RAFFLES JUNIOR COLLEGE H2 Mathematics 9740 2008 JC2 ...

RAFFLES JUNIOR COLLEGE H2 Mathematics 9740 2008 JC2 (Year 6)

____________________________ Preliminary Exam Paper 2 Solutions 1 [1]

x 2 ( x + 2) 2 − 1 = ( x 2 + 2 x) 2 − 12 = ( x 2 + 2 x + 1)( x 2 + 2 x − 1) = ( x + 1) 2 ⎡⎣( x + 1) 2 − ( 2) 2 ⎤⎦

(

= ( x + 1) 2 x + 1 + 2

[3]

1 ≥ x 2 , x ≠ −2 2 ( x + 2) ⇒ x 2 ( x + 2) 2 ≤ 1 since ( x + 2) 2 > 0

(

⇒ ( x + 1) 2 x + 1 + 2 ⇒

[2]

)( x + 1 − 2 ) (shown)

)( x + 1 − 2 ) ≤ 0

−2

− 1 − 2 ≤ x < −2 or − 2 < x ≤ −1 + 2

−1− 2

−1

−1+ 2

1 ≥ e2 x (e + 2) 2 Replacing x by e x in the above, we get −1 − 2 ≤ e x < −2 or − 2 < e x ≤ −1 + 2 x

(

Since e x > 0 , x ≤ ln −1 + 2

(

)

(

)⎦

Range of values of x is −∞, ln −1 + 2 ⎤ . 2 [2]

(i) To produce $1 of Y, $0.25 of X is needed. ∴ To produce $ y of Y, $0.25 y of X is needed. Similarly, to produce $x of X and $z of Z, $0.1x and $0.2z of X are needed respectively. For product X, Worth of internal provision = $(0.1x + 0.25 y + 0.2 z ) Worth of production = $x Worth of external demand = $ 50000. Since internal provision + external demand = production, we have

0.1x + 0.25 y + 0.2 z + 50000 = x ⇒ 0.9 x − 0.25 y − 0.2 z = 50000 --- (1)

[4]

(ii) Similarly, for products Y and Z, we have and

−0.3 x + 0.6 y − 0.5 z = 75000 --- (2) −0.1x − 0.15 y + 0.9 z = 125000 --- (3)

Solving (1), (2) and (3) using GC, we obtain x = 229921, y = 437795, z = 237402 (to nearest integer) _________ Page 1 of 6

3 [2] [2] [2]

(i) z − a = a

(ii) z + 2a = z

(iii)

⇒ z − (−2a) = z − 0

Im

Let −2a be represented by point B .

A O

1 arg ( z − a ) = arg ( a ) + π 2 Im

Im

a

A

Re

O

Re

B A O

4 [1]

(i)

[2]

(ii)

[3]

(iii)

Re

1 is decreasing for x ∈ (4, ∞ ) . f(x) The graph of y = f '( x ) is below the x -axis for x ∈ ( −∞, 2) ∪ (2, 4) . The graph of y =

y

y 2 = f ( x) √2

y = √2 0

2

x

4

y = −√2

−√2

x=2 [3]

(iv)

y

y = f( x)+ 2 4 (−4,2) −2

x=−2

y=4

2 (4,2) 0

2

x

x=2 _________ Page 2 of 6

5(a) [3]

n

⎡⎣1 + (2 x + 3 x 2 ) ⎤⎦ = 1 + n(2 x + 3 x 2 ) +

n( n − 1) (2 x + 3 x 2 ) 2 + ... + (2 x + 3 x 2 ) n 2

n(n − 1) (4) = 105 2 ⇒ 2n 2 + n − 105 = 0 ⇒ (2n + 15)(n − 7) = 0 15 (Rejected since n ∈ ⇒ n=− 2

Coefficient of x 2 = 3n +

5(b) [4]

+

) or n = 7

−1

⎡ ⎛ x2 x4 ⎞ x2 ⎤ 1− 4x 1 x x = − + = − (1 4 ) 3(1 ) (1 4 ) ⎜1 − + + ... ⎟ ⎢ ⎥ 2 3+ x 3 ⎦ 3 3 9 ⎣ ⎝ ⎠ 2 3 4 1⎛ x 4x x ⎞ ≈ ⎜1 − 4 x − + + ⎟ 3⎝ 3 3 9 ⎠ The expansion is valid for x 2 < 3

⇒ − 3 1)

= 2P( X = 1) [1 − P( X ≤ 1)]

= 0.00263 (3 s.f.) [3]

⎛

(ii) Since n = 50 is large, by Central Limit Theorem, X ∼ N ⎜ 0.15, ie. X ∼ N ( 0.15, 0.003) approximately.

⎝

0.15 ⎞ ⎟ approximately. 50 ⎠

P( X > 0.3) = 0.00309 (3 s.f.) The t -test should be used because the sample size is small and the population variance is unknown.

9 [2]

(i)

[5]

(ii) Let X be the distance thrown by the shot putter, in metres. H 0 : μ = 8.92 vs H1 : μ > 8.92 Perform a 1-tail test at 5% level of significance. Assume that X follows a normal distribution. Under H 0 , T =

X − μ0 ∼ t (n − 1) , where μ0 = 8.92, s = 0.099783, x = 8.983, n = 10 S/ n

Using a t -test, p -value = 0.0385 (3 s.f.) Since the p -value = 0.0385 < 0.05, we reject H 0 and conclude that there is sufficient evidence, at 5% level of significance, that the shot putter has improved. _________ Page 4 of 6

10

Given P(win) =

[1]

(i) P(1 draw and 1 defeat) = P(1 draw, 1 defeat, 2 wins)

1 1 1 and P(defeat) = ⇒ P(draw) = 2 6 3 2

1 1 ⎛ 1 ⎞ 4! = × ×⎜ ⎟ × 3 6 ⎝ 2 ⎠ 2! 1 = (shown) 6 [3]

(ii) P(wins the first match and goes on to win exactly one other match)

1 = × P(wins one other match and loses or draws in the other 2) 2 2 1 ⎡ 1 ⎛ 1 ⎞ 3! ⎤ 3 = ⎢ ⎜ ⎟ ⎥= 2 ⎢⎣ 2 ⎝ 2 ⎠ 2!⎥⎦ 16 [4]

(iii) P(wins exactly one match | obtains four pts)

=

P(wins exactly one match and obtains four points) P(obtains four points)

=

P(1 win, 1 draw, 2 defeats) P(1 win, 1 draw, 2 defeats) + P(4 draws) 2

1 1 ⎛ 1 ⎞ 4! × ×⎜ ⎟ × 2 3 ⎝ 6 ⎠ 2! = 2 4 1 1 ⎛ 1 ⎞ 4! ⎛ 1 ⎞ × ×⎜ ⎟ × + ⎜ ⎟ 2 3 ⎝ 6 ⎠ 2! ⎝ 3 ⎠ 1 9 = 18 = 1 1 11 + 18 81 11 [3]

(i)

Let X be the number of defective articles, out of 10. Then X ∼ B(10, 0.05) .

P( X = 1) = 0.31512 (5 s.f.) = 0.315 (3 s.f.) (shown) [5]

(ii) Let Y be the number of samples, out of 80, that will each contain exactly one defective article. Then Y ∼ B(80, 0.315) . Since n = 80 is large such that np = (80)(0.315) = 25.2 > 5 and

n(1 − p ) = (80)(1 − 0.315) = 54.8 > 5 , Y ∼ N ( 25.2,17.262 ) approximately P(25 ≤ Y < 35) = P(24.5 < Y < 34.5) (by continuity correction) = 0.554 (3 s.f.) _________ Page 5 of 6

12 (a)

[4]

Given that the diameter of a randomly chosen lid is X cm and the diameter of the top of a randomly chosen container into which the lid is pressed is Y cm, where X ∼ N(12.1, 0.032 ) and Y ∼ N(12.0, 0.032 ) .

X 1 + X 2 + X 3 + 2Y ~ N( 3 × 12.1 + 2 × 12.0, 3 × 0.032 + 22 × 0.032 ) ie. X 1 + X 2 + X 3 + 2Y ~ N(60.3,0.0063)

P( X 1 + X 2 + X 3 + 2Y < 60.5) = 0.994 (3 s.f.) (b) [3]

(i) X − Y ∼ N(0.1, 2 × 0.032 ) ie. X − Y ∼ N(0.1, 0.0018) Probability that a pairing is accepted = P(0.02 < X − Y < 0.17) = 0.921 (3 s.f.)

[2]

(ii) Number of trials required to obtain 2000 accepted pairings

= 13 [1] [3]

2000 ≈ 2171.9 = 2172 (nearest integer) or 2170 (3 s.f.) 0.92085

(i) Value of the product moment correlation coefficient is −0.807 (3 s.f.). (ii)

y The value found in (i) suggests there is some form of negative linear correlation between x and y but the scatter diagram shows that the relationship between x and y is non-linear. Thus, the value of the product moment correlation coefficient is not a good indication of the linearity between x and y.

x [2]

(iii) From the scatter diagram, we see that y decreases as x increases, which is the case for

b , b > 0 ). x For models A ( y = a + bx 2 ) and C ( y = a + b ln x ), y increases as x increases, since b > 0 . Hence model B is appropriate.

model B ( y = a +

[1]

(iv) Value of the product moment correlation coefficient between y and w is 0.976 (3 s.f.).

[3]

(v) Equation of the regression line of y on w is y ≈ 0.86652 + 3.2758w (5 s.f.), i.e. y = 0.867 + 3.28w (3 s.f.). When y = 1.4 , 1.4 ≈ 0.86652 + 3.2758w ⇒ w ≈ 0.16285 ⇒ x ≈ 6.14 ≈ 6 . An estimate of the number of pens ordered when a unit price of $1.40 is quoted is 6. This estimate is highly reliable since y = 1.4 is between the range of values of y from 0.80 to 1.50, and the product moment correlation coefficient between y and w is 0.976 (3 s.f.), which suggests a strong positive linear correlation between y and w. _________ Page 6 of 6