ACSI 2022 Y3 Core Mathematics Paper 1 Solution

1. [Maximum mark: 3] Express 2 24 11 2 3 2 2 x x x x +−+ − − as a single fraction in its simplest form. ( )( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )( ) ( )( ) 2 24 2 2 22 11 2 3 2 2 11 23 2 1 1 11 3 1 2 1 1 2 2 3 2 1 3 1 1 2 1 3 1 1 2 1 3 x x x x x xx xx x x x x xx x x x x x x x xx +−+ − − +=− +− +− =− + − + − + − += + + − −= + + − ++ 2. [Maximum mark: 4] Given that 3 2 5 132,3 4 1 2 1 0 a a b b     −=           find the value of a and of .b 2 5 __(1) 3 4 13 __(2) (2) (1) 2 : 3, 1 ab ab ab −= −= −  = =− • Students should not try to combine the fractions without even factorising the denominators. • 2 1x + should be cancelled out immediately to keep the working simple. • Some students forgot about the 2 in the denominator when combining the two fractions. • As the question asked for a fraction in its simplest form, 1x− must be cancelled in the final step. • Some students ended up with 3 2 13.ab−= • Careless mistakes were made when solving. These could be rectified by substituting the answers into the original equations for checking.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 2 3. [Maximum mark: 5] A right cone has base diameter 6 cm and height of 4 cm. (a) Find, in terms of , the total surface area of the cone. [3] (b) A solid hemisphere has the same surface area as the cone. Find the radius of the hemisphere in the form 2a cm, where a is a constant. [2] (a) Slant height 223 4 5= + = cm Total surface area of cone 2(3 ) (3)(5) 24  = + = (b) 23 ( ) 24r = 8 2 2r== cm 2a= 4. [Maximum mark: 5] (a) Solve 5 4127 81 .3 x  = [2] (b) Simplify 4 2log log 64 log 4.aa− [3] (a) 5 4 35 127 813 3 (3 ) 3 35 2 x x x x −  = = −= =− (b) 4 2log log 64 log 4 lg lg 64 4lg 4 lg 3lg 4 4lg 4 1 aa a a − =  − =− =− • The diameter was given in the question, not the radius. Students need to divide by 2 first. • As this is a solid hemisphere, the base area must be included in the calculations. • Many students used the wrong formulas for this question, sometimes even mixing up with volume. • Many misconceptions in the use of indices laws were observed such as: 3 1 23 (3 ) 3 .xx− = • Another common mistake was: 127 9.3 x  = • If students want to use shortcuts like 1log log a b b a= or 2 2log log ,a abb= they must be aware of potential pitfalls. For example, 4 13log 4 . 3log a a • Change of base law must be clearly shown in working. It cannot be taken for granted that 4log 4 log 1.a a=

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 3 5. [Maximum mark: 6] (a) Solve the simultaneous equations 16xy = and 3.yx= [3] (b) On the same axes, sketch the graphs of 16xy = and 3,yx= labelling the axes-intercept(s) and point(s) of intersection clearly. [3] (a) 16y x= Substitute into 3.yx= 316 xx = 4 16x = 2, 8 2, 8 xy xy == =− =− (b) • Remember to solve for y after obtaining the x values. • Many students forgot about . This should be rectified once students sketch the graph in the next part and see 2 intersections. 0y= 0x= 16y x= 3yx= ( )2,8 ( )2, 8−− • Students should take note to indicate the origin, which is both the x and y intercept of 3.yx= • As there are multiple graphs on the same axes, the equation of each graph should be clearly labelled. • 16y x= must be clearly shown to tend to but not touch the asymptotes. Students should also label the equations of the asymptotes.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 4 6. [Maximum mark: 7] (a) Simplify 2 108 5 ,43 1 3 −+ + leaving your answer in the form 3,ab+ where a and b are constants. [4] (a) 2 108 5 43 1 3 −+ + ( ) ( ) 5 1 32 3 6 3 34 1 3 1 3 − = − + +− 2 3 6 3 5 5 3 3 4 2 −= − + − 8 3 18 3 30 30 3 12 − − += 30 20 3 12 −+= 55 323=− + 55,23ab=− = (b) Given that 1 , 3 t = express 1 21 t t − − the form 3,pq+ where p and q are constants. [3] (b) 1 1 3 2 1 3 − − 13 3 23 3 − = − 13 23 −= − 1 3 2 3 2 3 2 3 −+= −+  2 3 3 43 −−= − 13=− − 1, 1pq=− =− • Rationalising the denominators first will make it easier to combine the fractions. • Final answers should be fully simplified. • A common denominator should be obtained within both the numerator and denominator. • Students should be extra careful when negative signs are involved in the working.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 5 7. [Maximum mark: 8] (a) Given that a and b are integers such that 25 a−   and 3 7,b−   find the (i) greatest possible value of 2 2 ,2 ba − [2] (ii) smallest possible value of ( ) 2 1. ab a− [2] (b) Solve the inequalities 4 17 5 12 3 50.x x x−  −  + [4] (a)(i) 24 0 16−= (b)(ii) 2 (2)( 3) 6(2 1) − =−− (c) 4 17 5 12 and 5 12 3 50x x x x−  − −  + 5 and 2 62xx−  -5 31x 8. [Maximum mark: 10] (a) If A is an obtuse angle and 12cos , 13A=− find the value of each of following: (i) tan ,A [2] (ii) sin cos .AA+ [2] (a)(i) tan tan(180 ) 5 12 AA=− − =− (a)(ii) 5sin 13A= 5 12sin cos 13 13 7 13 AA+ = − =− • Take note of the inequality. a cannot be 5. • 0 is considered an integer. • a must be the same number in both the numerator and denominator. • The trick is to make the denominator as small as possible, while keeping the result negative. • Some students were confused when dealing with 5,x− ending with 5.x • Avoid writing 31 5.x By convention, the smaller number appears on the left of an inequality. 5 A 13 12 • A simple diagram will make solving this question easier. • As this is an obtuse angle, students should remember to include the negative sign for tan .A • Since ( ) ( )sin 180 sin ,− = sin A is positive in this case.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 6 (b) The diagram below shows a n 8 metres by 14 metres rectangular assembly area where ,A ,B C and D are points on level ground. Two flagpoles stand at E and ,F such that DE FC= and t he flagpoles stand 2 metres apart from each other . T he height of each flagpole is 7 metres and the top of the flagpole at F is denoted as .G Find (i) the length BG , giving your answer in surd form, [3] (ii) tan , GBF [1] (iii) cos . DFB [2] (b)(i) 14 2 62CF −== 2268 10 BF =+ = 2210 7 149 BG=+ = (b)(ii) 7tan 10GBF = (b)(iii) cos cos(180 ) 6 10 3 5 DFB CFB=− − =− =− 2m F E 14m 8m D B C A • Students are recommended to add on to the given diagram to aid in visualization. Additional diagrams are also useful once the main diagram gets too cluttered. • Students whose answers contain a surd in the fraction, have mistakenly use the wrong side as the opposite/adjacent. • Again, this method is more clearly visible if students use the provided diagram wisely. • It is acceptable, but much more tedious to use cosine rule to solve this question.

ACS(Independent)/Y3IPCoreMathP1/2022/FinalExamination 7 9. [Maximum mark: 15] A company uses the function 2 10 21P n n=− + − to model its profits, P , in thousands of dollars, where n hundred units of Product A are produced and sold. (a) Explain the meaning of " 21"− in the context. [1] Starting cost of $21000 / Loss of $21000 when no goods were produced yet (b) The company makes a profit by producing 400 units of Product A. A director suggests doubling the number of units produced. Explain if this is advisable, justifying your answer clearly. [2] The company makes a profit of $3000 when producing 400 units, but a loss of $5000 when