數學幾條不懂得做1

1.
Sol
設(2014 -x^2)^(0.5)+(2004 - x^2)^(0.5)=2p
(2014 - x^2)^(0.5)=p+1
2014-x^2=p^2+2p+1
(2004 - x^2)^(0.5)=p－1
2004-x^2=p^2－2p+1
10=4p
p=5/2
x^2=2014-49/4=8007/4
x=+/-√8007/2

2.
Sol
Sinx+Cosx=7/13
1+2SinxCosx=49/169
SinxCosx=－60/169
0<=x<=180度
Sinx>0，Cosx<0
設p=Sinx－Cosx>0
2Sinx=p+7/13
2Cosx=7/13－p
4=(p^2+14p/13+49/169)+(p^2－14p/13+49/169)
4=2p^2+98/169
2p^2=578/169
p^2=289/169
p=17/13
2Cosx=7/13－17/13=－10/13
Cosx=－5/13
Cos x+Cos ^3 x +Cos ^5 x+...
=Cosx/(1－Cos^2 x)
=－5/13/(144/169)
=－65/144

3.
Sol
x^2+3y^2=6x+7
(x^2－6x+9)+3y^2=16
設 x－3=4Sinw，√3y=4Cosw
x^2+y^2
=(3+4Sinw)^2+(16/3)Cos^2 w
=16Sin^2 w+24Sinw+9+(16/3)Cos^2 w
=(32/3)Sin^2 w+24Sinw+43/3
=(32/3)[Sin^2 w+9Sinw/4+81/64]+43/3－81/6
=(32/3)(Sinw+9/8)^2+5/6
－1<=Sinw<=1
1/8<=Sinw+9/8<=17/8
1/64<=Sinw+9/2)^2<=289/64
1/6<=(32/3)(Sinw+9/2)^2<=289/6
1<=(32/3)(Sinw+9/2)^2+5/6<=49
1<=x^2+y^2<=49

4.
Sol
(a+c)(a+d)=1=(b+c)(b+d)
a^2+ad+ac+cd=b^2+bd+bc+cd
a^2+ad+ac=b^2+bd+bc
(a^2－b^2)+(ad－bd)+(ac－bc)=0
(a－b)(a+b+d+c)=0
a<>b
a+b+c+d=0
a+c=－b－d
(a+c)(b+c)
=(－b－d)(b+c)
=－1

5.
Sol
[a_(n+2)]=3[a_(n+1)]－2(a_n)
[a_(n+2)]－2[a_(n+1)]=[a_(n+1)]－2(a_n)
(a_2014)－2(a_2013)=2014－2*205=1604

6.
Sol
Sin^2 x－4Sinx+m
=(Sin^2 x－4Sinx+4)+m－4
=(Sinx－2)^2+m－4
－1<=Sinx<=1
－3<=Sinx－2<=－1
1<=(Sinx－2)^2<=9
m－3<=(Sinx－2)^2+m－4<=m+5
m－3<=y<=m+5
m－3=－8/3
m=1/3
極小值=(1/3)^(16/3)=3^(－16/3)

7.
Sol
1<Tan x<3
1/3<1/Tan x<1
30度<90度/Tan x<90度
90度/Tan x在第一象限
Tan (90度Tan x)] 在第三象限
[Tan(90度/Tan x)][Tan(90度Tan x)]=1
Tan(90度/Tan x)=Cot(90度Tan x)
Cot(90度－90度/Tan x)=Cot (90度Tan x)
So
90度－90度/Tan x+180度=90度Tan x
3－1/Tan x=Tan x
3Tanx－1=Tan^2 x
Tan^2 x－3Tanx+1=0
Tanx=(3+√5)/2 負不合

1.(2014 - x^2)^(0.5) - (2004 - x^2)^(0.5) =2
.((2014 - x^2)^(0.5) - (2004 - x^2)^(0.5))((2014 - x^2)^(0.5) + (2004 - x^2)^(0.5))=2((2014 - x^2)^(0.5) + (2004 - x^2)^(0.5))
2014 - x^2-(2004-x^2)=2((2014 - x^2)^(0.5) + (2004 - x^2)^(0.5))
10=2((2014 - x^2)^(0.5) + (2004 - x^2)^(0.5))
((2014 - x^2)^(0.5) + (2004 - x^2)^(0.5))=5

2014-09-07 21:54:29 補充：
2.cos x +cos ^3 x + cos ^5 x +...
=cos x/(1-(cosx)^2)
=cos x/(sin x)^2
As it is given that cos x +sin x=7/13 ,you should be able to calculate what x is.

2014-09-07 22:05:48 補充：
3.x^2 + 3y^2=6x+7
x^2 + 3y^2=6x+7
x^2 -6x+(3y^2-7)=0
D>=0
36-4(3y^2-7)>=0
9>=3y^2-7
16/3>=y^2
3y^2+(x^2-6x-7)=0
D>=0
0-4(3)(x^2-6x-7)>=0
x^2-6x-7<=0
(x-7)(x+1)<=0
7>=x>=-1
x^2 +y^2 的極大值=7^2+16/3

2014-09-07 22:29:48 補充：
4.Let (a+c)=1/x,a+d=x,b+c=1/y,c+d=y
c-d=1/x-x=1/y-y
-1/x+1/y-+x-y=0
(x-y)(1-1/xy)=0
1=1/(xy) (x is not equal to y because the four values are unequal)
(a+c)(b+c)=1/xy=1

2014-09-07 22:38:54 補充：
5.[a_(n+2)]=3[a_(n+1)]-2(a_n)
.[a_(n+2)-2[a_(n+1)]
=[a_(n+1)]-2(a_n)
........................
=a_2-2(a_1)

2014-09-07 22:42:06 補充：
6
y=sin^2 x -4 sin x +m
=(sinx-2)^2+m-4
m-4=-8/3
m=-8/3+4
Ans:m^y=(-8/3+4)^-8/3)

y=sin^2 x - 4 sin x + m
y=(sin x - 2)^2 + m-4

-1≤sin x≤ 1
當sin x=1,y為-8/3
-8/3=(1-2)^2 + m-4
m=-8/3 + 3=1/3

y=(sin x - 2)^2 - 11/3

極大值=(-1 - 2)^2 - 11/3=9 - 11/3=16/3

m^y
=(1/3)^(16/3)
=3^(-16/3)

2014-09-07 19:39:48 補充：
Let a=2014 - x^2 , b=2004 - x^2
a^(0.5) - b^(0.5) =2
(a - b)/[a^(0.5) + b^(0.5)] =2
a^(0.5) + b^(0.5)= (a - b)/2

(2014 - x^2)^(0.5) + (2004 - x^2)^(0.5)
=( 2014 - x^2 - 2004 + x^2 )/2
=10/2
=5

2014-09-07 20:01:40 補充：
∵90°/tan x在象限 I 及 90° tan x在象限 III
∴tan x>2

90° tan x=180° + (90° - 90°/tan x)
90° tan^2 x=270° tan x - 90°
tan^2 x - 3 tan x + 1 = 0
tan x=[3+√(9-4)]/2 或 [3-√(9-4)]/2 (捨去，因為<2)
tan x=[3+√(5)]/2

2014-09-07 20:43:19 補充：
(a+c)(a+d)=(b+c)(b+d)
a^2 + a(c+d)= b^2 + b(c+d)
a^2 - b^2=b(c+d) - a(c+d)
(a+b)(a-b)=(b-a)(c+d)
(a+b)=-(c+d)
a+b+c+d=0
a+c=-(b+d)

(a+c)(b+c)
=-(b+d)(b+c)
=-1

2014-09-07 20:58:02 補充：
(a_1)=205
(a_2)=2014

[a_(n+2)]=3[a_(n+1)]-2(a_n)
[a_(n+2)]-2[a_(n+1)]=[a_(n+1)]-2(a_n)
∴a_(2014)-2[a_(2013)]
=a_(2013)-2[a_(2012)]
=a_(2012)-2[a_(2011)]
=a_(2011)-2[a_(2010)]
=...
=(a_2)-2(a_1)
=1604