Author Topic: Math  (Read 6216 times)

Offline IGSTUDENT

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Re: Math
« Reply #15 on: November 14, 2009, 04:36:35 am »
Thanks  a lot for the help. I still do not really understand Q2.
For Q1. It was actually show that an=3^(n-1) for all positive integers. sorry for they typo.


I also had another question.
how do you solve   5logx  +2 is greater than 0. where x is the base
and 3+lnx is greater than e^x?


Thanks.


. let a1,a2,a3,...be a sequence defined by
    a1=1,an=3an-1; n?1

    Show that an=3^n-1 for all positive integers n.


 Prove the following statement.
    n
2. ? 1/(2i-1)(2i+1)=n/2n+1       for each positive integer n.
    1

3. Use mathematical induction to prove that (5^n)+(9^n)+2 is divisible by 4,for n?Z+.

Thanks.
1. a_1=3*1-1=3^1 -1=2

SO p(1) is true. Suppose p(k) is true, prove p(k+1) trues
a_{n+1} =3(3^{k} -1) -1=3*3^k-3-1=3^{k+1} -4 Are you sure this question is right?

2.\sum_1^n 1/(2i-1)(2i+1) =1/2 \sum_1^n 1/(2i-1) -1/(2i+1)=1-1/(2n+1)=1/2(2n/(2n+1))=n/(2n+1)

3.p(1) 5+9+2=16 hence p(1) is true
suppoose p(k) is true then (5^k)+(9^k)+2 is divisible by 4
p(k+1)-p(k) (5^(k+1))+(9^(k+1))+2-(5^k)-(9^n)-2  =5^k(5-1) +9^k(9-1)=4*5^k +8*9^k which is divisible by 4
« Last Edit: November 14, 2009, 10:03:41 am by IGSTUDENT »

Offline astarmathsandphysics

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Re: Math
« Reply #16 on: November 14, 2009, 09:37:48 am »
Will look when i get home

Offline IGSTUDENT

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Re: Math
« Reply #17 on: March 20, 2010, 02:47:00 pm »
i needed some help with 2 questions

1. The quadratic equation ax^2+bx+c=0 has roots x=alpha and x=beta
    a) Express the product of roots, alpha*beta in terms of a and c

2. Find a quadratic function in the form y=x^2+bx+c the satisfies the given functions:
    The function has zeros of x=1/2 and x=3 and its graph passes through the point (-1,4)

Thanks
« Last Edit: March 20, 2010, 03:54:32 pm by IGSTUDENT »

nid404

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Re: Math
« Reply #18 on: March 20, 2010, 03:05:54 pm »
I didn't quite get the first question...but i got the second one...so here it is

y=0 when x=1/2 or x=3


3eqns 3 variables...solve simultaneously
x=3
9y+3b+c=0
x=1/2
0.25y+0.5b+c=0
x=-1
1y-1b+c=4
y=2/3 b=-7/3 c=1
y=2/3x2-7/3x+1



Offline IGSTUDENT

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Re: Math
« Reply #19 on: March 20, 2010, 03:55:55 pm »
I didn't quite get the first question...but i got the second one...so here it is

y=0 when x=1/2 or x=3


3eqns 3 variables...solve simultaneously
x=3
9y+3b+c=0
x=1/2
0.25y+0.5b+c=0
x=-1
1y-1b+c=4
y=2/3 b=-7/3 c=1
y=2/3x2-7/3x+1

Thanks, i made a mistake in the 1st Q. it is alpha *beta and not delta will that help?


Offline IGSTUDENT

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Re: Math
« Reply #20 on: March 27, 2010, 06:55:52 am »
I have 3 question in maths,

1. Given that the roots of the equation x^3-9x^2+bx-216=0 are consecutive terms in a geometric sequence, find the value of b and solve the equation

2. The polynomial p(x)=(ax+b)^3 leaves a remainder of -1 when divide by x+1 and a remainder of 27 when divided by x-2. Find the values of the real numbers a and b.

3. Prove that when a polynomial p(x) is divided by ax-b the remainder if p(b/a)

Thanks

Offline astarmathsandphysics

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Re: Math
« Reply #21 on: March 27, 2010, 07:34:38 am »
1.It must factorise as (x-a)(x-b)(x-c) with a b and c all integers
Multiply this to givex^3-x^2(ab+bc+ac)+x(a+b+c)-216 (1)
If the smallest root is t then the other roots are rt and r^2t so product of roots r^3t^3=216
rt=6 so r=1,t=6 or r=6,t=1 or r=2,t=3 or some variation of these with plus or minus too.
I think a typo x^3-90x^2+bx-216=0 I get (x+3)(x+6)(x-12) b=-3
2.(-a+b)^3=-1 so -a+b=-1
(2a+b)^3=27 so 2a+b=3
a=4/3 b=1/3
will have to wait for q3

Offline IGSTUDENT

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Re: Math
« Reply #22 on: March 27, 2010, 09:40:46 am »
thanks!!!
« Last Edit: April 14, 2010, 03:39:23 pm by IGSTUDENT »

Offline IGSTUDENT

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Re: Math
« Reply #23 on: April 14, 2010, 03:45:56 pm »
Some more doubts...

1. Consider the trigonometric curve y=sin(2x-pi/2)
a) Find dy/dx and d2y/dx2

2. Find an equation for a line that is tangent to the graph of y=e^x that passes through the origin

3. Find the derivatice of y with repect to x, dy/dx by implicity differentiation: xy(x+y)^1/2=1

4. Find the derivative of y with repect to x, dy/dx: ln(1+x^2)^1/2=xarctanx

Thanks!

Offline astarmathsandphysics

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Re: Math
« Reply #24 on: April 14, 2010, 03:57:21 pm »
One mo

Offline astarmathsandphysics

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Re: Math
« Reply #25 on: April 14, 2010, 04:10:49 pm »
Some more doubts...

1. Consider the trigonometric curve y=sin(2x-pi/2)
a) Find dy/dx and d2y/dx2

2. Find an equation for a line that is tangent to the graph of y=e^x that passes through the origin

3. Find the derivatice of y with repect to x, dy/dx by implicity differentiation: xy(x+y)^1/2=1

4. Find the derivative of y with repect to x, dy/dx: ln(1+x^2)^1/2=xarctanx

Thanks!
a)dy/dx=2cos(2x-pi/2) and d2y/d2x =-4sin(2x-pi/2)
b)dy/dx =e^x so at x=0 gradient is e^0 =1 and y=1
y-1=1(x-0) so y=x+1
c)o sh*t
y(x+y)^1/2 +x(x+y)^1/2 dy/dx +1/2xy(x+y)^-1/2 (1+dy/dx) =0
dy/dx(x(x+y)^1/2+1/2xy(x+y)^-1/2 )=-y(x+y)^1/2 -1/2xy(x+y)^-1/2
dy/dx =(-y(x+y)^1/2 -1/2xy(x+y)^-1/2)/(x(x+y)^1/2+1/2xy(x+y)^-1/2 ) =(-y(x+y)-1/2xy)/(x(x+y)+1/2xy) =(-y^2-3/2xy)/(x^2 +3/2xy)
4. Where is y here

Offline IGSTUDENT

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Re: Math
« Reply #26 on: April 15, 2010, 01:53:45 am »
thanks for the help