2011 H2 Math 9740 Paper 2 - WordPress.com

2011 H2 9740 Paper 2 Solution 1.

Affinity Education Place

Im z

2+5i P

π

P

6+i

4

Re

O

2 + 5i = 29

29 − 3 Maximum | z | = 29 + 3 π The half line arg ( z ) = represents portion of line y = x in Cartesian coordinates and it passes through 4 Minimum | z | =

2 + 2i and 5 + 5i This means possible P is 2 + 2i or 5 + 5i. Maximum distance =

2.

17 .

(i) V = ( 2n − 2 x )( n − 2 x ) x = 2n 2 x − 6nx 2 + 4 x 3

dV = 2n 2 − 12nx + 12 x 2 dx dV = 0. At stationary of V, dx ⇒ 2n 2 − 12nx + 12 x 2 = 0 ⇒ n 2 − 6nx + 6 x 2 = 0

(ii)

⇒x= =

6n ± 36n 2 − 24n 2 12 n 3± 3

(

)

6

However, x < n Therefore, x =

(

n 3+ 3 2

and

(

n 3− 3 6

).

6

)>n. 2

Copyright © AFFIITY Education Place 2011 H2 Math 9740 Paper 2 Solution

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Given f ( x ) = ln ( 2 x + 1) + 3 , x > −

(i) Let y = ln ( 2 x + 1) + 3

Affinity Education Place 1 2

ln ( 2 x + 1) = y − 3 1 y −3 1 e − 2 2  1  −1 Range of f ( x ) is  − ,∞   2  −1 Domain of f ( x ) is ( −∞ ,∞ ) x=

y x=−

1 2

1 x −3 1 e − , x∈ℝ . 2 2

f −1 ( x ) =

y = f ( x)

(ii) Refer to the graph (iii)The graphs of y = f ( x ) and y = f −1 ( x ) intersect along the line y = x.

( 3,0)

Hence the points of intersection satisfy

f ( x) = x

⇒ ln ( 2 x + 1) + 3 = x

y = f −1 ( x )

⇒ ln ( 2 x + 1) = x − 3

(

Solve the equations by GC, 2the x-coordinates are -0.4847and 5.482 (4 s.f.)

e−3 −1 2

O

,0

)

x

( 0,3)

( 0, ) e−3 −1 2

4.

(a) (i)

y=−

1 2

n

2 −2 x ∫0 x e dx n

n  1  =  − x 2 e −2 x  + ∫0 xe −2 x dx  2 0 n

=−

1 2 −2 n  1 −2 x  n1 n e +  − xe  + ∫0 e −2 x dx 2 2  2 0 n

1 1  1  = − n 2 e −2 n − ne −2 n +  − e −2 x  2 2  4 0 1 1 1 1 = − n 2e −2 n − ne −2 n − e −2 n + 2 2 4 4

(ii) As n → ∞ , ne −2 n and n 2 e −2 n → 0 and e −2 n → 0 .


∞

1

2 −2 x ∫0 x e dx = 4

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2011 H2 9740 Paper 2 Solution (b) y =

4x x +1 2

16 x 2

1

Volume of the solid formed = π ∫0

x = tan θ , 1

π ∫0

2

+ 1) π

= 16π ∫04 π

(x

2

+ 1)

2

dx

dx = sec 2 θ = 1 + x 2 dθ

16 x 2

(x


2

dx

tan 2 θ dθ ( sec2 θ )

= 16π ∫04 sin θ dθ 2

π

Volume of the solid formed = 16π ∫04 sin θ dθ 2

π

= 8π ∫04 1 − cos 2θ dθ π

sin 2θ  4  = 8π θ − 2  0  π 1 = 8π  −  = 2π (π − 2 ) unit3.  4 2 5.

X ~ ( µ ,σ 2 )

P ( X < 40 ) = 0.05 ⇒

40 − µ

P ( X < 70 ) = 0.975 ⇒

σ

= −1.64485

70 − µ

σ

= 1.95996

Solving µ − 1.64485σ = 40 and µ + 1.95996σ = 70 µ = 53.7 (3 s.f.) and σ = 8.32 (3 s.f.)

6.

(i) Consider different age groups of residents, such as 6 to 16 years old, more than 16 to 25 years old, more than 25 years old to 35 years old, more than 35 to 45 years old and above 45 years old. The surveyors interview a given number of residents base on their preference. (ii) The selection of interviewees is biased and there is lack of fair representation of residents from different age groups. (iii) Using stratify sampling method. The number of interviewees from each age group is proportional to the total number of residents within the age group. However, this method will not be realistic because it may be difficult to determine the exact number of residents in each age group.


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(i) Any two assumptions: (1) The probability of successfully contact a friend remain constant at 0.7. (2) The n calls are independent. In another words, when the outcome of a call is known this will not affect the outcome of another call. (3) There are only 2 possible outcomes of a call. That is, successfully contact a friend or fail to contact a friend.

(ii) The probability of success in contacting a friend may not be constant for different people. Alternatively, the calls may not be independent. For example, knowing one of his friends cannot be contacted will lead to knowing another friend is not contactable. (iii) Given n = 8. R ~ B(8, 0.7)

P ( R ≥ 6)

= 1 − P ( R ≤ 5 ) = 0.552 (3 s.f.) (iv) n = 40. R ~ B(40, 0.7), p = 0.7 Since np = 28 > 5 and n(1 – p) = 12 > 5, R can be approximated by Normal distribution with mean 28 and variance 8.4. That is, R ~ (28, 8.4) approximately.

P ( R < 25 )

= P ( R < 24.5 ) (by c.c.) = 0.114 ( 3 s.f) 8.

y

(i)

O

x

(ii) For y = c + dx, r = -0.99231 (5 d.p.) Although the | r | is close to 1, the scatter diagram of y on x shows quadratic correlation (or non-linear correlation). Hence, linear correlation model may not be the best to describe the correlation between x and y. y

O

x

2 (iii) For y = a + bx , r = -0.99998 (5 d.p) (a = 22.23029, b = -0.85621)

The new | r | value is closer to 1, hence, it shows that y = a + bx 2 is a better model. Copyright © AFFIITY Education Place 2011 H2 Math 9740 Paper 2 Solution

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2011 H2 9740 Paper 2 Solution


(iv) Equation of the least-squares regression line of y on x2 is y = 22.23029 – 0.85621x2. When x = 3.2, y = 13.5 (3 s.f) 9.

(i) (a) P(a selected lens is faulty) = 0.6 × 0.05 + 0.4 × 0.07 = 0.0580 or

0.05

29 500

Faulty

A 0.6

0.95

(b) P(a selected lens from A | it is faulty)

0.6 × 0.05 = 0.058

Not Faulty Faulty

= 0.517 (3 s.f) or

0.07

0.4

15 29

B 0.93

(ii) (a) P(exactly one of 2 lens selected is faulty)

Not Faulty

 2 1

=   ( 0.058 )(1 − 0.058 ) = 0.109272 = 0.109 (3 s.f) (or state that X ~ B(2, 0.058) where X is the number of faulty lens out of 2 lens selected) (b) P(Two selected lens are both from A | exactly one is faulty) = =

P ( Two lens from A are selected and exactly 1 is faculty

)

P ( exactly 1 out of 2 lens is faulty )

0.6×0.05×0.6×0.95×2 0.109272 0.0342 0.109272

= = 0.313 (3 s.f.)

10. T ~ (38, 25) (i) Let µ be the mean time taken to install a component. Null hypotheses, Ho: µ = 38 Alternate hypotheses, H1: µ < 38 (ii) Level of significance = 5%. Since T follows normal distribution and population variance is known, using ztest.

 

Under Ho, T ~  38,

(

25  . 50 

)

To reject Ho, P T < t ≤ 0.05 .

  t − 38   ⇒ PZ <  ≤ 0.05 . 25   50   t − 38 ⇒ ≤ −1.6449 ⇒ t ≤ 36.83688 . 25 50

{

}

Therefore, the set of values of t is t ∈ ℝ | t ≤ 36.8 ( 3s.f.) .


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2011 H2 9740 Paper 2 Solution


  37.1 − 38   > 0.05 . (iii) Given t = 37.1 . When the null hypothesis is not rejected, ⇒ P Z <  25   n   37.1 − 38 ⇒ > −1.6449 . (no equality sign because P ( Z < −1.6449 ) = 0.05 ) 25 n − 0 .9 n ⇒ > −1.6449 ⇒ n < 83.5 (3 s.f) 5 Therefore, the set of values of n is {n ∈ ℤ | 0 < n ≤ 83} . 11. (i) P (R = 4) =

C418C612 816 = or 0.0941 ( 3 s.f.) C1030 8671

(ii) Given P (R = r) > P (R = r + 1) where r is the most probable number of women to be selected.

Cr18C1012− r Cr18+1C912− r > C1030 C1030 18 ! 12 ! 18 ! 12 ! ⇒ > r ! (18 − r ) ! (10 − r ) ! ( 2 + r ) ! ( r + 1) ! (17 − r ) ! ( 9 − r ) ! ( 3 + r ) !

⇒

⇒ ( r + 1) ! (17 − r ) ! ( 9 − r ) ! ( 3 + r ) ! > r ! (18 − r ) ! (10 − r ) ! ( 2 + r ) ! (shown) Further simplify above inequality, we get ( r + 1)( 3 + r ) > (18 − r )(10 − r ) .

⇒ 32 r > 177 ⇒ r > 5.53 Consider r ( > 5.5) P (X = r) 6 0.306 7 0.233 8 0.096 Therefore, r = 6. 12. (i) Let Y be the number of people joining an airport check-in queue in 4 minutes. Y ~ Po(4.8) P (Y ≥ 8) = 1 – P (Y ≤ 7) = 0.113 (3 s.f.)

 1 .2  t.  60  P (Yt ≤ 1) = 0.7

(ii) Let Yt ~ Po 

⇒ e −0.02 t + 0.02te −0.02 t = 0.7 From GC, t = 54.867 Therefore t = 55 seconds. (to the nearest whole number) (iii) Let X be the number of people leaving the queue in 15 minutes. X ~ Po(27) Since E(X) = 27 > 10, X ~ N(27, 27) approximately. P(X ≤ 11) = P(X < 11.5) (by c.c.) = 0.00143 (3 s.f.) (iv) Poisson distribution may not be valid because number of people check-in is not constant over several hours period. Copyright © AFFIITY Education Place 2011 H2 Math 9740 Paper 2 Solution

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