數學--需簡單解題過程

9(i) you may use the website http://fooplot.com/ to plot the graph

(ii) 4k^x = 20k^2

log4 + xlogk = log20 + 2logk

xlogk = log5 + 2logk

x = log5/logk + 2

x = log_k(5) + 2

(iii)(a) ∫ 4k^x dx

~ (1/4)(4 + 8k^0.5 + 4k)

(b) (1/4)(4 + 8k^0.5 + 4k) = 16

2k^0.5 + k = 15

Let y = k^0.5

y^2 + 2y - 15 = 0

(y + 5)(y - 3) = 0

y = 3 or -5

k^0.5 = 3

k = 9


2011-12-17 12:01:41 補充：
http://en.wikipedia.org/wiki/Trapezoidal_rule

你入黎淨係留句問老師
咁你入黎做咩

9i.
intersects with x-axis, => put y = 0 into y = 4k^x
you finally cannot find any intersection with x-axis
intersects with y-axis, => put x = 0 into y = 4k^x
you finally can find the intersection point (0,4) with y-axis.

9ii.
y = 4k^x
y = 20k^2
4k^x = 20k^2
k^x = 5k^2
logk (k^x) = logk (5k^2)
x logk k = logk 5 + logk (k^2)
x logk k = logk 5 + 2 logk k
x = logk 5 + 2

9iiia.
when x = 0 => y = 4k^0 = 4
when x = 1/2 => y = 4k^(1/2) or 4sqrt(k)
when x = 1 => y = 4k^1 = 4k
by trapezium rule
Int0>1 [4k^x] dx (imagine it is the area bounded by y=4k^x and x-axis from x=0 to x=1)
= (1/2)[4+4sqrt(k)]/2 + (1/2)[4sqrt(k)+4k]/2
= [1+sqrt(k)] + [sqrt(k)+k]
= 1 + 2sqrt(k) + k
= [1 + sqrt(k)]^2

9iiib.
16 = [1 + sqrt(k)]^2
4 = 1 + sqrt(k)
k = 9

問老師.....................................