2010 GCE A Level H2 Maths Solution Paper 2 Section A: Pure Mathematics. 1i) x 2 6x + 34 = 0 6 ± x = 2

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1 00 GCE A Level H Maths Solution Paper Section A: Pure Mathematics i) x 6x ± x 6 ± 0i 3 ± 5i (ii) Since the coefficients are all real, another root of the equation is x i. [ x ( + i) ] [ x ( i) ] (x + i)(x + + i) x + x ix + x + ix + i x + 4x + 5 By comparing coefficients: x 4 + 4x 3 + x + ax + b (x + 4x + 5)(x 4) x 4 + 4x 3 + x 6x 0 So a 6, b 0. The other three roots of the equation are i,,. i) Let P n be the statement: n r(r + ) 6 n(n + )(n + 7). r When n : LHS (3) 3 RHS 6 ()(9) 3 LHS P is true. Assume that P k is true for some k Z + k i.e. r(r + ) 6 k(k + )(k + 7). r Prove that P k+ is also true k+ i.e. r(r + ) 6 (k + )(k + )(k + 9). r LHS k r(r + ) + (k + )(k + 3) r 6 k(k + )(k + 7) + (k + )(k + 3) 6 (k + )[ k(k + 7) + 6(k + 3) ] It is possible to use long division to find the other factor, but comparing coefficients is quicker. Some students make the mistake of writing Assume that P k is true for all k Z +. It should be either some k or a k. Clearly, if you assume that it is true for all k, then there is nothing to prove. Students are advised to take

2 6 (k + )(k + 3k + 8) 6 (k + )(k + )(k + 9) RHS Since P is true and P k true P k+ true, hence by Math Induction, P n is true for all n Z +. (iia) By coverup rule, r(r + ) r (r + ). n r(r + ) n [ r (r + ) ] r r (3) + () (4) + (3) (5) + (4) (6) + (n ) : : (n + ) (n + ) + n + () (n + ) (n + ) 3 4 (n + ) (n + ). (n + ) and (n + ) 0. (b) As n, Hence r(r + ) is a convergent series, and the value of the r sum to infinity is i) dy dx x + + x (x + )/ x x + + (x + ) + x 3x + 4 For the curve to have a turning point, dy dx 0 out the common factor 6 (k + ) instead of expanding everything and then factorising later. Clearly a waste of effort. If student prefers not to use the coverup rule, he can let r(r + ) A r then find A and B by substitution. B (r + ) and

3 3x x Since there is only one value of x for which 3x + 4 0, there is only one value of x for which the curve has a turning point. x 4 3 (iia) y ±x x + dy dx ± 3x + 4 When x 0, dy dx ± ± (b) 4/3 (iii) f (x) 3x + 4 4/3 Students are to draw the curve so that it approaches the asymptote x without touching it. x 4i) x x (ii) If the domain of f is restricted to x 0, then any horizontal line will cut the graph at most once, so f is one one and f exists. Hence the least value of k is 0. (iii) fg(x) f( x 3 ) (x 3) 3

4 (iv) (x 6x + 9) (x 3) (x 3) x + 6x 8 (x 3) (4 x)(x ) (x 3) (4 x)(x ) > Some students may wish to use a graphical solution. < x < 3 or 3 < x < 4. (v) x fg x 4 3 y From the graph of fg, we see that range of fg is (, ) (0, ). Section B: Statistics 5i) In order to use stratified sampling, we need to know the composition of the spectators according to strata e.g. race, gender or agegroups. But we do not have this information. Hence it would be difficult to use a stratified sample. (ii) We line up all the spectators as they arrive at the competition. Pick a random starting number from to 00, e.g. 5. We sample every 00th spectator starting from the 5th spectator in the queue, i.e. 5, 05, 05, 305, 405,... 6) Unbiased estimate of the population mean x Unbiased estimate of the population variance s ( ).584 H 0 : μ 4 H : μ 4 Students must know that they need to use the t-test here since sample size is 4

5 small and population variance is unknown. From GC, pvalue Since < 0., we reject H 0. There is sufficient evidence at the 0% level to say that there has been a change in the mean time required by an employee to complete the task. 7i) P(A B ) P(A B ) P(B ) P(A B ) P(A B ) P(B ) 0.8 ( 0.6) 0.3 (ii) P(A B) P(A B ) + P(B) Hint to students: A Venn diagram can help you to solve this quickly. A B A B (iii) P(B A) (iv) P(A C) P(B A) P(A) P(A ) P(C) since A and C are independent A and C are independent ( 0.7) (v) P(A B C) 0.5 8i) 3, 4 or 5 4 ways 3 ways ways way st digit nd digit 3rd digit 4th digit 5th digit Probability ! 3 5 (ii) 3 ways ways way or 4 way st digit nd digit 3rd digit 4th digit 5th digit 5

6 Probability 3 5! 0 (iii) Case : The first digit is 3 or 5 3 or 5 3 ways ways way ways st digit nd digit 3rd digit 4th digit 5th digit Probability 3 5! 5 Hint to students: The number of ways of choosing the last digit depends on whether the first digit is 3, 4 or 5. Hence we need to consider the cases separately. Case : The first digit is ways ways way 3 ways st digit nd digit 3rd digit 4th digit 5th digit Probability 3 3 5! 3 0 Hence required probability i) Y X ~ N(400 80, ) N(40, 700) P(Y > X) P(Y X > 0) 0.68 (ii) 0.X Y ~ N( , ) N(4.6,.96) P(0.X Y > 45) 0.34 (iii) Let X, X number of peak-rate calls over the two three-month periods. 0.(X + X ) ~ N(0. ( ), 0. ( )) N(43., 5.9) P(0.(X + X ) > 45) 0.36 Students must remember to square 0. and 0.05 when computing the variance of 0.X Y. 6

7 0i) F 33.9 (ii) 0 36 v (a) r (b) r (iii) Since the product moment correlation coefficient is closer to for (b), the model F c + dv is the better model. (iv) From GC, the regression of F on v is F 0.044v v v 30.7 Since v is the independent or control variable, neither the regression line v on F nor the regression line v on F should be used. i) Let X no. of calls in a period of 4 minutes. X ~ Po(4 3) Po() P(X 8) (ii) Let length of time t. 7

8 Let Y no. of calls in a period of t minutes. Y ~ Po(3t) P(Y 0) 0. e 3t 0. 3t ln 0. t minutes 3 seconds (iii) Let W no. of calls in a working day of hours. W ~ Po( 60 3) Po(60) Since λ 60 > 50, W ~ N(60, 60) approximately. P(W > 00) P(W > 00.5) by continuity correction (iv) Let V no. of busy days out of 6 working days. V ~ B(6, 0.976) P(V ) 0.35 (v) Let U no. of busy days out of 30 working days. U ~ B(30, 0.976) Since n 30 is large, np > 5, nq 4.47 > 5, U ~ N(5.758, ) approximately. P(U < 0) P(U < 9.5) by continuity correction

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

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