Extension One Revising Game (2 Viewers)

addikaye03 · Jan 31, 2009

New Question Time, Someone?

bored.of.u · Feb 12, 2009

addikaye03 said:
New Question Time, Someone?

hmm ok i found this question kinda fun to do (easy question):

find the sum and product of the roots of the parabola ax^2 + bx + c = 0

gurmies · Feb 12, 2009

$\sum \alpha = -\frac{b}{a}$

$\prod \alpha = \frac{c}{a}$

Is that it? =S

tommykins · Feb 12, 2009

Trebla · Feb 12, 2009

Timothy.Siu · Feb 12, 2009

Trebla said:
$$Prove that $\\\displaystyle\sum_{k=1}^{n}k = \dbinom{n+1}{2}$

LHS=1+2+3+.....+n=n(n+1)/2 (arithmetic series sum)

RHS=(n+1)!/2!(n-1)!=n(n+1)/2=LHS

umm...

Show that sin A+sin B=2sin[(A+B)/2]cos[(A-B/2)]

azureus88 · Feb 24, 2009

sin(a+b) + sin(a-b) = 2sin(a)cos(b)
let A=a+b and B=a-b
sin A+sin B=2sin[(A+B)/2]cos[(A-B/2)]

New Q: Factorise a^2 + 3a + 2 and hence find the coefficient of a^4 in (a^2 + 3a + 2)^6

Trebla · Feb 24, 2009

azureus88 said:
sin(a+b) + sin(a-b) = 2sin(a)cos(b)
let A=a+b and B=a-b
sin A+sin B=2sin[(A+B)/2]cos[(A-B/2)]

New Q: Factorise a^2 + 3a + 2 and hence find the coefficient of a^4 in (a^2 + 3a + 2)^6

$(a^2 + 3a + 2) = (a + 1)(a + 2)\\(a^2 + 3a + 2)^6 = (a + 1)^6(a + 2)^6\\=\left(\displaystyle\sum_{r=1}^{6}\binom{6}{r}a^r\right)\left(\displaystyle\sum_{k=1}^{6}\binom{6}{k}a^k.2^{6-k}\right)\\=\displaystyle\sum_{r=1}^{6}\displaystyle\sum_{k=1}^{6}\binom{6}{r}\binom{6}{k}a^{r+k}.2^{6-k}\\$To obtain co-efficient of $a^4$ we require $r + k = 4\\$Consider$\\r=0,k=4 \Rightarrow \binom{6}{0}\binom{6}{4}2^2a^4 = 60a^4\\r=1,k=3 \Rightarrow \binom{6}{1}\binom{6}{3}2^3a^4 = 960a^4\\r=2,k=2 \Rightarrow \binom{6}{2}\binom{6}{2}2^4a^4 = 3600a^4\\r=3,k=1 \Rightarrow \binom{6}{3}\binom{6}{1}2^5a^4 = 3840 a^4\\r=4,k=0 \Rightarrow \binom{6}{4}\binom{6}{0}2^6a^4 = 960a^4\\$Adding them up gives: $ 9420a^4\\\therefore $Co-efficient of $a^4$ is $ 9420$

New Question:

A function f (x) is such that f (x) > 0 for all real x and f (a + b) = f (a). f (b) for any real numbers a and b.

(a) Show that f(0) = 1

(b) Show that f(-x) = 1/f(x)

(c) Use mathematical induction to prove that for all integers n: f(nx) = [f(x)]ⁿ

(d) Name one basic type of function that exhibits the properties shown above.

kurt.physics · Feb 24, 2009

Trebla said:
New Question:

A function f (x) is such that f (x) > 0 for all real x and f (a + b) = f (a). f (b) for any real numbers a and b.

(a) Show that f(0) = 1

(b) Show that f(-x) = 1/f(x)

(c) Use mathematical induction to prove that for all integers n: f(nx) = [f(x)]ⁿ

(d) Name one basic type of function that exhibits the properties shown above.

for (a)

$f(0 + 0) = f(0) \times f(0)$

$f(0) = f(0)^2$

$\frac{f(0)^2}{f(0)} = \frac{f(0)}{f(0)}$

$\therefore f(0) = 1$

cyl123 · Feb 25, 2009

a) kurt.physics is right but just note that f(0) > 0 so division by f(0) is allowed

b)f(x+(-x))=f(x)f(-x)
f(0)=f(x)f(-x) and noting f(0)=1 and that f(x)>0 so division by f(x) is allowed
1=f(x)f(-x)
f(-x) = 1/f(x)

c)
If statement is true for all integer n>=0 and then doing:
f(nx)=[f(x)]^n
1/f(nx)=f(x)^-n (noting that both sides are greater than 0)
using b) gives f(-nx)=f(x)^-n
since n>=0, then -n<0, so the statement is also valid for integer n<0 ONLY IF it is true for integer n>=0

step 1: Prove true for n=0
LHS= f((0)x)=f(0)=1
RHS=f(x)^0=1=LHS
Thus true for n=0

step 2: Assume true for n=k
ie. f(kx)=[f(x)]^k

step 3: Prove true for n=k+1
ie prove f((k+1)x)=[f(x)]^(k+1)

LHS=f((k+1)x)
=f(kx+x)
=f(kx)f(x) by definition of f(x)
={[f(x)]^k}f(x)
=[f(x)]^(k+1)=RHS

Thus the statement is true for all integer n>=0. From the first part, since it is true for all integer n>=0, then it follows on that it is true for all integer n<0. Thus the statement is true for all integer n.

d)f(x)=e^x

ayehann · Mar 9, 2009

two circles, with centres O and I touch externally aat P. PTQ is a straight line through T terminated at the circumference by P and Q. prove that the tangents drawn at P & Q qill be parallel.

kurt.physics · Mar 9, 2009

ayehann said:
two circles, with centres O and I touch externally aat P. PTQ is a straight line through T terminated at the circumference by P and Q. prove that the tangents drawn at P & Q qill be parallel.

I don't really understand the highlighted part... but if what i think is right, then isn't it a case of AA similarity?

ayehann · Mar 9, 2009

kurt.physics said:
I don't really understand the highlighted part... but if what i think is right, then isn't it a case of AA similarity?

i didnt get it either haha... thats y i asked.. i gues the questions wrong?

gigglinJess · May 2, 2009

2009 3U Maths Marathon

As there doesn't seem to be a 2009 thread.....

The velocity of a particle is given by v = 5/(3x^2) and the particle is initially 3 metres to the left of the origin. Find its displacemnet after 3 seconds.

studentcheese · May 2, 2009

Re: 2009 3U Maths Marathon

If tanx = $\sqrt[2]{2}$ , calculate sin2x

AnandDNA · May 2, 2009

Re: 2009 3U Maths Marathon

If t=tan0.5x and 2sinx + cosx =2 show that 3t^2-4t+1=0
hence solve 2sinx+ cosx=2 for 0<(or equal to) x< (or equal to) 360

Drongoski · May 2, 2009

Re: 2009 3U Maths Marathon

If I read correctly tan x = sqr(2):

$\100dpi sin\ 2x\ \ =\ \ 2sin\ x\ cos\ x\ \ =\ \ 2\times \frac{\sqrt{2}}{\sqrt{3}}\times \frac{1}{\sqrt{3}}\ \ =\ \ \frac{2}{3}\sqrt{2}$

nerdsforever · May 2, 2009

Re: 2009 3U Maths Marathon

2sinx + cosx =2
$2.\frac{2t}{1+ t^2} + \frac{1-t^2}{1+t^2} = 2$
$\frac{4t + 1 - t^2}{1+t^2} = 2$
$4t + 1 - t^2 = 2 + 2t^2$
$3t^2 - 4t + 1 = 0$

(3t -1) (3t-3)/3 = 0
(3t -1)(t-1) = 0
t = 1/3 OR t =1
@/2 = 45 degrees.
@ = 90 degrees
@/2 = 18 degrees 26 mins
@ = 36 degrees, 52 mins

Test tan0.5x = 180 degrees
2sin180 + cos 180
= 0 -1
= -1
=/= 2
therefore, x = 180 is not a solution

nerdsforever · May 2, 2009

Re: 2009 3U Maths Marathon

Given the equation $sin^2x + sin2x = m + 2cos^2x$
Transform the equation using the substitution t = tanx

gurmies · May 2, 2009

Re: 2009 3U Maths Marathon

gigglinJess said:
As there doesn't seem to be a 2009 thread.....

The velocity of a particle is given by v = 5/(3x^2) and the particle is initially 3 metres to the left of the origin. Find its displacemnet after 3 seconds.

Lol, I haven't done this topic yet, but I just read up on it. Maybe I can do it

v = dx/dt = 5/3x^2

dt/dx = 3x^2/5

t = x^3/5 + C

When t = 0, x = -3

0 = -27/5 + C

C = 27/5

t = x^3/5 + 27/5

When t = 3

3 = x^3/5 + 27/5

x^3/5 = -12/5

x^3 = -12

x = (-12)^(1/3)

nerdsforever said:
Given the equation $sin^2x + sin2x = m + 2cos^2x$

Transform the equation using the substitution t = tanx

$\sin^{2}x + \sin2x = m + 2\cos^{2}x$

$\sin^{2}x + 2\sin x\cos x = m + 2\cos^{2}x$

$\tan^{2}x + 2\tan x = m\sec^{2}x + 2$

$\tan^{2}x + 2\tan x = m(1 + \tan^{2}x) + 2$

$\tan^{2}x + 2\tan x = m + m\tan^{2}x + 2$

$(m-1)\tan^{2}x - 2\tan x + 2 + m$

$(m-1)t^{2} - 2t + 2 + m$

Don't think I need to go any further...

Extension One Revising Game (2 Viewers)

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