F.4 mathematics M1 question

YTC

2014-06-09 6:53 am

1. Let y be the amount (in suitable units) of suspended particulate in a laboratory.
It is given that
(E): y= 340/[2+ e^(-t) -2e^(-2t)] when t≥0
where t is the time (in hours) which has elapsed since an experiment started.
a) will the value of y exceed 171 in the long run? Justify your answer.
b) find the greatest value and least value of y.
ci) rewrite (E) as a quadratic equation in e^(-t)
ii) it is known that the amounts of suspended particulate are
the same at the time t=α and t=3-α. Given that 0≤α<3-α, find α.

Please help! Thank you

回答 (1)

用戶10012

2014-06-09 5:38 pm

✔ 最佳答案

(a) When t tends to infinity, e^(-t) and e^(-2t) tend to zero, y = 340/2 = 170.
So y cannot exceed 171 in the long run.
(b)
dy/dt = 340[ - e^(-t) + 4e^(-2t)]/[ 2 + e^(-t) - 2e^(-2t)]^2
Put dy/dt = 0
- e^(-t) + 4e^(-2t) = 0
e^(-t) [ - 1 + 4e^(-t)] = 0
t = infinity or ln 4.
when t = 0, y = 340
when t = ln 4, y = 340/(2 + 1/4 - 2(1/4)^2) = 340/(2 + 1/4 - 1/8) = 160
when t = infinity, y = 170
so greatest value of y is 340, least value of y is 160.
(c)
(i) Let e^(-t) = x
y = 340/(2 + x - 2x^2)
2y + yx - 2yx^2 = 340
the quadratic equation is :
2yx^2 - yx + 340 - 2y = 0
(ii)
For the 2 roots to be e^(-α) and e^(-(3- α))
Product of roots = e^(-α) x e^(-3 + α) = e^(-3) = (340 - 2y)/2y
e^(-3) + 1 = 340/2y
y = 340/2[ e^(-3) + 1] = 161.9376
The quadratic equation becomes :
323.8752x^2 - 161.9376x + 16.1248 = 0
x = 0.137248 or 0.362752
= e^(-1.014) or e^(-1.986) = e^(- (3 - 1.014))
so α = 1.014

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