有a.maths 唔識做

有 2 個辦法
Differentiation 同 Supp. Angel

1. diff
Let g = sin y – 2cos y +3
dg / dy = cos y + 2 sin y
d2g / dy2 = -sin y + 2 cos y

put dg / dy = 0
and we have tan y = -1/2
that is we got 2 pair of answers

(cos y, sin y) = (2/sqrt5, -1/sqrt5) or (-2/sqrt5, 1/sqrt5)

Check with d2g / dy2, we have
(cos y, sin y) = (2/sqrt5, -1/sqrt5) is the min pt
and (cos y, sin y) = (-2/sqrt5, 1/sqrt5) is the max pt

so the required value = 1/sqrt5 – 2* (-2/sqrt5) + 3 = 3 + sqrt5

2. supp. angel
Let sin y – 2cos y = r cos (y + z), where z is a constant

r^2 = 1^2 + 2^2
that is r = sqrt 5
[ 原本可以計埋 z 出黎, 不過題目無要求就算喇 ]

sin y – 2cos y +3 = sqrt5 * cos (y + z) + 3
咁 max value 當然係cos (y + z) = 1 既時候
亦即 3 + sqrt 5

Let f(y)=siny- 2cos y + 3

f'(y)=cosy+2siny
f''(y)=-siny+2cosy

Let f'(y)=0
cosy+2siny=0
tany=-1/2
we got 2 pair of answers

(cos y, sin y) = (2/sqrt5, -1/sqrt5) or (-2/sqrt5, 1/sqrt5)

2008-05-12 20:55:45 補充：
Check with f''(y), we have
(cos y, sin y) = (2/sqrt5, -1/sqrt5) is the minimum point
and (cos y, sin y) = (-2/sqrt5, 1/sqrt5) is the maximum point

so the maximum value = 1/sqrt5 – 2* (-2/sqrt5) + 3 = 3 + sqrt5
minimum value=-1/sqrt5 – 2* (2/sqrt5) + 3 = 3 - 5/sqrt5