2009 GCE ‘A’ Level H1 Mathematics Solution (Contributed by Hwa Chong Institution)
1) x + 2y = 3 Þ x = 3 – 2ySubstitute x = 3 – 2y into x2 + xy = 2:
(3 – 2y)2 + (3 – 2y)y = 2
9 – 12y + 4y2 + 3y – 2y2 = 2
2y2 – 9y + 7 = 0
(2y – 7)(y – 1) = 0
y = , 1
x = –4, 1 / Hint to students:
Students have a wide choice of substitutions to use:
· From equation (1): x = 3 – 2y
· From equation (1): y =
· From equation (2): y =
All these substitutions work equally well.
2i)
(ii) ò dx
= + c
= x3/2 + c
ò x dx
= x2 + c
(iii) Area = – x dx
= [ x3/2 – x2 ]
= – 4
= / Hint to students:
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.
3i) Let y = ex. Then x = ln y
h(x) = f –1 (x + 2) = ln (x + 2)
(ii)
(iii) ln (x + 2) = –x + 2
From GC, x = 0.926
4i)
/ Hint to students:
The y-axis is a vertical asymptote to the curve. Do not draw the curve as if it touches the y-axis, even though it may appear so on the GC in some views. Zoom in to get a better view.
(ii) = 1 +
When x = 2, = .
Therefore gradient of normal is – .
(iii) When x = 2, y = .
Therefore equation of normal is
y – = – (x – 2)
10y – 15 = –8(x – 2)
8x + 10y – 31 = 0
(iv) When x = 0, y = . So N = (0, ).
Equation of tangent at P is y – = (x – 2)
i.e. 4y – 6 = 5x – 10
4y = 5x – 4
When x = 0, y = –1. So T = (0, –1).
NT = – (–1) =
Area of triangle PTN = ´ 2 ´
= / Hint to students:
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.
5i) = 6x2 – 10x – 4
= 6x2 – 10x – 4
= 2(3x2 – 5x – 2)
= 2(3x + 1)(x – 2) = 0
x = – , 2
y = , –9
\coordinates of stationery points = (– , ) and (2, –9). / Hint to students:
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)
(iii) 2x3 – 5x2 – 4x + 3 > 0
Þ –1 < x < or x > 3
2e3x – 5e2x – 4ex + 3 > 0
Þ –1 < ex < or ex > 3
Þ ex < or ex > 3
Þ x < ln or x > ln 3.
6i) 0.2 ´ 0.7 = 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)
= 0.14 + 0.3 ´ 0.6 + 0.5 ´ 0.8
= 0.14 + 0.18 + 0.4
= 0.72
(iii) P(call is for C | researcher being called not in office)
=
=
=
= 0.357
7i) P(A È B) = P(A) + P(B) – P(A Ç B)
= + – P(A Ç B)
P(A Ç B) = + – =
(ii) P(A) P(B) = ´
= ¹ = P(A Ç B).
Hence A and B are not independent.
(iii) Method 1:
P(A¢ È B) = + = / Hint to students:
Drawing a Venn diagram is the preferred way of solving such problem.
Method 2:
P(A¢ È B) = P(A¢) + P(A Ç B)
= 1 – P(A) + P(A Ç B)
= 1 – +
=
8i) Let X = lifetime of a component.
X ~ N(120, 182)
P(X > 144) = 0.091211 = 0.0912
(ii) 2! P(X > 144) P(X < 144)
= 2 ´ 0.091211 ´ (1 – 0.091211)
= 0.166 / Hint to students:
It is necessary to multiply by 2! since there are 2! ways of arranging these 2 components.
H0 : m = 120
H1 : m > 120
Since p–value = 0.058 > 0.05, we do not reject H0 . There is insufficient evidence at 5% level to say that the mean lifetime is longer than 120 days. / Hint to students:
It is necessary to give the conclusion in the context of the problem, instead of a generic conclusion.
9i)
/ Hint to students:
Students are reminded to label and indicate 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.
(ii)
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 + 15.5
(iv) Estimated weight
= 0.012329(135) + 15.4866
= 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.
10i) Let X = no. of students out of 10 who failed the piano examination.
X ~ B(10, 0.2)
P(X = 2) = 0.302
(ii) Let Y = no. of students out of 10 who will be awarded distinction.
Y ~ B(10, 0.15´0.8) = B(10, 0.12)
P(Y < 2)
= P(Y £ 1)
= 0.658
(iii) Let W = no. of students out of 50 who failed the examination.
W ~ B(50, 0.2)
Since n = 50 is large, np = 10 > 5, nq = 40 > 5, W~ N(10, 8) approximately.
P(W £ 12)
= P(W £ 12.5) by continuity correction
» 0.812 / Hint to students:
Some students mix up the use of nq and npq.
Also it is common for students to forget to do continuity correction.
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.
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) = + 1000 = 1044.33
s2 = [ 8 282 000 – ] = 67614.7 / 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 q if E(T) = q.
(iii) H0 : m = 1000
H1 : m ¹ 1000
H0 is rejected
Þ p–value = 0.0618 <
Þ a > 6.18.
12a) Let X = mass of a plum.
X ~ N(m, s2)
P(X < 22) = 30%
P(Z < ) = 0.3
= –0.52440
22 – m = –0.52440s –––(1)
P(X > 29) = 20%
P(X < 29) = 80%
P(Z < ) = 0.8
= 0.84162
29 – m = 0.84162s –––(2)
(2) – (1) Þ 7 = 1.36602s
Þ s = 5.1244 = 5.12
Þ m = 24.7
(b) Let A, N = mass of an apple and a nectarine respectively.
A ~ (0.15, 0.032)
N ~ (0.07, 0.022)
(i) A1 + A2 – N1 –...– N4
~ N(2´0.15 – 4´0.07, 2´0.032 + 4´0.022)
~ N(0.02, 0.0034)
P(A1 + A2 > N1 +...+ N4)
= P(A1 + A2 – N1 –...– N4 > 0)
= 0.634 /
Hint to students:
There is no need to square 2 and 4 when computing variance since we are dealing with sums of normal variables.
(ii) Let Y = total cost of 2 apples & 4 nectarines.
E(Y) = 9´2´0.15 + 12´4´0.07 = 6.06
Var(Y) = 92´2´0.032 + 122´4´0.022 = 0.3762
P(5 < Y < 6)
= 0.419 / Hint to students:
Students need to square 9 and 12 when computing variance since we are dealing with the multiples of normal variables.
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