2009 GCE A Level H1 Mathematics Solution

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Alicia Short
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1 2009 GCE A Level H1 Mathematics Solution 1) x + 2y = 3 x = 3 2y Substitute x = 3 2y into x 2 + xy = 2: (3 2y) 2 + (3 2y)y = y + 4y 2 + 3y 2y 2 = 2 2y 2 9y + 7 = 0 (2y 7)(y 1) = 0 y = 7 2, 1 x = 4, 1 Students have a wide choice of substitutions to use: From equation (1): x = 3 2y From equation (1): y = 3 x 2 From equation (2): y = 2 x2 x All these substitutions work equally well. 2i) y = x/2 y = x (0, 0) (4, 2) (ii) 1 2 x dx x dx = x3/2 3/2 + c = 2 3 x3/2 + c = 1 4 x2 + c (iii) Area = 0 4 x 1 2 x dx = [ 2 3 x3/2 1 4 x2 ] 4 0 = Even though the integration has to be done manually because the question specifies Without using a calculator, it is helpful to use the GC to check that the answer obtained is numerically correct. = 4 3 3i) Let y = e x. Then x = ln y h(x) = f 1 (x + 2) = ln (x + 2) 1

2 (ii) x = 2 (0, ln 2) ( 1, 0) (iii) ln (x + 2) = x + 2 From GC, x = i) The y-axis is a vertical asymptote to the curve. Do not draw the curve as if it touches ( 1, 0) the y-axis, even though it may appear so on the GC in some views. Zoom in to get a (1, 0) better view. (ii) dy dx = x 2 When x = 2, dy dx = 5 4. Therefore gradient of normal is 4 5. (iii) When x = 2, y = 3 2. Therefore equation of normal is y 3 2 = 4 5 (x 2) y 15 = 8(x 2) 8x + y 31 = 0 (iv) When x = 0, y = So N = (0, ). 2

3 T N P It is helpful to draw a diagram even though it is not asked for in the question. This will help you to see how the area of triangle PTN can be calculated. Equation of tangent at P is y 3 2 = 5 4 (x 2) i.e. 4y 6 = 5x 4y = 5x 4 When x = 0, y = 1. So T = (0, 1). NT = ( 1) = Area of triangle PTN = = 41 5i) dy dx dy dx = 6x2 x 4 = 6x 2 x 4 = 2(3x 2 5x 2) = 2(3x + 1)(x 2) = 0 x = 1 3, 2 y = 0 27, 9 coordinates of stationery points = ( 1 3, 0 27 ) and (2, 9). The derivative have to be computed manually since the questions requires exact coordinates. However it is helpful to use the GC to check that the answers are numerically correct. (ii) ( 1 3, 0 27 ) ( 1, 0) (0, 3) ( 1 2, 0) (3, 0) (iii) 2x 3 5x 2 4x + 3 > 0 1 < x < 1 2 or x > 3 (2, 9) 2e 3x 5e 2x 4e x + 3 > 0 3

4 1 < e x < 1 2 or ex > 3 e x < 1 2 or ex > 3 x < ln 1 2 or x > ln 3. 6i) = 0.14 (ii) P(call for A and A is in office) + P(call for B and B is in office) + P(call for C and C is in office) = = = 0.72 (iii) P(call is for C researcher being called not in office) P(call is for C and C is not in office) = P(researcher being called not in office) = = = i) P(A B) = P(A) + P(B) P(A B) = P(A B) P(A B) = = 1 6 (ii) P(A) P(B) = = = P(A B). Hence A and B are not independent. (iii) Method 1: A B P(A B) = = 5 6 Drawing a Venn diagram is the preferred way of solving such problem. 4

Method 2: P(A B) = P(A ) + P(A B) = 1 P(A) + P(A B) = 1 1 3 + 1 6 = 5 6 8i) Let X = lifetime of a component. X ~ N(120, 18 2 ) P(X > 144) = 0.091211 = 0.0912 (ii) 2! P(X > 144) P(X < 144) = 2 0.

5 Method 2: P(A B) = P(A ) + P(A B) = 1 P(A) + P(A B) = = 5 6 8i) Let X = lifetime of a component. X ~ N(120, 18 2 ) P(X > 144) = = (ii) 2! P(X > 144) P(X < 144) = ( ) = It is necessary to multiply by 2! since there are 2! ways of arranging these 2 components. H 0 : µ = 120 H 1 : µ > 120 Since p value = > 0.05, we do not reject H 0. There is insufficient evidence at 5% level to say that the mean lifetime is longer than 120 days. It is necessary to give the conclusion in the context of the problem, instead of a generic conclusion. 9i) y Students are reminded to label and indicate 18 the scale on the axes. It is advisable to draw the scatter diagram to scale and to copy what appears on the screen of the GC as closely as possible. 15 (ii) x 5

r = 0.931. The scatter diagram shows the data lying close to a straight line. This agrees with the value of r which is close to 1. (iii) Regression line of y on x is y = 0.0123x + 15.

6 r = The scatter diagram shows the data lying close to a straight line. This agrees with the value of r which is close to 1. (iii) Regression line of y on x is y = x (iv) Estimated weight = (135) = 17.2 kg (v) Since y = 20 is outside the range of the data values, it is unsuitable to use the equation in part (iii) is the regression line of y on x to estimate the amount of liquid nutrient. i) Let X = no. of students out of who failed the piano examination. X ~ B(, 0.2) P(X = 2) = (ii) Let Y = no. of students out of who will be awarded distinction. Y ~ B(, ) = B(, 0.12) P(Y < 2) = P(Y 1) = (iii) Let W = no. of students out of 50 who failed the examination. W ~ B(50, 0.2) Since n = 50 is large, np = > 5, nq = 40 > 5, W ~ N(, 8) approximately. P(W 12) Some students mix up the use of nq and npq. Also it is common for students to forget to do continuity correction. 6

= P(W 12.5) by continuity correction 0.812 11ai) 72 8 = 9. We first determine the sampling interval = 72 8 = 9. Then choose random starting number between 1 to 9 (inclusive), e.g. 5.

7 = P(W 12.5) by continuity correction ai) 72 8 = 9. We first determine the sampling interval = 72 8 = 9. Then choose random starting number between 1 to 9 (inclusive), e.g. 5. Then we sample claim no. 5, 14, 23, 32, 41,... (ii) Systematic sampling is a better indication since the first 8 claims received may all come from the same area. (bi) x = = s = 119 [ ] = Teaching Point: The formula for the unbiased estimate of the variance is found in MF15. There is no need for students to memorise it. (ii) A sample statistic T is an unbiased estimate of a population parameter θ if E(T) = θ. (iii) H 0 : µ = 00 H 1 : µ 00 H 0 is rejected p value = < α 0 α > a) Let X = mass of a plum. X ~ N(µ, σ 2 ) P(X < 22) = 30% 7

8 P(Z < 22 µ ) = 0.3 σ 22 µ = σ 22 µ = σ (1) P(X > 29) = 20% P(X < 29) = 80% P(Z < 29 µ ) = 0.8 σ 29 µ = σ 29 µ = σ (2) (2) (1) 7 = σ σ = = 5.12 µ = 24.7 (b) Let A, N = mass of an apple and a nectarine respectively. A ~ (0.15, ) N ~ (0.07, ) (i) A 1 + A 2 N 1... N 4 ~ N( , ) ~ N(0.02, ) P(A 1 + A 2 > N N 4 ) = P(A 1 + A 2 N 1... N 4 > 0) = (ii) Let Y = total cost of 2 apples & 4 nectarines. E(Y) = = 6.06 Var(Y) = = P(5 < Y < 6) = There is no need to square 2 and 4 when computing variance since we are dealing with sums of normal variables. Students need to square 9 and 12 when computing variance since we are dealing with the multiples of normal variables. 8

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

2010 GCE A Level H2 Maths Solution Paper 2 Section A: Pure Mathematics. 1i) x 2 6x + 34 = 0 6 ± x = 2 00 GCE A Level H Maths Solution Paper Section A: Pure Mathematics i) x 6x + 34 0 6 ± 36 36 x 6 ± 0i 3 ± 5i (ii) Since the coefficients are all real, another root of the equation is x i. [ x ( + i) ] [