2016 VJC JC2 Prelim Paper 2 Solutions/Comments

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Abigail Glenn
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1 6 VJC JC Prelim Paper s/comments Qn i Since speed is decreasing and v is positive, dv kv, where k is a positive constant dt dv k v dt d v k d t v ln v kt C v v Be kt When t = s, v = D m s - B = D Let k = p, hence v= De pt, where p is a positive constant. v, D C v C t Stretch C parallel to the t-ais, factor e, v-ais is invariant. y = e cos dy e cos e sin cos = d e cos y e sin = y dy y = y e sin d Alternative y = e cos y = e cos dy y = e cos e sin cos d = y e sin i dy d y dy + y y e sin e cos = d d d When =, y = e cos =

2 dy dy = = d d d y d y 3 + = = 3 y 4! d d e cos 8 = + + ( )( ) 3 = + ( ) ! 3 = = + + 3a From the diagram, π arg 4 + i z + = arg + 5i z i 4 i z 5iz ( ) ( ) 4i iz 5i z i z i z Im A,5 B z D z C 4, Re bi 4 5 Midpoint of AC is,,3 Let z iy Since the diagonals of a square bisect other, Midpoint of BD is,3 y,,3 3, y6 z 3 6i i θ * iθ Alternatively Let ze z e 5 * z z z 5 z = z 6 * z zz z 6 z = 6 z 6 kπi z = e, k kπi 3 z = e, k =,,,3, 4,5 π π 4π 5π i i i i =,e,e,,e,e

3 kπ Since < arg(z) < π, i i 3 k 3 z = e z = e π π kπ π kπ i i i ( + i) = e.e = e k z ( + i) π kπ π 3π If k is purely imaginary, =±, ±, z 4 3 π kπ π 3π Since k is positive, =,, 4 3 kπ 3 7 = π π,, kπ 3π Smallest positive k when = 3 4 4a Smallest positive k = C = ( a + b), D = b CD = b ( a + b) = ( 5a b) Since line m passes through D and is parallel to CD, r D μcd 3 b μb5a b μ5ab r bλ5a b, λ 5 Equation of m is r = 3 b+ λ( 5a b), λ. 5 F is a point on m 3 F = b + λ( 5a b) for a value of λ 5 F is perpendicular to l F ( 5a b) = 3 5a b b+ λ ( 5a b) ( 5 ) 5 a b = F 6 3( a b) ( b b) + λ 5( a a) ( a b) + 4( b b) = 5 6 3( a b) b + λ 5 a ( a b) + 4 b = 5 Since ab = ab cos 6 = 5 = 5 6 3( 5) ( 5) + λ 5( ) ( 5) + 4( 5) = 5 3 λ =

4 3 3 3 F = b + ( 5a b) = ( 5a + b) 5 Alternative Method DF D 5a b 5ab 5 a b 5 a b b a b a b 5a b 5 6 3ab b 5 5ab 5ab 5ab ab 5 a ab 4 b 5 5ab D 5a b F 3 5 a b F = b + ( 5a b) = ( 5a + b) 5 b The equation of the plane π is 3+ y+ 5z= 45. ( ) pp,, lies in π 3p+ p+ = 45 p= q is perpendicular to q = q= q= 3 Since v is perpendicular to both u and n, 3 5 u n= 3 = r = 9 + λ 3 + µ, λµ, 3 Alternative method to find v Let v y z 3 y = and y 3 = z 5 z

5 5i 3 + y + 5z = and 3y = 5 z, y z, z z 3 3 Let z = 3 (any non-zero number will work) 5 v 3 He will not get to survey the students who do not go to the school gymnasium. Hence, the sample obtained is biased. He can obtain a numbered list of all the students (labelled to N) in the school. Using a random number generator, he generates 3 distinct numbers. He will survey the students corresponding the numbers generated. Alternatively. Let the total number of students be N N Sampling interval = 3 He can obtain a numbered list of all the students (labelled to N) in the school. N Using a random number generator, select a starting number k where k. He can interview the 3 9 students corresponding to the numbers k, k N, k N,, k N i Number of 4-digit numbers = 5 4 = 65 Case: Starts with No. of ways = (3!) = 6 Case : starts with No. of ways = 3! = 6 6 Total number of ways = 8 i Case : XXXXYZ No. of ways = 5 C 3 ( 3 C ) = 3 Case : XXXYYZ No. of ways = 5 C 3 ( 3 C )( C ) = 6 Case 3: XXYYZZ No. of ways = 5 C 3 = 7i Total number of ways = 5 3 C 3 C P(Mr Wong is correct) = C C C C C C C 5 5 C3 C P(Mr Tan is correct) = Alternative method: 87 3 P(Mr Tan is correct) 3!!

6 i P(Mr Wong s guess is right, given that Mr Tan s guess is wrong) P(Mr Wong is correct and Mr Tan is wrong) = P( Mr Tan is wrong).465 =.469 =.8 8i Let T be the total number of refrigerators sold in a 4-week period. T Po((.3 +.) 4) T Po(9.6) P(T ) = P(T 9) =.494 =.49 (3sf) Let X be number of good periods out of 5. X B(5,.49) or X B ( 5,.494) Since np = 5.53 > 5 and np( p) = > 5 X N(5.53,.996) appro. or X N ( 5.539,.996) appro. i P(5 < X 3) = P(5.5 < X 3.5) =.477 (or.478) We need to assume that the sales of all the refrigerators are independent of one another. We also need to assume that the average rate of refrigerators being sold is constant. The first assumption may not hold as the two brands of refrigerator are in the same price range and they can be competing in terms of sales. R The average rate of refrigerators sold is unlikely to be a constant due to sale, festive seasons, economic conditions etc. 9 Let A kg and B kg be masses of a randomly chosen grade A and grade B durian respectively. A N(.96,.4 ) and B N(., σ ) P(B >.8) = P Z > =.95 σ.8. P Z =.5 σ. = σ =.59. σ i A N(.96,.4 ) and B N(., σ ).4 A~ N.96, and 5

7 B N(., (. ) or ( ) B N(.,.59 ~.4,.6688 or A B~ N.4,.69 A B N P AB.436 (or.435) Central limit theorem is not needed because the masses of grade A durians follow a normal distribution. Let Y be the number of grade B durians with a mass of more than.8 kg out of 5 durians. Y B(5,.95) np = 5.95 = 47.5 > 5 and n( p) = 5.5 =.5 < 5 Let Y be the number of grade B durians with a mass.8 kg out of 5 durians. Y ' Po (.5) appro. P(Y > 47) = P( 5 Y ' > 47) = P( Y ) =.544 i t and t9.8, 54.5 Hence, (9.8, 54.5) lies on the regression line on t. t /C o r =.934 (3s.f.) i iv From the scatter diagram, increases by decreasing amounts as t increases. Hence, a quadratic model might be more appropriate. By GC, a = (3sf), b = 79 (3sf) v Substituting t = 3., ( ) = = 7.388

8 Epected number of cups of ice cream sold is 7. H : µ = 4 H : µ 4 Level of significance: 5% X 4 Test Statistic: When H is true, T = S / 5 Computation: ν = 5 = 4. By GC, = 39.34, s=.97, p-value =.57 (3sf) Conclusion: Since p-value =.57 >.5, H is not rejected at 5% level of significance. So there is insufficient evidence to conclude that the claim is invalid. It is assumed that the masses of loaves of Gardener wholemeal bread follow a normal distribution s H : µ = 4 H : µ 4 Level of significance: k% Test Statistic: When H is true, Z = Computation: By GC, = , p-value =.5 (3sf) X For H to be rejected at k% level of significance, k p-value k 5. k : k 5. Set of values = { } k k% significance level in this contet means there is a probability of (or k%) that the test will conclude that the mean mass of Gardener wholemeal bread is not 4g, when it is actually 4g. Let Y be the number of wrong conclusions out of n hypothesis tests Y B( n,.4) PY ( ) <.5 By GC, n PY ( ) Least n = 7 Alternatively

PY ( ) <.5 n n ( ) n( ) ( ).96 +.96.4 <.

9 PY ( ) <.5 n n ( ) n( ) ( ) <.5 y n n y.96.4n.96 y.5 Least n = 7 6.7,.5 n

GCE A Level H2 Mathematics November 2014 Paper 1. 1i) f 2 (x) = f( f(x) )

GCE A Level H2 Mathematics November 2014 Paper 1. 1i) f 2 (x) = f( f(x) ) GCE A Level H Mathematics November 0 Paper i) f () f( f() ) f ( ),, 0 Let y y y y y y y f (). D f R f R\{0, } f () f (). Students must know that f () stands for f (f() ) and not [f()]. Students must also