✔ 最佳答案
Find x.
1. (2^2x+3)+(2^x+1)-1=0
(2^3)(2^x)^2 + 2(2^x) - 1 = 0
8(2^x)^2 + 2(2^x) - 1 = 0
[4(2^x) - 1] [2(2^x) + 1] = 0
2^x = 1/4 or 2^x = - 1/2 (rejected)
2^x = 2 ^ - 2
x = - 2
2a. y=tan(180°-θ)[(2sin(180°-θ)cos(-θ)-tan(90°+θ)]
Simplify y.
y = (- tanθ) [ 2sinθ cosθ - (- 1 / tanθ) ]
y = (- sinθ / cosθ) [ 2sinθ cosθ + (cosθ / sinθ) ]
y = - 2(sinθ)^2 - 1
2b.Find the max. and min. values of the function y.
Since - 1 =< sinθ =< 1 ,
So 0 =< (sinθ)^2 =< 1 ,
y max. = - 2(0) - 1 = - 1
y min. = - 2(1) - 1 = - 3
3.It takes 1hour more for Peter to complete a 24km journey than John to complete a 20km journey.If Peter walks slower than John by 2km/h ,find the speed of each boy.
Let Peter walks x km/h , then John walks x+2 km/h ,
24/x - 20/(x+2) = 1
24(x+2) - 20x = x(x+2)
4x + 48 = x^2 + 2x
x^2 - 2x - 48 = 0
(x - 8)(x + 6) = 0
x = 8 or x = - 6 (rejected)
Peter walks 8 km/h while John walks 8+2 = 10km/h
2010-07-09 22:51:51 補充：
Corrections :
2a. y=tan(180°-θ) [(2sin(180°-θ)cos(-θ)-tan(90°+θ)]
Simplify y.
y = (- tanθ) [ 2sinθ cosθ - (- 1 / tanθ) ]
y = (- sinθ / cosθ) [ 2sinθ cosθ + (cosθ / sinθ) ]
y = - 2(sinθ)^2 - 1 , (θ =/= 0° , 90° , 180° since tan(180°- 90°), tan(90°+ 0°) ,
tam(90° + 180°) are undefined.
2010-07-09 22:51:57 補充：
2b.Find the max. and min. values of the function y.
Since - 1 =< sinθ =< 1 , but θ =/= 0° , 90° , 180° ,
So - 1 < sinθ < 1 ,
0 < (sinθ)^2 < 1
0 > - 2(sinθ)^2 > - 2
- 1 > - 2(sinθ)^2 - 1 > - 3
- 1 > y > - 3
2010-07-09 22:56:07 補充：
Missing θ =/= 270° since tan(180°-270°) is undefined.
2010-07-09 22:56:59 補充：
So sinθ cann't be - 1 , sorry for my mistakes!
2010-07-09 22:59:22 補充：
y = - 2(sinθ)^2 - 1 , (θ =/= 0° , 90° , 180° , 270°)since tan(180°- 90°), tan(90°+ 0°) ,
tam(90° + 180°) , tan(180°-270°) are undefined.
2010-07-10 14:05:01 補充：
Yes , the answer given by the book incorrect ,
put θ = 45° to
y = tan(180°-45°)[(2sin(180°-45°)cos(-45°)-tan(90°+45°)]
y = tan135° (2sin45° cos45° - tan135°)
y = (- 1) (1 - (-1))
y = - 2
But
2(sin45°)^2 +1 = 2
