maths 超難~答到堅係神人!

(1) y = cos 2θ + √3 sin 2θ - 2√3 cos θ - 2 sin θ
==> y = 2[(1/2) cos 2θ + (√3/2) sin 2θ] - 4[(√3/2) cos θ + (1/2) sin θ]
==> y = 2[cos π/3 cos 2θ + sin π/3 sin 2θ] - 4cos π/6 cos θ + sin π/6 sin θ]
==> y = 2 cos(π/3 - 2θ) - 4 cos(π/6 - θ)
==> y = 2 [2 cos(π/6 - θ)^2 - 1] - 4 cos(π/6 - θ)
==> y = 4 cos(π/6 - θ)^2 - 4 cos(π/6 - θ) - 2
==> y = [4 cos(π/6 - θ)^2 - 4 cos(π/6 - θ) + 1] - 3
==> y = [2 cos(π/6 - θ) - 1]^2 - 3
As -π/2 ≦ θ ≦ 0, so 2π/3 ≧ (π/6 - θ) ≧ π/6
When (π/6 - θ) = π/3, y = 0 - 3 = -3, therefore,
the smallest value of y is -3.

(2) x + log(3) x = 14
==> log(3) x = 14 - x
In order to solve x, draw two graph,
(i) y = log(3) x, and (ii) y = 14 - x
the intersection point of x-coordinate is between 11 and 12.
Using bisection method, you will get
x = 11.75678 (corr. to 5 d.p.)




2012-04-08 18:17:03 補充：
No way.
這條應該在國際學校高中的題目，他們都是用有繪圖的計算機做的，因為是書教的方法。

2012-04-08 18:19:00 補充：
香港以前學校都有教，就是我做的那個，叫半分法。

第2條:
log x=14-x
3
log x/log 3=14-x
log x/14-x=log 3
x^[1/(14-x)]=3
x=3^(14-x)
x* 3^x=3^14
計5落啦~