A. MATHS

2007-01-06 8:26 am
我有ANS
但希望有詳細STEPS
find the max. and min. value of y for all value of x
7〕 y = - ( 1/4 ) cos ( x/2 )
ans : 1/4, - 1/4
8〕y = 2 cos ^2 ( 兀+ x/2 )
ans : 2 , 0
9〕y= 4 sin ^2 ( x/3 ) - 3
ans : 1 , - 3
10〕y = 1 - ( 1/2 ) sin ( 2x + 兀/3 )
ans : 3/2 , 1/2
11〕if sin A and cos A both increase as A increase, in which quadrant does A lie ?
quad. 4
可以解釋給我嗎?因為快考試了,我想明白。謝謝!

回答 (2)

2007-01-06 8:59 am
✔ 最佳答案
第 7 至 10 題,其實每條問都是相似,有個 sin 和 cos 的實真你要先知道:

For all real θ,

-1 ≦ sinθ ≦ 1

-1 ≦ cosθ ≦ 1

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find the max. and min. value of y for all value of x
7〕 y = - ( 1/4 ) cos ( x/2 )
ans : 1/4, - 1/4

-1 ≦ cos (x/2) ≦ 1

1/4 ≧ (-1/4)cos(x/2) ≧ -1/4

1/4 ≧ y ≧ -1/4

So the minimum value of y is -1/4, the maximum value is 1/4

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8〕y = 2 cos ^2 ( π+ x/2 )
ans : 2 , 0

-1 ≦ cos (π + x/2) ≦ 1

0 ≦ cos^2(π + x/2) ≦ 1^2

0 ≦ cos^2(π + x/2) ≦ 1

0 ≦ 2 cos^2(π + x/2) ≦ 2

0 ≦ y ≦ 2

So the minimum value of y is 0, the maximum value is 2

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9〕y= 4 sin ^2 ( x/3 ) - 3
ans : 1 , - 3

-1 ≦ sin (x/3) ≦ 1

0 ≦ sin^2(x/3) ≦ 1^2

0 ≦ sin^2(x/3) ≦ 1

0 ≦ 4 sin^2(x/3) ≦ 4

0-3 ≦ sin^2(x/3)-3 ≦ 4-3

-3 ≦ sin^2(x/3)-3 ≦ 1

-3 ≦ y ≦ 1

So the minimum value of y is -3, the maximum value is 1

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10〕y = 1 - ( 1/2 ) sin ( 2x + π/3 )
ans : 3/2 , 1/2

-1 ≦ sin (2x + π/3) ≦ 1

1/2 ≧ -(1/2) sin (2x + π/3) ≧ -1/2

1/2 + 1 ≧ -(1/2) sin (2x + π/3) + 1 ≧ -1/2 + 1

3/2 ≧ 1 - (1/2) sin (2x + π/3) ≧ 1/2

3/2 ≧ y ≧ 1/2

So the minimum value of y is 1/2, the maximum value is 3/2


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11〕if sin A and cos A both increase as A increase, in which quadrant does A lie ?
quad. 4

Recall the range change of sin and cos in the 4 quadrant:

sinθ from 0 to 1 for θ = 0 to π/2 [increasing]
sinθ from 1 to 0 for θ = π/2 to π [decreasing]
sinθ from 0 to -1 for θ = π to 3π/2 [decreasing]
sinθ from -1 to 0 for θ = 3π/2 to 2π [increasing]

cosθ from 1 to 0 for θ = 0 to π/2 [decreasing]
cosθ from 0 to -1 for θ = π/2 to π [decreasing]
cosθ from -1 to 0 for θ = π to 3π/2 [increasing]
cosθ from 0 to 1 for θ = 3π/2 to 2π [increasing]

So when sinθ and cosθ both increasing, the quadrant is 4.


2007-01-06 01:02:20 補充:
小小補充:第 7-10 題,因為題目只是問 y 的最小及最大值,其實與角度無關,所以只要考慮 sinθ 及 cosθ 的範圍便可。
2007-01-06 9:10 am
呢d可以話係step既,
因為理解之後就可以出答案
老師講既,如果真係要既最多就係
例如7:
y = - ( 1/4 ) cos ( x/2 )
max:y=-(1/4)(-1)
y=11/4
mix:y=-(1/4)(1)

理解方面:
因為cos同sin既範圍值係1至-1之間(你按計算機or睇圖就明),而tan就係無範圍既(所以唔會出最大/最小既tan題目)
只要找cos/sin範圍內能令該方程達最大/最小的值
8〕y = 2 cos ^2 ( 兀+ x/2 )
ans : 2 , 0
cos2次的最大值都係1至0之間(因為2次唔會出負數)
所以令其最細既係0
下面既都係咁解
仲係唔得sd mail比我^^

2007-01-06 01:12:13 補充:
呢d係可以不用寫step既老師say的

2007-01-06 01:13:15 補充:
但我知你間學校係唔係可以問吓你既老師
參考: 老師/自己


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