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slvri
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« Reply #30 on: December 03, 2009, 04:37:27 PM » |
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zeroes of g? never heard of the zeroes of a function b4  |
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i hate A level...........
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astarmathsandphysics
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« Reply #33 on: December 03, 2009, 10:01:19 PM » |
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let g(x) = log5|2log3x|=0
2 log3x=1 so x=sqrt(3)
Have I got the question right? I can only see one root.
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IGSTUDENT
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« Reply #34 on: January 23, 2010, 06:19:25 AM » |
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Posting my queries:
1.prove that cos^6(x)+ sin^6(x)=1/8(3cos4x+5)
2.Prove that tan5x=(5tanx-10tan^3(x)+tan^5)/(1-10tan^2(x)+5tan^4(x)). By considering the equation tan5x=0, show that tan^2(pi/5)=5-2sqrt5.
For the 2nd question I was able to do the first part but not the second
Thanks
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astarmathsandphysics
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« Reply #35 on: January 23, 2010, 11:09:14 AM » |
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if tan5x=0 x=tan^-10= pi/5
0=(5tanp-10tan^3 p +tan^5 p)=tanp(5 -10tan^2 p+tan^4 p)
hence solve t^2-10t+5=0 where t=tan^2 p and p=pi/5
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cooldude
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« Reply #36 on: January 23, 2010, 11:35:46 AM » |
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wats GP
geometric progression |
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IGSTUDENT
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« Reply #37 on: January 24, 2010, 12:49:06 AM » |
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if tan5x=0 x=tan^-10= pi/5
0=(5tanp-10tan^3 p +tan^5 p)=tanp(5 -10tan^2 p+tan^4 p)
hence solve t^2-10t+5=0 where t=tan^2 p and p=pi/5
thanks a lot |
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sweet777
Newbie
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Posts: 40
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« Reply #38 on: January 30, 2010, 09:27:33 AM » |
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can u please help me solve this matrix question??
Write 5A2-6A=3I in the form AB=I and hence find A-1 in terms of A and I
please help fasssssssssttttttttttttt!!!!!!!!!!!!!!!!!!!!!!!
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~Alpha
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« Reply #39 on: January 30, 2010, 09:58:49 AM » |
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wats GP
Geometric Progression, I guess. |
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astarmathsandphysics
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« Reply #41 on: January 30, 2010, 01:52:10 PM » |
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Write 5A2-6A=3I in the form AB=I and hence find A-1 in terms of A and I
5A^2-6A+1=4I so (A-I)(5A-!)=4I so A-1= 4I(5A-1)^-1
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IGSTUDENT
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« Reply #42 on: April 21, 2010, 10:38:58 AM » |
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Need help in a few questions:
1)The polynomial p(x) + (ax+b)3 leaves a remainder of -1 when divided by x+1 and a remainder of 27 when divided by x-2.Find the values of the real numbers a and b.
2)Given that the roots of the equation x3-9x2 +bx -216 =0 are consecutive terms in a geometric sequence find the value of b and solve the equation.
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astarmathsandphysics
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« Reply #43 on: April 21, 2010, 11:27:44 AM » |
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1)x+1=0 so x=-1 p(-1)=(-a+b)^3 =-1 so -a+b=-1 (1)
x-2=0 so x=2 p(2)=(2a+b)=3 so 2a+b=3 (2)
(2)-(1) 3a=4 so a=4/3
b=a+1 =7/3
I did the other one already somewhere - I will try to find a link
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astarmathsandphysics
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« Reply #44 on: April 21, 2010, 11:28:45 AM » |
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The orrts are anyway 1,6,36
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