9740/02 October/November MATHEMATICS (H2) Paper 2 Suggested Solutions. Ensure calculator s mode is in Radians. (iii) Refer to part (i)

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1 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. MTHEMTICS (H) Paper Suggested Solutions. Topic : Functions and Graphs (i) 974/ October/November 8 (iii) (iv) f(x) g(x) <.5 e x sin x (x + x + x3 ) <.5 3 Ensure calculator s mode is in Radians. Refer to part (i).5 < e x sin x (x + x + x3 3 ) <.5 e x sin x (x + x + x3 3 ) +.5 > or ex sin x (x + x + x3 3 ).5 < x <.56 x >.96 (ii) f(x) = e x sin x = ( + x + x + x3 x3 + ) (x + ) 6 6 = x x3 + 6 x + x3 + = x + x + x3 + 3 g(x) = x + x + x3 3 Refer to formula list under Maclaurin s expansion.96 x / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. / 8

2 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version.. Topic : Definite Integrals (i) y = x( x) y = x ( x) 4 rea of R = x ( x) 4 dx = (.49944) = (3 sig. fig.) (ii) Let u = x du = dx When x =, u = When x =, u = Volume = π y dx Reversing limits: b f(x) dx a a = f(x) dx b = π x x dx = π ( u) u = π ( u = π u u 5 = π = π 5 units3 (iii) y = x x y dy dx y dy dx dy dx u 3 ) du du = x ( x) ( ) + x ( x) [ x + ( x)] = = 3x 4 y x To find max point, let dy = dx 3x = 4 y x 3x = x = 3 3. Topic : Complex numbers (polar form) (a) w = re iθ w * = re i(-θ) p = w w r e iθ = r e i( θ) p = e iθ p =, arg (p) = θ p 5 = (e iθ ) 5 = e iθ = cos θ + i sin θ Given that p 5 is positive and real, p 5 = cos θ + i () For a complex number z = x + iy, sin θ = when z is real, imaginary part y = asic = θ = π, 4π θ = π, 4π Given that < θ < π, = π, π < θ < 5π / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. / 8

3 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. (b) (i) 6 = rad (3 d.p.) greatest possible value of arg z = β Greatest arg z occurs at COD = β = CO + O + α = =.9.9 rad (3 d.p.) 4. Topic : Functions and inverse functions (i) Graph of f - is a reflection of graph of f in the line y = x. (ii) cos O = 5 6 O = cos 5 6 = sin OD = 3 5 OD = sin 3 5 =.6435 CO = O = least possible value of arg z = α = OD O = Smallest arg z occurs at OD = α / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 3 / 8

4 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. (ii) Let y = (x 4) + (x 4) = y x 4 = ± y x = ± y + 4 f - (x) = 4 + x, x > D f = (, ) (iv) Equation of the line y = x f(x) = f - (x) = x (x 4) + = x x 8x x = x 9x + 7 = 9+ 3 x = 5. Topic : Sampling x = 9 ± ( 9) 4()(7) = 9 ± 8 68 = 9 ± 3 = 9+ 3 Call an assembly of the 95 pupils. D f = R f or 9 3 (rej.) Common point of intersection of the 3 graphs Start from a randomly selected student and pick every 95 5 = 9th student. s the different pupils belong to different groups and organizations that will use different sports facilities for different activities, a stratified sample put together from random samples taken from each group will provide a more accurate representation of the pupil population without any possible bias towards any groups. 6. Topic : Hypothesis Testing x = x n = 6 5 = 68.4 ( x) n ] S = [ n x 6 = [ ] 4 5 = 56.5 = (.5) X N(78, S ) n X N(78, 56.5 ) 5 H : µ = 78 H : µ 78 From G.C, t =.654 p =. Since p =..5, we do not reject H. We conclude that at the 5% level, there is insufficient of evidence to suggest that the mean mass of calcium in a bottle has changed / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 4 / 8

5 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. 7. Topic : Probability 8. Topic : Correlation coefficient and linear regression (i) From G.C, r =.97 (3 sig. fig.) Since r is close to, this indicates that there is a strong positive linear correlation between x and t. st set nd set 3 rd set.6.4 (i) P( wins in the second set) =.6 (.7) +.4 (.) =.5 (ii) P ( wins the match) =.6 (.7) +.4 (.)(.7) +.6 (.3)(.) = = *Note that this is a conditional probability question. (iii) P( won the first set won the match) = P( wins st set and wins the match) P( wins the match) =.4(.)(.7).5 = from (ii) + (ii) t P(4.8, 7.6) (iii) s the remaining points scattering pattern appears to follow that of a logarithmic data set, the scatter diagram may be modeled by a straight line t = a + b ln x. x / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 5 / 8

6 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. (iv) Using G.C, t = ln x a =.4, b = 4.4 X + Y = total number of pianos sold in a week X + Y Po (.8 +.6) Po (4.4) P(X + Y = 4) =.97.9 (3 sig. fig.) (v) t = ln x When x = 4.8, t = ln 4.8 = 8.3 (3 sig. fig.) (vi) Since x = 8 is out of the given data set s range, the value of t obtained may not be accurate. 9. Topic : Poisson Distribution Let r.v. X = number of grand pianos sold in a week X Po(.8) P(X 4) = P(X 3) =.899 =.87.9 (3 sig. fig.) Remove coordinates of P and ln values of x. Mean number of pianos sold in a year = 5.8 = 9 Let = number of grand pianos sold in a year. Po(9) Since mean >, N(9, 9) approximately P( < 8) = P( < 79.5) [Continuity Correction] = (3 sig. fig.) Use continuity correction to approximate a discrete distribution (i.e Poisson) by a continuous distribution (i.e normal). The sales of the pianos are likely to follow demand trends across the different months of the year. Hence a piano sale should not be considered random event within the interval of a year. Let r.v. Y = number of upright pianos sold in a week, Y Po(.6) / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 6 / 8

7 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version.. Topic : Combinations (i) 3 C 4 C 3 5 C 3 = (ii) 9 C 8 = 9 (iii) 5 C 4 7 C C 5 7 C 3 = = (i) X + X N(5, 8 ) N(, 8) P(X + X > ) = (3 sig. fig.) (ii) X X N(, 8) Choose 8 from the total of 9 from L & M P(X > X + 5) = P(X X > 5) =.94 (3 sig. fig.) Choose from 3 K 3 from 4 L 3 from 5 M 4 from M and 4 from remaining 7 (K + L) OR 5 from M and 3 from remaining 7 (K + L) (iv) C 8 9 C 8 8 C 8 = = 485 C 8 : total no. of ways with no restriction 9 C 8 : group without K. Topic : Normal distributions 8 C 8 : group without L : group without M (impossible as K + L = 7) X = X N(μ, ) E(X + X ) = μ + μ = μ Var(X + X ) = + = E(X X ) = μ μ = Var(X X ) = + = Consider, Y N(μ, ) P(Y < 74) =.668 P( Z < 74 μ ) = μ =.5 () P(Y > 46) =.668 P(Y < 46) =.668 =.933 P(Z < 46 μ ) = μ =.5 () Change to standard normal X N(μ, ) z = X μ = X μ N(, ) () (), / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 7 / 8

8 GCE Level October/November 8 Suggested Solutions Mathematics H (974/) version. From qn, E(X) = 5 7 = 3 = 4 sub = 4 into (), 74 μ =.5(4) μ = μ =, = 4 E(Y) = Var(Y) = 4 = 576 E(aX + b) = ae(x) + b = 5a + b = b = 5a (3) Var(aX + b) = 4 a Var(X) = 576 a (64) = 576 a = 9 a = 3 or 3 (reject) sub a = 3 into (3), a = 3, b = 4 b = 5(3) = / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 8 ϕ exampaper.com.sg. ll rights reserved. 8 / 8