JPJC 9758 2023 Prelim P2 Solution

Jurong Pioneer Junior College H2 Mathematics JC2-2023 Prelim Paper 2 Solution 1 (i) Method 1 ( ) 2 2 2 2 22 2 22 2 ln 1 e ln 1 1 2 ln 2 2 ln 2 1 24 ln2 ln 1 24 24ln2 24 2 ln2 24 8 1ln2 28 x xx xx xx xx xx xx xx x x x −  + ≈ + −+  = −+ = −+  = + +− +  −+ = +− + − + = −+ − +  = −+ +    Method 2 ( ) ( )f ln 1 e xx −= + ( ) ( )11f ' e1e e 1 x xxx − −= −= −++ ( ) ( ) 2 f '' e 1 e xxx − = + When 0x= , ( )f 0 ln 2= , ( ) 1f ' 0 2=− , ( ) 1f '' 0 4= ( ) 211f ln2 ......2 4 2! xxx =−+ +  ( ) 21f ln2 ...... 28 xxx= −++ (ii)

2 ( )d e1ln 1 ed 1e 1e x x xxx − − − −+= = − ++ Using the series from part (i), 21d 1 ln 21e d 2 8x x xx ≈− − ++  = 12 11 28 24xx−− + = − . 2 d 9.8d v Rt = − ( ) d 9.8 (Since , where 0)d 1 d 1 d 9.8 1 d 1 d 9.8 1 ln 9.8 ln 9.8 (9.8 0) 9.8 e , where e When 0, 0 (helicopter is stati kt kC v kv R kv kt vtkv k vtk kv kv t Ck kv kt kC kv kv A A tv −− = −= > =− −−= − − −= + − = −− − > −= = = = ∫∫ ∫∫ 0 onary) 9.8 (0) e 9.8 9.8 9.8e 9.8(1 e ) kt kt kAA kv v k − − − = ⇒= ∴−= −= 1The terminal velocity of the para 9.8As , e chuti 0, st is 9.8 ms kttv k k − − →∞ → → ∴ 3 (a) Method 1 eee 2xxxyy y= →= −→= −+ Reflect about the x-axis. Translate by 2 units in the positive y-direction. Method 2 ( )e e2 e2 = e2xx x xyy y= →=−→= − − −+ Translate by 2 units in the negative y-direction. Reflect about the x-axis. (b)

3 After A : 36yx= + After B : ( )3 2 63yx x= − += After C : ( )1 33y xx= = (c) (i) Asymptotes : xa= , 1y d= Axial intercepts : ( ),0c , 10, b   (ii) Asymptotes : xd= , yc= Axial intercepts : ( )0,a , ( ),0b 4 (i) h= 2 3 ( ) 6 1 f (2 f(2 ) f (2 2 ) f (2 3 ) f (2 4 ) f (2 5 ) f (2 6 ) n n h h hh hh hh hh hh hh = + =+++ ++ ++ ++ ++∑ represents the sum of areas of six rectangles drawn and the rectangles are above the curve, so it is greater than area of A. (ii) ( ) 5 0 f (2 ) n nh h = +∑ (6, f(6)) y 2 O x 6 2+h 2+2h 2+3h 2+4h 2+5h

4 (iii) Upper bound of area of A = ( ) 266 2 3 11 22f(2 2 e 291433 n nn nh h n  + = = += + ≈  ∑∑ Lower bound of area of A = ( ) 255 2 3 00 22f(2 2 e 131133 n nn nh h n  + = = += + ≈  ∑∑ (iv) Exact area of A 6 2 66 2 2 662 2 e d e ed 6e 2e e x xx x xx xx= − =−−  ∫ ∫ = 625e e− units2 (v) 5 (a) (i) Given = ••rq pq . R represents any point on the plane that is perpendicular to q and containing the point P. (ii) Given = ××rq pq ×−×=rqpq 0 ( )− ×=rp q 0 ( ) ( ) // , , kk kk − −= ∈ ∴=+ ∈ rp q rp q rp q   R represents any point on the line containing the point P and parallel to q . Let ux= and d ed xv x = . Then d 1d u x = and exv= (6, g(6)) y = g(x) 2+h 2+2h 2+3h 2+4h 2+5h

5 (b) Given that : 3:2AC CB = . By Ratio Theorem, 3 += c 2ab 3⇒= −c b 2a . Consider ( ) ( ) 2 2 22−= − −ac ac ac  2 24− =−−+a c a a 2a c c 2a c c   22 2 2 44−= − +a c a ac c  (Since 22 , , = = =ac ca aa a cc c   ) ( ) 22 2 2 4 43−= − − +a c a a b 2a c  2 2 22 2 4 12 8−= − + +a c a ab a c  22 2o2 12 12 cos 60−= − +a c a ab c ( ) 22 2 12 12(1) 12(1)(1) 7 2ac −= − +  2 2 13ac−= 2 13 since 2 0ac ac∴ −= −≥ 6 (i) Let X be the mass of the badminton racket. Let µbe the mean mass of a badminton racket. H0: 800µ = H1: 800µ ≠ Under H0, since n is large, By Central Limit Theorem, 220~ N 800,X n   approximately. Test statistic, 0 807 800 7 20 20 x nz nn µ σ − −= = = At 2% level of significance, for a two-tail test, critical region is 2.3263 or 2.3263zz<− > Since H0 is rejected, z lies inside the critical region. 77 2.3263 or 2.326320 20 (no real solution) 6.6467 44.179 nn n n <− > > > Least 45n= (ii) There is no need to know anything about the distribution of the population since the sample size is large, by Central Limit Theorem, the sample mean distribution is approximately normal.

6 7 (i) Number of ways = 4 2C = 6 (ii) Case 1: All same colour 3 1 3C = (excludes green bricks) Case 2: 2 different colours 4 2 6C = (to choose 2 colours) 2 1 2C = (to choose the colour which will be repeated) Number of ways = 6 2 12×= Or 43 11 12CC×= Case 3: All different colours 4 3 4C = Total number of ways = 3 + 12 + 4 = 19 (iii) Case 1: 2R 1B 1G 5 42 211C CC×× = 80 Case 2: 2R 2Y 54 22CC× = 60 Case 3: 1R 2B 1Y 544 121CC C×× = 120 Case 4: 4B 4 4 1C = Total number of ways = 80 + 1 + 120 + 60 = 261

7 8 (i) P( ' ') 1 P( ) P( ) 1A B A B ab∩ =− − =−− (ii) or If 'A and 'B are mutually exclusive events, then P( ' ') 1 0A B ab∩ =−−= 1ab+= (iii) Since A and C are independent events, P( )A C ac∩= P( )ABC∪∪ =P( ) P( ) P( ) P( ) P( )A B C AC BC+ + −∩ −∩ 0.14abca c=++− − 0.2 0.3 0.2 0.14cc= + +− − 0.36 0.8 c= + (iv) P( ' ' )ABC∩∩ =P( ) P( ) P( )ABC A B∪∪ − − 0.36 0.8 0.2 0.3c= + −− 0.8 0.14c= − Or P( ' ' )ABC∩∩ =P( ) P( ) P( )C AC BC−∩ −∩ 0.14c ac= −− 0.2 0.14cc= −− 0.8 0.14c= − (v) P( ) 0.36 0.8 1ABC c∪∪ = + ≤ and P( ' ' ) 0.8 0.14 0ABC c∩∩ = − ≥ 0.8 0.64c≤ 0.8 0.14c≥ 0.8c≤ 0.175c≥ a b A B C 0.14 0.2c A B b a a b

8 0.175 0.8c≤≤ 9 (i) From the scatter plot, the points seem to lie close on a curve rather than a straight line. Or P decreases at a decreasing rate as w increases. Therefore, the relationship between P and w is unlikely to be well modelled by a linear equation of the form P aw b= + , where a and b are constants. (ii) Between P and w: 0.949r =− (3 sf) Between P and w : 0.969r =− (3 sf) Since the product moment correlation coefficient between P and w is closer to 1− as compared with that between P and w , there is a stronger negative linear relationship between P and w . Hence, the model given by P aw b= + is better. From GC, 569.66 4130.0Pw= −+ The equation of the line is 570 4130Pw= −+ . (iii) When 40w= , 569.66 40 4130.0 527P= − += As 40w= is outside the range of the given data, an extrapolation is being done. The linear relationship between P and w may not hold at this extrapolated range. Estimation may not be reliable. 27.5 w (%) P (KPa) 10 2500 1270

9 (iv) For the regression line to remain unchanged, ( )9 9,wP which resulted from the missing data point ( )99,wP must be equivalent to( ),wP for the 8 points. From GC, 91693.75PP= = Shear strength for the data point is 1694 KPa (nearest integer). (v) 1000 569.66 4130.0Pw= −+ 0.56966 4.130Pw= −+ 0.570 4.13Pw= −+ The product moment correlation coefficient would not differ as product moment correlation coefficient is not affected by a change in unit of the variables (or scaling of the graph). 10 (i) X = BMI of 18-year-old boys ( ) 2~ N 22.7,X σ ( )P 16.7 0.05 16.7 22.7P 0.05 6 1.64 3.6476 3.65 (shown) 49 X Z σ σ σ ≤= − ≤= − − = ≈   = (ii) ( )P 0.1Xm≥≤ Using GC, 27.4m≥ Minimum value of BMI is 27.4 (iii) Y = BMI of 15-year-old boys ( ) 2~ N 21.6, 3.49Y 2 XY+ ~ N 2222.7 21.6 3.6476 3.49, 24  ++   2 XY+ ~ N(22.15, 6.3713)

10 P 20.12 XY+ > ≈ 0.792 (iv) The BMI of a randomly chosen 15-year-old boy and a randomly chosen 18-year-old boy are independent of each other. 11 Since ( ) ( ) 1P 1P 4 6TT= = = = , then ( ) ( ) 4P 2P 3 6TT= += = . If ( ) ( ) 2P 2P 3 6TT= = = = , then the modes of T are 2 and 3, which contradicts the question. Hence, ( ) 31P2 62T = = = and ( ) 1P3 6T