2. Topic: Series (Mathematical Induction, Method of Difference) (i) Let P n be the statement. Whenn = 1,

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1 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. MATHEMATICS (H2) Paper 2 Suggested Solutions 9740/02 October/November 200. Topic:Complex Numbers (Complex Roots of Quadratic Equations) (i) x 2 6x + 34 = 0 x = 6± 36 4(34) 2 = 6± 00 2 = 6±0 2 = i = 6±0i 2 = 3 ± 5i 3 + 5i and 3 5i are the solutions. (ii) x 4 + 4x 3 + x 2 + ax + b = 0 () Since x = 2 + i is a root, by Factor Theorem, Both real and imaginary parts must be 0 ( 2 + i) 4 + 4( 2 + i) 3 + ( 2 + i) 2 + a( 2 + i) + b = i + 4( 2 + i) + 3 4i 2a + ai + b = i i + 3 4i 2a + ai + b = 0 2 2a + b + (6 + a) i = a = 0 2 2a + b = 0 a = 6 b = 2a + 2 = 20 a = 6 and b = 20 Factor Theorem: x a is a factor of f(x) f(a) = 0 Since x = 2 + i is a root and the coefficient of each term in () is real, then by Complex Conjugate Root Theorem, x = 2 i is also a root. [x ( 2 + i)] [x ( 2 i)]= (x + 2 i) (x +2 + i) = (x +2) 2 i 2 = x 2 + 4x = x 2 + 4x + 5 Factorizing (), x 4 + 4x 3 + x 2 6x 20 = 0 (x 2 + 4x + 5) (x 2 4) = 0 (x 2 + 4x + 5) (x 2) (x + 2) = 0 x = 2 + i, 2 i, 2, 2 The other roots are 2 i, 2 and Topic: Series (Mathematical Induction, Method of Difference) (i) Let P n be the statement n r= r(r + 2) = n(n + )(2n + 7), n 6 Z+. Whenn =, L.H.S. = r= r(r + 2) = ( + 2) = 3 R.H.S. = ()( + )(2 + 7) 6 = = 3 L.H.S. = R.H.S. Since L.H.S. = R.H.S., P is true Copyright 200 ϕ exampaper.com.sg. All rights reserved. / 20

2 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Assume P k is true for some k Z + i.e. k r= r(r + 2) = k(k + )(2k + 7) 6 To show that P k+ is also true i.e. k+ r= r(r + 2) = (k + )(k + 2)[2(k + ) + 7], 6 Bring out common factor (k + ) since it 6 appears on RHS. k+ L. H. S. = r= r(r + 2) = k r= r(r + 2) + (k + ) (k + 3) = k(k + )(2k + 7) + (k + ) (k + 3) 6 = (k + 6 )[2k2 + 7k + 6k + 8] = 6 (k + )[2k2 + 3k + 8] = (k + )(2k + 9)(k + 2) 6 = (k + )(k + 2)[2(k + ) + 7] 6 = R. H. S. Since L. H. S. = R. H. S., P k+ is true if P k is true. Since P is true and P k+ ifp k is true, P n is true n, n Z + by mathematical induction. From MF5: Partial fractions decomposition (Non-repeated linear factors): pp + q (aa + b)(cc + d) = A (aa + b) + B (cc + d) (ii) (a) Let = A + B A(r + 2) + Br = r(r+2) r r+2 r = 0 2A = A = 2 r = 2 2B = B = 2 = r(r+2) 2r 2(r+2) Cover-up Rule may be used directly to save time. (k + ) th term n n r= = r= 2r r(r+2) r(r+2) 2(r+2) = n 2 r r= = [ 2 = + 2 = + 3 = + 4 = = + = + = + n n 2 n (r+2) n n+ ] n+2 = n+ n+2 = 3 (shown) 4 2(n+) 2(n+2) (b) r= = lim n n r= = lim n 3 4 r(r+2) 2(n+) 2(n+2) As n, 0 and 0 2(n+) 2(n+2) lim n 3 = = 3 4 2(n+) 2(n+2) 4 4 r= = 3 r(r+2) 4 Using expression proven in 2(ii)(a) Since it converges to a constant value (with a sum to infinity of 3 4 ), this is a convergent series. Copyright 200 ϕ exampaper.com.sg. All rights reserved. 2 / 20

3 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version Topic: Differentiation (i) Given y = x p + 2 (b) Using G. C. (refer to Appendix for detailed steps), = x(p + 2) 2 dy = (p + dx 2) 2 + p (p + 2) 2 2 = p x 2 x+2 = 2(x+2)+x 2 x+2 3x+4 = 2 x+2 When dy dx = 0 3x+4 2 x+2 = 0 x = 4 3 Product Rule: d dv du (uu) = u + v dx dx dx 2 There is only one stationary point when x = 4 3. (ii) (a) Given y 2 = x 2 (x + 2) y = ±p p + 2 From Part (i), we have dy = ± 3x+4 dx 2 x+2 dy When x = 0, = ± 4 dx 2 2 = ± 2 2 = ± 2 Possible values of the gradient is 2 and 2. Copyright 200 ϕ exampaper.com.sg. All rights reserved. 3 / 20

4 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. (iii) Using G. C. (refer to Appendix for detailed steps), 4. Topic: Functions (i) Using G. C. (refer to Appendix for detailed steps), When x = 0, y = 2 y y = f(x) x = x = 0 x 2 x = x = x = (ii) For the function f to exist, f must be one-one over the given domain, where there exists only one value of x for each image of f. From the sketch in Part (i), f is one-one when x 0,x. Hence the least value of k is 0. Horizontal lines will cut the graph only once. x 0 Copyright 200 ϕ exampaper.com.sg. All rights reserved. 4 / 20

5 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. (iii) fg(x) = f[g(x)] = f x 3 (iv) fg(x) > 0 (x 3) 2 (4 x)(x 2) = = = = = (x 3)2 (x 3) 2 (x 3) 2 (x 3) 2 (x 2 6x+9) (x 3) 2 x 2 +6x 8 (x 3) 2 (shown) (4 x)(x 2) > 0 Since (x 3) 2 0,(4 x) (x 2) must be positive, By number line: < x < 4, x 3 ALTERNATIVE APPROACH Using G. C. to plot fg(x) (refer to Appendix for detailed steps), y x = 2 x = 4 x 0 x = 2 3 y = fg(x) From the graph, 2 < x < 4, x 3 for fg(x) > 0 (v) Given g(x) =, p R, p 2, p 3, p 4. From graph of g(x), (x 3) y R g = (, ) (,0) (0, ) (, ). When D f = (, ) (,0) (0, ) (, ) R f = (, ) (0, ) (from graph in (i)) Range of fg(x) =(, ) (0, ) x = 4 y = g (x) y = ALTERNATIVE APPROACH From the graph sketched in part (iv), range of fg(x) =(, ) (0, ) x x = 3 Copyright 200 ϕ exampaper.com.sg. All rights reserved. 5 / 20

6 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version Topic: Sampling (i) Due to the great cultural and regional variety of the international spectators, it would be difficult to divide them into appropriate the strata for a suitable analysis. Moreover, given the large size, multinational and mobile nature of the population of spectators, it will be tedious and time-consuming to accurately obtain the required representative sample % of spectators in each stratum. (ii) A systematic sample of % of the spectators could be obtained by first randomly interviewing a person leaving the premise of the catering facilities, and thereafter interviewing every 00 th person leaving the premise of the catering facilities. 6. Topic: Hypothesis Testing Given: n = t = t 2 = Stratified Sampling: Unbiased estimate of population mean, t = t n = = 4.3 Divide population into mutually-exclusive subgroups (strata), and then apply random or systematic sampling within each subgroup. Systematic Sampling: To obtain a systematic sample of size n from a population of size N, pick a random element from among the first k = N elements, and n thereafter picking every k th element. Unbiased estimate of population variance, s 2 = = n t ( t)2 n (454.3) Let µ be the mean time required by an employee to complete a task. To test H o : µ = 42.0 against H : µ 42.0 at 0% of significance Reject H o if p-value < 0.0. Applying t-test with t = 4.3, n =, s 2 =.584 using G. C. (refer to Appendix for detailed steps), From GC, the p-value = < 0.0, we reject H o. Hence, there is sufficient evidence at the 0% significance level that there has been a change in the mean time required by an employee to complete the task. From MF5 Testing for change in µ Two-tailed test Since population variance is not given and sample size is small (n < 30), a t-test is used. 5% 5% Copyright 200 ϕ exampaper.com.sg. All rights reserved. 6 / 20

7 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version Topic: Probability Given P(A) = 0.7, P(B) = 0.6, P(A B ) = 0.8 (i) P(A B ) = 0.8 P A B P(B ) = 0.8 P(A B ) = 0.8 P(B ) = 0.8 [ P(B)] = 0.32 (ii) P(A B) = P(A B ) + P(B) = = 0.92 (iii) P(B A) = P(B A) P (A) = = 6 35 Given P(C) = 0.5 and A andc are independent. (iv) P(A C) = P(C) P(A C) = P(C) P(A) P(C) = = 0.5 A A B A and C are independent P(A C) = P(A) P(C) Note: This also means P(C) P(A) P(C) = [ P(A)] P(C) P(A C) = P(A ) P(C) B (v) Max P(A B C) case (C subset of B): When C B A B C A C P(A B C) P(A C) P(A B C) 0.5 Min P(A B C) case (minimal intersection between B and C): P(A B C) This is just a -mark question. It should be sufficient to state either P(A B C) 0.5 or P(A B C) P(A B) + P(A B C) P(A B) + [P(A C) P(A B C)] P(A B C) P(A B C) P(A B C) P(A B C) 0.5 Copyright 200 ϕ exampaper.com.sg. All rights reserved. 7 / 20

8 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version Topic: Probability (i) (ii) st digit 2 nd digit 3 rd digit 4 th digit 5 th digit 3 ways 4 ways 3 ways 2 ways way (i.e. 3, 4, 5) P(number is greater than 30,000) = 3 4! 5! = 3 5 st digit 2 nd digit 3 rd digit 4 th digit 5 th digit 3 ways (i.e., 3, 5) 2 ways way 2 ways (i.e. 2, 4) P(last 2 digits are both even) = 3! 2! = 5! way (i.e. 4 or 2) (iii) Case ( st digit is 3 or 5): st digit 2 nd digit 3 rd digit 4 th digit 5 th digit 2 ways (i.e. 3, 5) 0 3 ways 2 ways way 2 ways (i.e., 5 or 3) Case 2 ( st digit is 4): st digit 2 nd digit 3 rd digit 4 th digit 5 th digit way (i.e. 4) 3 ways 2 ways way 3 ways (i.e., 3, 5) P(number is greater than 30,000 and odd) = 2 3! 2+ 3! 3 5! = 0.35 no.of ways for event A to occur P(A) = total no.of possible outcomes 9. Topic: Normal Distribution Let X and Y be the random variables such that Ken makes X minutes of peak-rate and Y minutes of cheap-rate telephone calls, respectively, over a 3-month period. Given X ~ N(80, 30 2 ) Y ~ N(400, 60 2 ) (i) E(Y 2X) = E(Y) 2E(X) = 400 2(80) = 40 Var(Y 2X) = Var(Y) Var(X) = = 7200 Y 2X ~ N(40, 7200) Using G. C. (refer to Appendix for steps to access the normal distribution functions), P(Y> 2X) = P(Y 2X> 0) = (3 sig.fig.) (ii) Let T be the random variable for the total cost (in dollars) of Ken s calls made over a three-month period T = 0.2 X Y E(T) = E(0.2X Y) = 0.2E(X) E(Y) = 0.2(80) (400) = 4.6 Var(T) = Var(0.2X Y) = Var(X) Var(Y) = T ~ N(4.6, 2.96) = 2.96 If X~N(μ X, σ 2 X ) and Y~N(μ Y, σ 2 Y ) are two independent normal distributions, aa ± bb~n(aμ X ± bμ Y, a 2 σ 2 X + b 2 σ 2 Y ) Remember to square the 0.2 & 0.05 when calculating the variance! Copyright 200 ϕ exampaper.com.sg. All rights reserved. 8 / 20

9 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Using G. C. (refer to Appendix for steps to access the normal distribution functions), 0. Topic: Correlation and Regression (i) Using G. C. (refer to Appendix for detailed steps), P(T > 45) = (3 sig.fig) (iii) Let W be the random variable for the total cost (in dollars) of Ken s peakrate calls made over two three-month periods (X and X 2 being the number of peak-rates calls in each period, respectively) F W = 0.2X + 0.2X 2 E(W) = 2(0.2) E(X) = 43.2 Var(W) = 2(0.2 2 ) Var(X) = W ~ N(43.2, 25.92) If X and X 2 are two independent observations of the random variable X where X ~ N(µ, σ 2 ), ax + ax 2 ~ N(2aµ, 2a 2 σ 2 ) Using G. C. (refer to Appendix for steps to access the normal distribution functions), v P(W > 45) = (3 sig.fig) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 9 / 20

10 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. (ii) (a) For F = a + bv, using G. C. (refer to Appendix for detailed steps) (iii) Since the scatter diagram reveals a non-linear relationship between F and v in part (i) and the correlation coefficient between v 2 and F yields a higher value of (as compared to for v and F) in part (ii), F = c + dv 2 is a better model. (iv) Sub a = and b = obtained by G. C. in part (ii)(b) into d and c respectively in F = c + dv 2, L3 = (L) 2 r = (4 decimal places) (b) For F = c + dv 2,, using G. C. (refer to Appendix for detailed steps) Required equation F = v 2 Sub F = 26.0, 26.0 = v 2 Note: Reg (ax + b) used in G. C. v = 30.7 As the wind speed is controlled, v is the independent variable and we are using the regression line F on v 2 to predict v. Since F is not the independent variable, we should not use the regression line of v on F or v 2 on F to estimate v.. Topic: Binomial, Poisson Distributions& Their Normal Approximation Let X be the random variable for the number of calls received in one minute. X Po(3). (i) Let X 4 be the random variable for the number of telephone calls received in a period of 4 minutes. Additive Property of X 4 = 4X Po(4 3) X 4 Po(2) Poisson Distributions: Using G. C. (refer to Appendix for detailed steps), n X i ~ Po λ i i= n i= r = (4 decimal places) P(X 4 = 8) = (3 sig.fig) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 0 / 20

11 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. (ii) Let n be the number of minutes and X n be the random variable for the number of telephone calls received in a period of n minutes. X n Po(3n) P(X n = 0) = 0.2 Note: 0! =! 3n (3n)0 e 0! = 0.2 e 3n = 0.2 3n = ln 0.2 Probability density function of X, where X Po(λ): P(X = x) = e λ λx x! n = ln = mins = seconds 32 seconds (nearest second) (iii) 2 hrs = 2 60 = 720 min Let X 720 be the random variable for the number of telephone calls received in 720 min. X 720 = 720X Po(720 3) X 720 Po(260) Since λ is large (>0), we use a normal distribution to approximate the Poisson distribution as follows X 720 N(260,260) approximately. Additive Property of Poisson Distributions When λ > 0, X Po(λ) N(λ, λ) (iv) Let Ybe the random variable for the number of busy working days out of 6 working days. Y B(6, 0.976) Using G. C. (refer to Appendix for detailed steps), P(Y = 2) = (3 sig.fig) Binomial Distribution: c.c X B(n, p) where n = no. of trials = 6 p = probability of success = 0.92 from part (iii) P(X 720 > 2200) P(X 720 > ) by continuity correction Using G. C. (refer to Appendix for detailed steps), Approximating a discrete distribution with a continuous distribution by Continuity Correction: P discrete (X>x) P continuous (X>x + 0.5) c.c P(X 720 > ) = (3 sig.fig) Copyright 200 ϕ exampaper.com.sg. All rights reserved. / 20

12 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. (v) Let W be the random variable for the number of busy working days out of 30 randomly chosen working days. W B(30, 0.976) np = = > 5 nq = 30 ( 0.976) = > 5 Since np > 5 and nq > 5, we use a normal distribution to approximate the Binomial distribution as follows W N(5.7528, ( 0.976)) W N(5.7528, ) approximately. P(0 W < 0) P( 0.5 < W < 9.5) by continuity correction. Using G. C. (refer to Appendix for detailed steps), When n is large and np> 5 and nq> 5, X N(n, p) N(np, npq) Since the number of days cannot be < 0, it should be P(0 X< 0) instead of P(X< 0) c.c c.c 0.5 P( 0.5 < W < 9.5) = (3 sig.fig) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 2 / 20

13 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Appendix: Detailed G. C. Steps (for those still trapped in G. C. limbo) Q3 (b)(ii), Q3 (iii), Q4 (i): Graph Sketching Ensure G. C. is in FUNC mode. 3(ii)(b) 3(iii) 4(i) 4(iv) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 3 / 20

14 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. 3(ii)(b) [SHIFT] [EXIT] [F6] 3(iii) [SHIFT] [EXIT] [F6] 4(i) [SHIFT] [EXIT] [F6] 4(iv) [SHIFT] [EXIT] [F6] Copyright 200 ϕ exampaper.com.sg. All rights reserved. 4 / 20

15 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q6: Hypothesis Testing (t-test with Data Summary) Select Stats input method. Enter the population/sample mean, (sample variance), sample size. Selectµ: µ o for two-tailed test. Select Calculate for results in summary view. Select Draw for results in graphical view. [Enter the parameters] Results (summary view) Results (graphical view) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 5 / 20

16 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q0 (i): Plotting Scatter Diagram Enter v and F values in L and L2 respectively Turn On Plot Copyright 200 ϕ exampaper.com.sg. All rights reserved. 6 / 20

17 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q0 (ii)(a): Finding Correlation Coefficient Note: The r value will not appear if you miss this step! Note: L contains the values of v (independent variable) and L2 the values of F(dependent variable) aspopulated in 0(a). Copyright 200 ϕ exampaper.com.sg. All rights reserved. 7 / 20

18 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q0 (ii)(b): Finding Correlation Coefficient Populate L3 with the square of values of v contained in L. N.B. Make sure L3 is highlighted when doing this! N.B. L3 contains the values of v 2 and L2 the values of F. The first argument contains the independent variable. Copyright 200 ϕ exampaper.com.sg. All rights reserved. 8 / 20

19 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q (i): Poisson Distribution Key in the parameters. Q (iv): Binomial Distribution Key in the parameters. Copyright 200 ϕ exampaper.com.sg. All rights reserved. 9 / 20

20 GCE A Level October/November 200 Suggested Solutions Mathematics H (9740/02) version 2. Q9 (i-iii), Q (iii), Q (v): Normal Distribution Key in the relevant parameters. Results shown are for Q9(i) Copyright 200 ϕ exampaper.com.sg. All rights reserved. 20 / 20