2012 DSE Maths Paper 1 Q19

19.
(a)(i)
A(1) = 254100
ab² = 254100 ...... [1]

A(2) = 254100
ab⁴ = 307461 ...... [2]

[2] / [1] :
b² = 1.21
b = 1.1 or a = -1.1(rejected)

Put into [1] :
a(1.1)² = 254100
a = 210000

The weight of the goods handled by X in the 4th year
= 210000 × 1.1⁸
= 450154 tonnes (to the nearesttonnes)

(a)(ii)
A(1) = 210000 × 1.1²
A(2) = 210000 × 1.1⁴
......
A(n) = 210000 × 1.1²ⁿ

A(1), A(2), A(3) ...... ,A(n), ...... is an geometric sequence.
First term, a = 210000 × 1.1² = 254100
Common ratio, r = 1.1² = 1.21

The total weight of goods handled by X in the first n years, S(n)
= a (rⁿ - 1) / (r - 1)
= 254100 × (1.21ⁿ - 1) / (1.21 - 1)
= 1210000 × (1.21ⁿ - 1)

(b)(i)
In the same year, m = n - 4
where m ≥ 1 and n ≥ 5

B(m)
= 2abᵐ
= 2 × 210000 × 1.1ᵐ

A(n) / B (n-4)
= (210000 × 1.1²ⁿ) / (2 × 210000 × 1.1ⁿ⁻⁴)
= (1/2) × 1.1ⁿ⁺⁴
≥ (1/2) × 1.1⁵⁺⁴
≈ 1.18
> 1

Since A(n) / B(n - 4) > 1 where n is an integer,
then A(n) > B(n-4)
Hence, I agree.

(b)(ii)
A(1) = 210000 × 1.1²
A(2) = 210000 × 1.1⁴
......
A(n) = 210000 × 1.1²ⁿ

A(1), A(2), A(3) ...... ,A(n), ...... is an geometric sequence.
First term, a = 2 × 210000 × 1.1 = 462000
Common ratio, r = 1.1

The total weight of goods handled by Y in the first n-4 years, S(n-4)
= a (rⁿ⁻⁴ - 1) / (r - 1)
= 462000 × (1.1ⁿ⁻⁴ - 1) / (1.1 - 1)
= 4620000 × (1.1ⁿ⁻⁴ - 1)

A(n) + B(n-4) > 20000000
1210000 × (1.21ⁿ - 1) + 4620000 × (1.1ⁿ⁻⁴ - 1) > 20000000
1210000 × 1.21⁴ × 1.21ⁿ⁻⁴ - 1210000 + 4620000 × 1.1ⁿ⁻⁴ - 4620000 > 20000000
2593742.46 × (1.1ⁿ⁻⁴)² + 4620000 × 1.1ⁿ⁻⁴ - 25830000 > 0

Let u = 1.1ⁿ⁻⁴
2593742.46u² + 4620000u - 25830000 > 0
(u + 4.170)(u - 2.388) > 0
u < -4.170 (rejected for u > 0) or u > 2.388

1.1ⁿ⁻⁴ > 2.388
(n-4) log1.1 > log2.388
n - 4 > 9.133
n > 13.133
Since n is an integer, n = 14

It is in the 14th year.

19(a)(i)

A(1) = ab^2 = 254100 ; A(2) = ab^4 = 307461So, b^2 = 1.21 => b = 1.1Substitute back into A(1) => a = 210000A(4) = 210000 * 1.1^8 = 450153.6501 ~ 450154 tonnes(ii) S(n) = A(1) + A(2) + ... + A(n)= ab^2 + ab^4 + ... + ab^2n= ab^2[1 + b^2 + b^4 + ... b^n]= ab^2 * [(b^n - 1)/(b^2 - 1)]= 1210000 * (1.21^n - 1)(b)(i) Generally speaking m = n - 4 where n >= 5So,at the same year A(n) = ab^(2n), B(m) = B(n - 4) = 2ab^(n - 4)A(n)/B(m) = ab^(2n)/2ab^(n - 4) = b^(n + 4)/2 >= (1.1)^9 / 2 > 1.17So, A(n) > B(m) and the claim is true(ii) By a(ii), S(n) = 1210000 * (1.21^n - 1)T(m) = T(1) + ... + T(2) + ... + T(n - 4)=2ab + 2ab^2 + ... + 2ab^(n - 4)= 2ab[(1 + b + .. + b^(n - 5)]= 2ab * [(b^(n - 4) - 1)/(b - 1)] = 4620000 * [1.1^(n - 4) - 1]Let S(n) + T(m) > 200000001210000 * (1.21^n - 1) + 4620000 * [1.1^(n - 4) - 1] > 200000001210000 * 1.21^n + 4620000 * 1.1^(n - 4) > 258300001.21^n + 2.60787 * 1.1^n > 21.3471071.1^(2n) + 2.60787 * 1.1^n - 21.347107 > 0 1.1^n > 3.49683n > 13.134555So, since X has been operated for 14 years, the new facilities be installed

時間關係，你是有答案的，可否闡明你不明白的地方，這樣會更有效率！！！

2015-04-12 14:44:44 補充：
https://s.yimg.com/rk/HA00430218/o/1116749654.png

2015-04-12 15:10:36 補充：
時間　　　Ａ　　　　Ｂ
　０　　Ａ（０）　－－－
　１　　Ａ（１）　－－－
　２　　Ａ（２）　－－－
　３　　Ａ（３）　－－－
　４　　Ａ（４）　Ｂ（０）
　５　　Ａ（５）　Ｂ（１）
　６　　Ａ（６）　Ｂ（２）
　７　　Ａ（７）　Ｂ（３）

題目問，是否在每一個時間Ａ的值都比Ｂ大。
〔留意Ｂ要在時間４才開始有。〕

即是問是否對於所有 t，
「A(t + 4) > B(t)」都成立呢？

2015-04-12 15:13:48 補充：
即是問，是否 ab^(2t + 8) > 2ab^t
即是問，是否 b^(2t + 8) > 2b^t
即是問，是否 b^(2t + 8 - t) > 2
即是問，是否 b^(t + 8) > 2
即是問，是否 1.1^(t + 8) > 2
即是問，是否 1.1^t × 1.1^8 > 2
即是問，是否 1.1^t × 2.14358881 > 2

答案是 Yes! （因為 t ≥ 0 時左方都大於2）