TPJC JC 2 H2 Maths 2011 Mid Year Exam Solutions

Tampines Junior College (JC 2 H2 Mathematics 9740/2 - Semestral Assessment Two) 1.

 x −  4

The region enclosed by the curve y = xe , the line x = 2 and the x-axis is denoted by R. Find the volume of revolution formed when R is rotated completely about the x-axis, giving your answer in exact form. [4]

Solution: Volume of revolution formed about the x-axis 2

2

= π ∫ y 2 dx = π ∫ xe − x /2 dx 0

0

2

= π  xe − x /2 (−2)  − π ∫ e − x /2 (−2) dx 0 0 2

= π ( −4e −1 ) − π  4e − x /2 

2 0

= π (−4e −1 ) − π (4e −1 − 4) = 4π − 2.

8π e

Functions f and g are defined by f : x  4 − ( x − 2) ,

x ∈ , x > a

g : x → ln x ,

x ∈ , x > 0

2

(i)

State the least value of a for which the function f −1 exists.

[1]

(ii)

Hence find f −1 ( x ) and state the domain of f −1 .

[3]

(iii)

Give a reason why the composite function gf does not exist. Hence, find the largest domain of f for which gf exists.

Solution: (i) (ii)

a=2 Let y = 4 – (x -2)2 x = 2± 4− y Since x > 2 , x = 2 + 4 − y , f -1 (x) = 2 + 4 − x D f-1 = (−∞, 4)

(iii)

R f = (−∞, 4) D g = ( 0, ∞ )

Since R f ⊄ D g , gf does not exist. For gf to exist, R f ⊆ D g i.e. R f = (0, 4) ⇒ domain of f is (2, 4 ) 1

[3]

3.

r2 (r − 1) 2 1 Given that f(r) = r , show that f(r) – f(r+1) = − r. 2 2 r +1 2  (r − 1) 2 1  r +1 − r ∑ 2 r =1  2 n

Hence, find

∞

Deduce the value of

  in terms of n. 

 (r − 1) 2 1  − r +1  . r +2 2   2

∑ r =2

Solution: r 2 (r + 1) 2 f(r) – f(r+1) = r − r +1 2 2 =

2r 2 − (r + 1) 2 2r +1

=

r 2 − 2r − 1 = 2r +1

=

(r − 1) 2 1 − r 2r +1 2

=

2r 2 − r 2 − 2r − 1 2r +1

(r − 1) 2 − 2 2r +1

n  (r − 1) 2 1    = − ∑ [f(r) – f(r + 1)] ∑  r +1 2 r  r =1 r =1  2 = f(1) – f(2) + f(2) – f(3) . . . + f(n – 1) – f(n) + f(n) – f(n + 1) = f(1) – f(n + 1)

n

=  (r − 1) 2 1  ∑  r + 2 − r +1  = 2  r =2  2 ∞

1 (n + 1) 2 − n +1 2 2  (r − 1) 2 1   1 ∑  r + 2 − r +1  −  −  2   4 r =1  2 ∞

1 n  (r − 1) 2 1   1  ∑  r +1 − r  +   n →∞ 2 2  4 r =1  2

= lim = lim

n →∞

1 2

 1 (n + 1) 2  1 1 (n + 1) 2 = since → 0 as n → ∞ − + 2 n +1  n +1 2 4 2 2   2

[7]

4.

ax 2 + bx + c The curve C has equation y = , where a, b and c are constants. It is given x +1 that y = x + 2 is an asymptote of C. (i)

Find the values of a and b.

[2]

(ii)

Given also that C has a turning point at x = 1, find the value of c.

[3]

(iii)

Find, algebraically, the set of values of y for which there are no points on C.

[4]

Solution: (i)

(ii)

(iii)

ax 2 + bx + c c −b+ a y= = ax + (b − a) + x +1 x +1 Given asymptote is y = x + 2 ⇒ y = x + 2 = ax + (b − a ) a = 1, 2 = b – a ⇒ b = 3 dy −(c − b + a ) = a+ dx ( x + 1) 2 c−2 = 1− ( x + 1) 2 dy When x = 1, = 0, dx c−2 1 ⇒= ⇒ = c 6 4 x 2 + 3x + 6 y= x +1 ⇒ yx + y = x 2 + 3 x + 6 ⇒ x 2 + (3 − y ) x + 6 − y = 0 For no real x, D < 0 ⇒ (3 − y ) 2 − 4(6 − y ) < 0 ⇒ 9 − 6 y + y 2 − 24 + 4 y < 0 ⇒ y 2 − 2 y − 15 < 0 ⇒ ( y − 5)( y + 3) < 0 ⇒ −3< y < 5 Solution set is {y ∈  : −3 < y < 5}

3

5.

The equations of two planes p 1 and p 2 are 2 x + 2 y − z = 6 and −2 x − 2 y + z = 18 respectively. The points A and B have position vectors 2i + k and 3i + j + 4k respectively. (i)

Find the position vector of the foot of the perpendicular from A to p 1. Hence find the perpendicular distance from A to p 1. [4]

(ii)

A right cylinder has one of its circular ends contained in p 1 and the other in p 2. Find the height of the cylinder. [2]

(iii)

Find the acute angle between the line AB and the plane p 2

(iv)

The plane p 3 contains the origin and the points A and B. Find the line of and p 3. intersection between p1 [4]

Solution:

(i)

2   6 p1 : r ⋅  2  =  −1    −2  p 2 : r ⋅  −2  =18 1   2 →   → = OA = 0  , OB 1  

 3   1  4  

Let foot of perpendicular from A be N. 2 + 2µ   2 2 →          AN / / n1 ⇒ AN = µ n1 ⇒ ON −  0 = µ  2  ⇒ ON = 2µ   ~ ~ 1  −1  1− µ         2 + 2µ   2  1 Since N lies on p 1 ,  2 µ  ⋅  2  = ⇒µ= 6 3  1 − µ   −1    

4

[3]

8 → 3 ∴ ON =  23   2  3

Perpendicular distance from A to p 1 = AN 8 2  2  → →  3     3  AN =  23  −  0  =  23  ⇒ AN =AN =1  2   1   1   3    − 3

6 (ii)

(iii)

Height of cylinder =

22 + 22 + 12

+

18 22 + 22 + 12

= 8

 3  2 1  →        AB = OB − OA =  1  −  0  =  1   4   1   3       1  2      1 ⋅ 2   3   −1 1     θ θ 5.8 sin= = ⇒= 11 9 3 11

(iv)

 2   3   −1          normal to p 3 is OA × OB =  0  ×  1  =  −5  1  4  2         −1    Equation of p 3 : r ⋅  −5  =0  2  

i.e. − x − 5 y + 2 z = 0

To find line of intersection: 2x + 2 y − z = 6 − x − 5 y + 2z = 0  2 2 −1 6  The augmented matrix is    −1 −5 2 0  5

rref

→

1 15   1 0 − 8 4     0 1 −3 − 3    8 4 

⇒

1 15 x− z = 8 4 3 3 − y− z = 8 4

Let z = λ 15 1 + λ 4 8 3 3 y =− + λ 4 8

x=

 x   ∴  y = z  

 15   4  1    − 3  + γ  3    4 8     0    

 15   4  1   3  Hence, r = − + γ  3  , γ ∈  is the line of intersection of p 1 and p 3 .  4 8     0    

6.

In a presidential election, the governing party wishes to find out whether the citizens will vote for Candidate A or Candidate B. A list of households, in order of surnames, in a particular estate is generated and every 50th household will be picked and interviewed to get their response. (a)

Name the type of sampling method used and explain why the sample is biased. [2]

(b)

Suggest a better sampling method, stating your reason.

[2]

Solution: (a)

Systematic sampling. The sample is biased as only households in a particular estate are interviewed.

6

(b)

7.

A better sampling method is stratified sampling as it gives a good representative of the population. [The population in the country who are eligible to vote can be divided into non-overlapping strata such as race, income and age group. Citizens from each stratum will be randomly selected to give a better representation of the population.]

Three cards are to be drawn without replacement from a pack of ten cards numbered 11 to 20. Find the probability that (a)

all 3 cards drawn have consecutive numbers,

[3]

(b)

every card drawn has an even number on it, given that the product of the three numbers drawn is even. [3]

Solution: (a)

{11,12,13}, {12,13,14}, {13,14,15}, {14,15,16}, {15,16,17}, {16,17,18}, {17,18,19}, {18,19,20} 8 1 Probability = 10 = c3 15 5

c3 1 c3 1 P (all even) Probability = = = 12 = 5 11 11 c 1 − P (all odd ) 1 − 10 3 c3 12 10

(b)

8.

Blood pressure is the pressure exerted by circulated blood onto the walls of blood vessels. During every heartbeat, blood pressure varies between a maximum (systolic) and a minimum (diastolic) pressure. The table shows the systolic and the diastolic pressure, in suitable units, of a patient taken at two hourly intervals during a day.

Reading

1

2

3

4

5

6

7

8

9

10

11

12

Systolic pressure (x)

92

103

103

104

114

120

122

125

135

142

142

a

Diastolic pressure

70

68

62

60

78

64

72

74

96

81

90

b

7

(y)

It is given that the equations of the estimated least square regression line of y on x and of x on y for the data are y = −13.66 + 0.7539 x and = x 47.64 + 0.9492 y respectively. (i)

Giving your answers correct to the nearest integer, show that a = 161 and find the value of b. [4]

(ii)

Calculate the linear product moment correlation coefficient and state what this indicates about the relation between x and y over the data range. [3]

(iii)

Obtain an estimate of the systolic pressure when the diastolic pressure is 110 units. Comment on the reliability of the estimate, giving a reason. [3]

Solution: y= −13.66 + 0.7539 x

(i)

= x 47.64 + 0.9492 y

(1) (2)

Solving (1) and (2), = x 121.92 = and y 78.256 1302 + a = x ⇒ = a 121.92 ×12 − 1302 = 161.04 = 161 (nearest integer) 12 815 + b = y ⇒= b 78.2558 ×12 − 815 = 124.06 = 124 (nearest integer) 12 (ii)

r 2 = bd = ⇒ r 2 (0.7539)(0.9492) = 0.71560 ⇒= r 0.84593 = 0.846 (3 s.f.)

Alternative: Using G.C., r = 0.84594 = 0.846 (3 s.f.) The value is close to 1 indicating a strong positive linear correlation between x and y over the data range. (iii)

When y = 110, x= 47.64 + 0.9492 y = 47.64 + 0.9492(110) = 152.052 = 152 (nearest integer) The estimate is reliable since y = 110 lies within the data range of y values given.

8

9.

(a)

A motorist records the time taken, T, minutes, to drive a particular stretch of road on each of 64 occasions. Her results are summarized by

∑ t = 876.8,

(b)

∑t

2

= 12657.28 .

(i)

Calculate unbiased estimates of population mean and variance.

(ii)

Test, at the 5% significance level, whether the mean time for the motorist to drive the stretch of road is greater than 13.1 minutes. Explain whether it is necessary to use the Central Limit Theorem in your test. [7]

A machine is set to fill the phials containing a drug with a mean volume of 14 ml and the volume dispensed is normally distributed. Random samples of 6 phials were selected and the volume of drug in each was measured in milliliters, as shown below. 13.1

14.4

15.0

13.4

16.1

17.4

Using a 10% significance level, test whether the mean volume dispensed by the machine is 14 ml, given that the population standard deviation of the volume dispensed is unknown [4] Solution: (a)

t ∑ =

876.8 = 13.7 n 64 1  (876.8) 2  Unbiased estimate of population variance, s2 = − = 12657.28 10.24 63  64 

(i)

Unbiased estimate of population mean, = t

(ii)

Let T be the time to drive along a stretch of road. H0 : µ = 13.1 H1 : µ > 13.1 Perform 1-tailed test at 5% level of significance. 10.24   T ~ N 13.1,  under H0 64   T −µ Test Statistic: Z = ~ N(0,1) s/ n 13.7 − 13.1 Z test = = 1.5 10.24 / 64 Using G.C., µo = 13.1, t = 13.7, s 2 = 10.24, n = 64, Z test = 1.5, p − value = 0.0668 Since p-value = 0.0668 > 0.05, we do not reject H 0 and conclude that there is insufficient evidence at 5% level of significance that the mean time is greater than 13.1 mins. Yes, it is necessary to use Central Limit Theorem since distribution is not normal. 9

(b)

Let X be the the volume of drug in a phial. H0 :

µ = 14

H1 :

µ ≠ 14

Perform a 2-tailed test at 10% level of significance Use t-test as n = 6 is small and σ2 is unknown. Test Statistic:

= t

T=

X − 14 ~ t (5) under H0 S/ 5

14.9 − 14 = 1.345 1.6395 / 5

Using t-test, by GC, n = 6, x = 14.9 , s = 1.6395, t = 1.345, p-value = 0.2365 Since p-value = 0.2365 > 0.10, we do not reject H0 . We conclude that there is insufficient evidence, at the 10% significance level, to support the claim that the mean volume dispensed by the machine in not 14 ml.

10.

On average 2% of a particular model of laptop computer are faulty. Faults occur independently and randomly. (i)

Find the probability that exactly 1 of a batch of 10 laptops is faulty.

(ii)

State the conditions under which the use of a Poisson distribution is appropriate as an approximation to a Binomial distribution. [1]

(iii)

A school buys a batch of 150 of these laptops. Use a suitable approximation to find the probability that

(iv)

[2]

(a) there are no faulty laptops in the batch,

[3]

(b) there are more than the expected number of faulty laptops in the batch.

[3]

A large company buys a batch of 2000 of these laptops for its staff. (a) State the exact distribution of the number of faulty laptops in this batch.

[1]

(b) Use a suitable approximation to find the probability that there are at most 50 faulty laptops in this batch. [4] Solution: 10

Let X be the number of faults. (i)

X ~ B(10, 0.02) P(X = 1) = 0.16675 = 0.167 (3 s.f.)

(ii)

The number of faults is independent. n is large, p is small and np < 5

(iii)

(a)

X ~ B(150, 0.02) Since n = 150 is large, p = 0.02 is small, np = 150(0.02) = 3 < 5, X ~ P 0 (3) approximately P(X = 0) = 0.049787 = 0.0498 (3 s.f.)

(b)

expected number = 3 P(X > 3) = 1 – P(X ≤ 3) = 0.352768 = 0.353 (3 s.f.)

(iv)

(a)

X ~ B(2000, 0.02)

(b)

Since n = 2000 is large, p = 0.02 is small, np = 2000(0.02) = 40 > 5, n(1 – p) = 2000(1 – 0.02) = 1960 > 5, np(1 – p) = 39.2, X ~ N(40, 39.2) approximately c .c .

P(X ≤ 50) = P(X < 50.5) = 0.953

11.

In the vegetable section of a supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 4. The weights of unprepared leeks are modelled by the random variable X which has the Normal distribution with mean 260 grams and standard deviation 24 grams. The prepared leeks have had 40% of their weight removed, so that their weights, Y, are modelled by Y = 0.6X. (i)

Find the probability that a randomly chosen prepared leek weighs more than 175 grams. [3]

(ii)

The probability that the total weight of a randomly chosen pack of 4 prepared leeks is greater than k is 0.975. Find the value of k, giving your answer to 1 decimal place. [4] 11

One Sunday, Mary decided to make soup for dinner using 3 unprepared leeks and 2 onions. The weights of onions are modelled by the Normal distribution with mean 150 grams and standard deviation 18 grams. (iii)

Find the probability that the total weight of the ingredients is more than 1000 grams. [3]

A large consignment of unprepared leeks is delivered to the supermarket. A random sample of 100 of them is taken and placed in the vegetable section. (iv)

Find the probability that the mean weight of the leeks from this consignment differs from the weight of an onion by at most 80 grams. [5]

Solution: X ~ N(260, 242) (i)

E(Y) = E(0.6X) = 0.6E(X) = 156 Var(Y) = Var(0.6X) = 0.62 Var(X) = 207.36 Y ~ N(156, 207.36) P(Y > 175) = 0.093510 = 0.0935 (3 s.f.)

(ii)

Y1 + Y2 + Y3 + Y4 ~ N(624, 829.44) P( Y1 + Y2 + Y3 + Y4 > k ) = 0.975 k − 624   P Z < 0.025 = 829.44   k − 624 = −1.95996 829.44 k = 567.55 k = 567.6 (1 d.p.)

(iii)

Let W be the weight of onions. W ~ N(150, 182) X 1 + X 2 + X 3 + W1 + W2 = S ~ N(1080, 2376) P(S >1000) = 0.94962 = 0.950 (3 s.f.)

(iv)

 242  n = 100, X ~ N  260,  100   12

X − W ~ N(110, 329.76)

(

)

P | X − W |≤ 80= P(-80 ≤ X − W ≤ 80) = 0.0493

End of Paper

13