TKGS 2024 S4 Math Prelim P2 Solutions

2024 TKGS PRELIM MATH P2 MARK SCHEME Qn Solution Content/Success Criteria 1 (a) 3 2 1 (2 1)73 x x− =− 3(3 2 ) 7(2 1)xx− = − 9 6 14 7xx− = − 20 16x= 40.8 / 5x = I can multiply LCM to remove the denominators on both sides of a linear equation involving fractions. I can solve a linear equation. Content L Complexity L Context L Response Strategy Simple Assessment Objective AO1 (b) (i) 22 2 25 9 15 10 6 9 ab a ab b b − − + − 22(5 ) (3 ) 5 (3 2 ) 3 (2 3) ab a b b b −= − + − (5 3 )(5 3 ) 5 (3 2 ) 3 (3 2 ) a b a b a b b b −+= − − − (5 3 )(5 3 ) (5 3 )(3 2 ) a b a b a b b −+= −− 53 32 ab b += − I can factorise algebraic expressions by identity and grouping. Content L Complexity L Context L Response Strategy Simple Assessment Objective AO1 (ii) 3 2 3 4 12 cc ab ab 2 3 3 12 4 c a b ab c= 2 3 36 4 a bc abc= 2 9a c= I can take reciprocal to convert division to multiplication. I can apply laws of indices to simplify algebraic expressions. Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1

4052/S4Prelim/P2/2024 Qn Solution Content/Success Criteria (c) (i) 2 790 2xx− + − = 2 790 2xx− + = 22 9 7 9 02 2 2x   − + − − =       2 9 67 024x − − = 2 9 67 24x −= 67 9 42x =+ 8.59268 or 0.407324x= 8.59 or 0.41x= I can complete the square. I can solve equation using complete the square method Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1 (ii) 22 7 679 ( 4.5) 16.75 /24x x x− + − = − − + Since the maximum value of 2 79 2y x x= − + − is 16.75/ 67 4 < 18. There is no solution when 18.y = I know that 2 79 2y x x= − + − has max. value. I can find max. value with complete square form of quadratic equation. Content L Complexity M Context M Response Strategy Routine Assessment Objective AO3 OR 2 2 79 18 2 4390 2 xx xx − + − = − + − = Discriminant 2 43(9) 4( 1) 2 = − − −  5=− Since discriminant < 0, 2 79 18 2xx− + − = has no solution.

Qn Solution Content/Success Criteria 2 (a) P(man who passed test) 1 3 3 7 3 21 ==  the number of men who passed the test 3= Let the number of women who failed the test be x P(both women failed the test) 11 21 20 10 xx −=  = 2 420 10xx−= 2 42 0xx− − = ( 7)( 6) 0xx− + = 7 or 6 (rejected)x=− Women who passed the test 21 9 7 3= − − − 2= Passed Failed Men 3 9 Women 2 7 I can find the probability of a single event. I can use multiplication of probabilities for simple combined events. Content M Complexity M Context M Response Strategy Unfamiliar Assessment Objective AO2 Qn Solution Content/Success Criteria 2 (b) (i) A set of students who only listen to pop music but not classical music. I can represent elements of set notation in words. I can find the number of elements in any sets. (ii) 10 Content L (iii) 4 Complexity M (iv) 9 Context M Response Strategy Simple Assessment Objective AO2

4052/S4Prelim/P2/2024 17.5 20 –5 x 0 y 7.5 5 15 10 2.5 12.5 x 5 10 15 20 25 30 x x x x Qn Solution Content/Success Criteria 3 (a) 2− I can find the value by substitution. (b) See graph below I can draw graph that passes through all points plotted smoothly. Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1 x x (10, 2.5) (0, –5 ) y = 7 y = x – 2 202 15yx x= − +

Qn Solution Content/ Success Criteria (c) 202 22x x+ 202 15 22 15x x+ −  − 202 15 7x x+ −  Draw a line of 7y = on the graph 1 10x   I can solve inequality using graph. Content M Complexity M Context L Response Strategy unfamiliar Assessment Objective AO2 (d) Draw a tangent at 4x = 2.5 ( 5)gradient = 10 0 0.75 ( )accept 0.2 −− − = I know how to find tangent on a graph. Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1 (e) 2 13 20 0xx− + = 2 13 20 0xx x x x x− + = 2013 0x x− + = 202 15 2xx x− + = − Draw 2yx=− on the graph 1.783 or 11.217x= (accept 0.15 ) I can solve quadratic equation using graph. Content L Complexity M Context M Response Strategy Routine Assessment Objective AO2

4052/S4Prelim/P2/2024 Qn Solution Content/Success Criteria 4 (a) Using sine rule, 7.5 sin 40sin 95ST = 4.83932= 1 4SX ST= 1.209831= 1.21 m= I can use sine rule to find unknown length Content L Complexity L Context L Response Strategy simple Assessment Objective AO1 (b) 180 40 95RSX = − − (angle sum of triangle) 45= Using cosine rule, 221.209831 7.5 2(1.209831)(7.5)cos 45XR= + − 6.69937= 6.70 m= I can apply cosine rule to find unknown length Content L Complexity L Context L Response Strategy simple Assessment Objective AO1 OR Using sine rule, 7.5 sin 45 sin 95TR = 5.3236 m= 3 4.839324XT = 3.62949 m= Using cosine rule, 223.62949 5.3236 2(3.62949)(5.3236)cos95XR= + − 6.69937= 6.70 m= (c) By Pythagoras’ Theorem, 222.5 4.8393PT =+ 5.4469= 5.45 m= I can use Pythagoras’ Theorem to find unknown length Content L Complexity L Context L Response Strategy simple Assessment Objective A01 P S T 2.5 4.8393 Essential step Essential step Essential step Essential step

Qn Solution Content/Success Criteria (d) Let the point directly below the bird be B. 90SBX= for XB to be the shortest distance sin 45 1.209831XB= 0.8554797= m 2.5tan 0.8554797 = 1 2.5tan 0.8554797 − =  71.10947= 71.1 (1 d.p)= I know that shortest distance gives largest angle of elevation I can use trigo ratio to find unknown angle Content L Complexity M Context M Response Strategy taught Assessment Objective A02 OR 11 7.5 1.20983 7.5 sin 4522 h  =    0.85547h= 1 2.5tan 0.8554797 − =  71.10947= 71.1 (1 d.p)= S B X 1.209831 45o Bird B X 2.5 Essential step Essential step

4052/S4Prelim/P2/2024 Qn Solution Content/Success Criteria 5 (a) (i) ABD ACD = ( in same segment)s 30= I can apply circle properties to find angle (ii) ACB OBC = (base of isosceles )s 40= 40 30BCD = + 70= 180 70BAD = − ( in opposite segments)s 110= I can apply circle properties to find angle Content L Complexity L Context L Response Strategy taught Assessment Objective AO1 (iii) Let r be the radius of the circle, 140DOB= ( at centre 2 at circumferene) =  140 2 11360 r= 11 7 218 r  =  4.5= I can find radius given angle Content L Complexity L Context M Response Strategy taught Assessment Objective AO2 OR Let r be the radius of the circle, 140DOB= ( at centre 2 at circumferene) =  11r = 11 140 180 r =  4.5= (b) (i) 70BTC BCD = = (found in (a(ii)) 90OBT= (tan radius)⊥ 90 40CBT = − 50= 180 (40 2)BOC = −  (base s of isosceles ) 100= 100 2BDC= ( at centre 2 at circumference) =  50= 50CBT BDC = = is similar to by AA similary test CTB BCD  I can apply circle properties to find angle to prove similarity Content L Complexity M Context M Response Strategy taught Assessment Objective AO2

Qn Solution Content/Success Criteria