The required region is shown in the diagram above. The

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Rudolf Harrington
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1 011 GCE A Level H1 Maths Solution SECTION A (PURE MATHEMATICS) 1 For x k x k 1 0 to be true for all real x, the discriminant That is, k k k 8k 0 k k8 0 D b 4ac must be negative The critical points are k 0 and k 8. From a number-line diagram: Note: When a quadratic expression is always positive or negative, this means the corresponding quadratic graph is not intersecting the x -axis. Thus there is no real roots for such a case D 0. (i) (ii) (iii) Therefore the answer is 0k 8. Thus, the set of values is k : 0 k 8. From the graphic calculator, the x -coordinates are and correct to 4 d.p. y y x Note: The expression 0.6 x belongs the exponential family, such as e x and y e x has a horizontal asymptote y 0. Thus it is expected that the graph of y 0.6 x will also have a horizontal asymptote. A TI84 Plus screenshot of the sketch is shown in the next part. A TI84 Plus screenshot showing the computed result of one of the intersection points. x The required region is shown in the diagram above. The area is given by x dx.615 correct to d.p by graphic calculator. x 1 x e dx e C, where C is an arbitrary constant. (i) x MATHEMATICS DEPARTMENT PAGE 1

2 (ii) 9 1 x dx 4 x 9 1 x x Remark: Though a calculator is not allowed in this question, we can always use it to check the answer: 4 From the question s description, the box will have base area PS SR and height x. Thus the volume is given by V x x x 4x 1 x 4x 1 x x 4x 8x 4x. (Shown!) dv Find 1x 16x 4 dx and let d V 0 dx x 4x x 1 x 1 or. d V Compute 4x 16 dx. Sub d V x 1 0. dx 1 d V Sub x dx 1 Thus x gives the maximum V which has value m. 7 y x ln x 1. 5(i) Given Therefore, d y 1 dx x 1. Let d y 0 dx 1 x x. MATHEMATICS DEPARTMENT PAGE

3 This is the only one answer from calculation, which coincides with only one turning point on the diagram. This means it should give the minimum point on the curve When x, y ln 1 ln. 1 1 Thus, the required coordinates is, ln. 5(ii) Find the equation of the normal to the curve at P : At y ln 1 ln5 and x, dy 1 dx Therefore, gradient of normal. dy dx Thus, equation of normal is given as y y m x x y 5 16 y x ln 5. ln 5 x At A, y 0 0 x ln5 x ln Thus, coordinates of A are ln 5, At B, x 0 y 0 ln 5 y ln Thus, coordinates of B are 0, ln 5. Area of triangle OAB 1 OA OB ln 5 16 ln ln 5. 0 That is, p 16 and q. MATHEMATICS DEPARTMENT PAGE

4 011 GCE A Level H1 Maths Solution SECTION B (STATISTICS) 6 Using the result P A B P A PB P A B and substituting P AB 0.46 and A B 0.46 ab 0.04 ab (1), P 0.04 : Since A and B are also independent events, this means P A B =P A P B 0.04 ab () Combining (1) and () to remove b : 0.04 a 0.5 a a 0.5a which should be required quadratic equation in a. 7(i) 7(ii) 7(iii) Solve this quadratic equation: a a a 0.4 or 0.1. That is, P A 0.4 or 0.1 A random sample here means the sample is obtained by selecting 100 students from the 000 students in such a way that each of the 000 students will have an equal probability of of being selected to do the survey. The three strata are By car, By bicycle and On foot. 100 The sample size under By car is The sample size under By bicycle is The sample size under On foot is Stratified sampling would ensure that each group (stratum) in the population is represented, while random sampling may have a chance of missing out an important group completely, or may end up with a certain group overly represented in the sample. Instead of the mentioned strata, a better way could be MATHEMATICS DEPARTMENT PAGE 4

to further divide the sub-population in each stratum according to students year-groups. This means there will be 6 strata such as By car, Year 1, By car, Year etc.

5 to further divide the sub-population in each stratum according to students year-groups. This means there will be 6 strata such as By car, Year 1, By car, Year etc. 8(i) The scatter diagram is shown below: T The scatter diagram for T against H generated by the TI84 Plus is shown below: H 8(ii) From the graphic calculator, the product moment correlation coefficient is correct to s.f. This value indicates a strong negative linear relation between T and H. This means that the temperature decreases at almost a constant rate as we ascend up the mountain (altitude increases). Note: From the context of the question, it is more likely that a diagram for T against H is sketched then one for H against T. The evidence is apparent from the phrase recorded at each of 8 locations in the question. From the TI84 Plus screenshot: 8(iii) From the graphic calculator, the regression line of T on H is given by T H 7.0. T 7 TI84 Plus screenshot with the regression line drawn: H Remark: There should only be one diagram drawn (right) in the solution to this Question 8. Note that a scatter diagram should only consist of just the observed points. The regression line should not be part of the scatter diagram. MATHEMATICS DEPARTMENT PAGE 5

8(iv) Using the equation of the regression line (with more d.p), T 0.0147 H 7.0098 and substitute H 1000 into it: T 0.0147 1000 7.0098 1.898. Thus, the air temperature at this instant is 1.

6 8(iv) Using the equation of the regression line (with more d.p), T H and substitute H 1000 into it: T Thus, the air temperature at this instant is 1. C, correct to s.f. Since the input value H 1000 is within the data range for H : 00 H 1550, and the value of r calculated previously indicated a strong linear trend, we can safely conclude that this estimate is reliable. 9 Let denote the mean lifetime of the light bulbs in the batch. TI84 Plus screenshot showing the input values for the test: Null hypothesis H 0 is Alternative hypothesis H 1 is Since n is large, we carry out a z -test. Given that the sample mean is 11500, and the population standard deviation is 1400, from the graphic calculator, the p -value is which means p 0.01 (1% level of significance). The results: Thus, we reject H 0 and conclude that there is evidence at 1% level of significance that the batch of light bulbs is substandard. Given that the sample mean in another batch is T hours and the test is not significant at 5% level of significant means p -value is greater than That is, if L represents the lifetime of a light bulb from P LT this batch, then L T Standardizing: P L1000 T 1000 P T P Z TI84 Plus screenshot showing the use of inverse normal function: Note: To use the InvNorm function in the graphic calculator, we must ensure the probability is P X or of the form P X, with the inequality sign being less than or less than or equal. MATHEMATICS DEPARTMENT PAGE 6

7 10(i) T T Thus, this concludes that the least possible value of T is correct to s.f. Let X denote the number of puzzles Jon will complete X ~ B 7, 0.8. in a week of 7 days. Then 10(i)(a) The required probability is P X Note: This answer is exact. Thus we do not need to correct it to 10(i)(b) The required probability is P X 5 1 P X 4 10(ii) 10(iii) correct to s.f. Let Y denote the number of weeks Jon will complete at Y ~ B 10, least 5 puzzles. Then The required probability is PY correct to s.f. Let W denote the number of puzzles Jon will complete W ~ B 70, 0.8. in a period of 10 weeks. Then Since the number of trials n 70 is large, np 56 5 and np1 p 14 5, ~ N56,11. W approximately. Here, the mean of the approximation is 56, while the variance is 11.. The required probability is PW 50 cc. PW correct to s.f. significant figures. Remark 1: The purpose of using more decimal places for the success probability ( ) is to minimize rounding off errors. Remark : This question can also be answered by thinking in terms of 10 consecutive weeks, all of which Jon is to complete at least 5 puzzles in each of them. That is, the probability is simply Note: Variance np1 p. TI84 Plus screenshots in calculating the probability: MATHEMATICS DEPARTMENT PAGE 7

8 11(i)(a) A tree diagram is drawn as shown: Remark: In the second branch of the probability tree, the number such as 0.4 represents the 0.5 Red conditional probability of A 0.4 Green choosing a red ball given that Box Yellow A is selected B 0.75 Red 0.5 Green 11(i)(b) 11(i)(c) The required probability P A, R P B, R The required probability P selecting Box A a red ball is chosen P selecting Box A and a red ball is chosen P a red ball is chosen Note: 0.18 means with the decimal numbers alternating between 1 and 8. Decimal numbers with such behaviour can always be written as a fraction. 11(ii) Required probability P A, R, R P A, G, G P B, R, R P B, G, G or correct to s.f Let B and G denote the masses of a boy and of a girl 1(i) 1(ii) respectively. Then B ~ N60,1 and ~ N50,10 The required probability P 50 B correct to s.f. The required probability P B G P BG 0 Compute the distribution of B G: BG~ N 60 50,1 10. G. Note: Again, the answer to this question is exact. If we did not realize the answer is exact, we will give the second best answer rounded to s.f. TI84 Plus screenshots on calculating the probability: MATHEMATICS DEPARTMENT PAGE 8

9 That is, B G~ N10, 44. Here, the mean is 10 and the variance is 44. Thus, by graphic calculator computation, the probability is correct to s.f. 1(iii) Let B1, B, B denote the masses of the three boys, and G1, G denote the masses of the two girls. Then, 1 1 B B B G G ~ N 60 50, 1 10 That is, B B B G G ~ N80, Here the mean is 80 and the variance is 6. TI84 Plus screenshot on the inputting of values to calculate the required probability: 1(iv) Finally, the probability required is P B B B G G correct to s.f. Let W denote the masses of the six boys. Then, ~ N6 60, 6 1 W. That is, ~ N60, 864 Given W L P W. By graphic calculator, L correct to s.f. TI84 Plus screenshots on finding L : MATHEMATICS DEPARTMENT PAGE 9

u x y reduces the differential equation

u x y reduces the differential equation CATHOLIC JUNIOR COLLEGE H MATHEMATICS 06 JC PRELIM Paper (i) Prove that the substitution (ii) (i) Given u x y, du dy x y dx dx du dy x y ----------- (I) dx dx Substitute (I) & u x y and into D.E: we get