2016 H1 A Level Mathematics Solution edited 2020 (SAJC)

1 2016 H1 A Level Mathematics Solution Section A Q1. (i)         2 22 d 2 122ln 3 4 6d 3 4 3 4 xxxx xx     (i)         3 23 d 1 1 3 2 3 1 3d2 2 1 3 1 3 xx xx        Q2. Substituting xue into inequality 22 9 3xxee  22 9 3xxee  becomes 22 9 3uu    22 3 9 0 3 2 3 0 uu uu       33 or 2 33 or e 2 xx uu e     Since 0 xe  for x , 3 xe  is rejected. Hence, 3 2 3ln 2 xe x    Q3. (i) y x (0,1) (0.703,0) -3 3/2

2 (ii) From GC, gradient of C when x = 0.5 is -1.61 ( 3 sig. fig. ) (iii) OUT of SYLLABUS when x = 0.5, gradient of the normal to C at the point = 1 0.6224591994 0.622 (3 sig.fig.)1.606531    At point (0.5, 0.3565307) Equation of the normal C when 0.5x  0.35653 0.62246 0.5 0.622 0.0453 (3 s.f) yx yx     (iv)   2 0 k xe x dx  3 0 3 0 3 3 0 33 1 3 k x k k xe kee ke                  Q4. (i) 23 2 1 6 3 4 d 6 6 12d y x x x y xxx        For stationary point, 2 d 0d 6 6 12 0 1 or 0.5 y x xx xx       When 1 , 4xy  When 0.5 , 2.75xy Coordinates of stationary points on the curve are ( -1, -4) and (0.5, 2.75). (ii) 2 2 d 6 24d y xx   For 1x

3   2 2 d 6 24 1 18d y x     ( > 0) (-1, -4) is a minimum point. For 0.5x   2 2 d 6 24 0.5 18d y x    ( < 0 ) (0.5, 2.75) is a maximum point. (iii) The coordinates of the points where the curve crosses the x-axis are ( -1.59, 0 ) , ( -0.157, 0) and (1, 0). (iv)

4 Numerical value of the area under C between x = 0.5 and x =1 1 2 3 2 0.5 1 6 3 4 d 0.9375 unitx x x x     Q5. Since DEF and is an equilateral triangle, 60oEFD Area of the remaining shape ABEDFCA = Area of triangle ABC – Area of triangle EDF       22 22 112 sin 60 sin 60 properties of equilateral triang les22 33 4 xy xy     Alternative Height of Triangle ABC = 22(2 ) 3x x x [Pythagoras Theorem] Height of Triangle EDF= 2 2 3() 22 yyy  Area of the remaining shape ABEDFCA = Area of triangle ABC – Area of triangle EDF   22 1 1 3(2 )( 3 ) ( )2 2 2 33 4 yx x y xy   Given that Area of the remaining shape ABEDFCA = 23 cm2. cos 60 / 2yy  2 cos60xx  Area of triangle = ½ absinC Area of triangle = ½ (base)(height)

5   22 22 22 33 2 3 4 1 24 4 8 shown [Equation 1] xy xy xy    (ii) Given that perimeter of the shape ABEDFCA = 10 cm    2 2 2 2 10 4 2 2 10 6 10 x y x y x y x y xy          Substitute 10 6yx into Equation 1     22 22 2 2 4 10 6 8 4 100 120 36 8 32 120 108 0 32 120 108 0 1.5 or 2.25 xx x x x xx xx xx                For 1.5 , 1xy For 2.25 , 3.5 (reject as length cannot be less than 0 )x y y  The values of x and y are 1.5 and 1xy Section B Q6. (i) OUT of SYLLABUS Select the number of male and female students according to the table below. Gender Sample size Male 1260 80 422400 Female 1140 80 382400 Within each category of students, use simple random sampling. For example in the category of male students, number the male students from 0001 to 1260, using 4-digit random numbers to generate a sample of 42 students.

6 (ii) OUT of SYLLABUS Stratified sampling method gives a sample that is more representative of the students expenditure on music annually compared to simple random sampling method. (iii) 312Unbiased estimate for population mean, 3 .9 80 x   2 2 1Unbiased estimate for population variance, 79 3121328 80 1611 1.41 (3 s.f)395 s     

7 A B 0.05 0.20 0.55 0.20 A Q7. (i) (iia) P (at least one of A or B occurs) =        P A B P A P B P A B 0.6 0.25 0.05 0.8          (iib) P (exactly one of A or B occurs) = 0.55 + 0.20 = 0.75 (iii)       ' ' ' P A B 0.55 11P A|B 1 0.25 15PB      Q8. (i) P(both balls are red) = 1 3 2 1 2 10 9 30   (ii) P(both balls are different) (2 diff coloured Balls are fm Box A)+ ( 2 diff coloured Balls are fm Box B) = P(Box A & BR +BG+ RG + GR+GB+RB) + P(Box B & BG +GB) 1 5 3 5 2 3 2 3 2 2 5 3 5 1 4 2 2 4 2 10 9 10 9 10 9 10 9 10 9 10 9 2 6 5 6 5 PP                   11 18    Alternative Method P(both balls are different) = 1 – P(both balls are same colour) = 1 – P(Both are Red) – P(Blue balls fm Box A) – P(Green balls fm Box A) – P(Blue balls fm Box B) – P(Green balls fm Box B) 1 1 5 4 1 2 1 1 4 3 1 2 11 30 2 10 9 2 10 9 2 6 5 2 6 5 11 18                                     (iii) P(both balls are red given that that they are the same different)     1 P both balls are red 330 11P both balls are the same colour 35 1 18    

8 Q9. Let X be the r.v. number of batteries that have a lifetime of less than two years out of 8 batteries.  ~ 8,0.6XB (ia) Required probability =  8 0.01679616 0.0168PX    (ib) Required probability =      4 1 4 1 3 0.8263296 0.826P X P X P X         Let Y be the r.v. number of packs of batteries that have at least half of the batteries have a lifetime of less than two years out of 4 packs of batteries.  ~ 4,0.8263296YB (ii) Required probability =  2 0.1417924285 0.142PY    (iii) OUT Of SYLLABUS Let T be the r.v. number of batteries that have a lifetime of less than two years out of 80 batteries.  ~ 80,0.6TB Since n = 80 is large, np = 80(0.6) = 48 > 5, nq = 80(0.4) = 32 > 5,  ~ 48,19.2TN approximately E(T) = 48, Var(T) = 19.2 (iii) Required probability =    40 39.5 0.974ccP T P T     Q10 . (i) Let X be the random variable denoting the mean top speed of cheetahs in km/h and  be the population mean top speed of cheetahs. H0 : = 95 H1 : 95  Under H0, 2_ 4.1~ (95, ) 40XN . Using a two tailed z-test at 5% level, 96.3x  gives 2.01calz  and p value =0.0449 < 0.05 We reject Ho and conclude that there is sufficient evidence at 5% level of significance that the mean top speed of cheetahs is not 95 km/h. (ii) H0 : = 95 H1 : 95  Under H0, 2_ 4.1~ (95, ) 40XN .  2 ~ 0,1XZN n    Using a one tailed z-test at 5% level. Critical value = 1.64485 Critical region: 1.64485z 2 95 4.1 40 cal xz  Since H0 is not rejected, 1.64485calz   

9 2 95 1.64485 4.1 40 x  _ 96.1x The required set of values of mean top speed is __ | 96.1xx  Q11 (i) NOT tested in 2020 ‘A’ level Exam (ii) The product moment correlation coefficient, r is 0.903 (iii) The required equation is 0.294 1.89yx (iv) When 16.9x , 3.08y . Therefore the estimated time is 3.08 minutes. This is a reliable estimate as r is close to 1 and that x = 16.9 is within the given data range (interpolation). (v) The new product moment correlation coefficient, r is 0.568. (vi) (ii) is more likely to represent the correlation between swimming and running times for all the members of the club as r = 0.903 is closer to 1 which is supported by the the scatter diagram which shows an increaing trend; as x increases, y increases. The timings of the new member are likely to be outliers. Q12 Let X be the random variable denoting the mass of a biscuit.   2~ 20,1.1XN (i) Required probability =  19 0.182PX  Let T be the random variable denoting the mass of a box.   2~ 5,0.8TN 15 18.2 2.5 3.5 y x

10     1 12 22 E 20 12 5 245 Var( ) 1.1 12 0.8 15.16 ~ 245,15.16 Y X X T Y Y YN             (ii) Required probability =  248 0.221PY  Cost of a box of biscuits,  1 120.6 0.2C X X T         