily, and did not account for their removal subsequently.

2020 Year 5 H2 Mathematics Promotion Examination: Questions and Solutions with comments Page 4 of 25 3 Given that 2f ( ) e sinxxx= , show that f() 4 f() 5 f ()x xx′′ ′ = − . [2] Hence find the first three non-zero terms in the Maclaurin series for f( )x . [3] Verify that the same result is obtained if the standard series expansions for 2e and sinx x are used. [2] Solution Comments 3 [2] 2f ( ) e sinxxx= [ ] 22 2 22 f ( ) 2e sin e cos 2f ( ) e cos f ( ) 2f ( ) 2e cos e sin 2 f () 2f () 2 f () f () 4f ( ) 5f ( ) (shown) xx x xx x xx xx xx x x x x xx xx ′ = + = + ′′ ′ = +− ′′=+ −− ′= − Most were able to show the result required. However, majority used inefficient methods and hence wasted time. As with all “show” questions, clear working must be shown. [3] (3)f () 4 f() 5 f ()x xx ′′ ′= − When 20, e 1, sin 0, cos 1xx xx= = = = (3) f(0) 0 f (0) 1 f (0) 4 f (0) 4(4) 5(1) 11 = ′ = ′′ = =−= The Maclaurin series for f( )x is 23 3 2 11f ( ) 4 11 ... 2 ...2! 3! 6xx xxx xx =+ + +=+ + +   Students should differentiate the result shown in the previous part ie to obtain (3)f () x . Students who did not penalized themselves by spending extra time on this. [2] ( ) ( ) 2 23 3 33 2 3 23 3 2 f( ) e sin 221 2 ... ...2! 3! 3! 41 2 2 ... ... 36 2 2 ... 6 112 ... (shown)6 xxx xx xxx xxxx x xxx x xxx =  =++ + + −+      =++ + + −+       = + + −+ = ++ + This was generally well done. Do note that since we are only interested in terms up to 3x , there is no need to spend time evaluating terms involving higher powers of x. f() 4 f() 5 f ()x xx′′ ′ = −

2020 Year 5 H2 Mathematics Promotion Examination: Questions and Solutions with comments Page 5 of 25 4 Do not use a calculator in answering this question. (i) Given that x is real, prove that 2 2xx−+ is always positive. [1] (ii) Solve the inequality 32 2 8 132xx xxx++ ≥−−+ . [4] (iii) Hence solve the inequality 3 2 2 8 132xx xxx ++ ≥−−+ . [2] Solution Comments 4(i) [1] 22 2 2 112 2 22 17 24 xx x x   −+= − −− +     = −+ Since 2 1 02x −≥ x∀∈ , 2 17 024x − +> . So 2 2xx−+ is always positive. Alternative: Discriminant = ( ) 2 1 4(2) 7 0− − = −< Since coefficient of 2x is positive and discriminant is negative, the graph of 2 2yx x= −+ is always above the x-axis. So 2 2xx−+ is always positive. Majority were able to do this. However, students who chose the alternative method often omitted the checking of the coefficient of 2x . (ii) [4] ( ) ( ) ( ) ( ) ( ) ( )( ) 32 2 32 2 32 2 2 32 3 2 2 2 2 2 2 8 132 8 1032 8 1 32 0 32 8 3 2 32 032 5 5 10 0 32 52 012 xx xxx xx xxx xx x x x xx x x x x xx x xx xx xx xx xx ++ ≥−−+ ++ −−≥−+ + +− − − + ≥−+ + +− − + − + − ≥−+ −+ ≥−+ −+ ≥−− Since 2 2xx−+ is always positive (from i), the inequality reduces to This was generally well done. However, when simplifying the inequality, some students did not realize that they have shown that 2 2xx−+ is always positive in (i). Some went on to check that discriminant for 2 20xx−+= is negative and thus has no real roots. They then concluded that ( )( )1 20− −>xx . This argument is flawed.

2020 Year 5 H2 Mathematics Promotion Examination: Questions and Solutions with comments Page 6 of 25 ( )( )1 20− −>xx 1 or 2xx∴< > A number forgot to exclu