Author Topic: 3U Maths Question Thread (Read 1242012 times) Tweet Share

legorgo18 · « **Reply #1770 on:** April 05, 2017, 05:40:47 pm »

I kinda need some help with this bionomial q

Simplify n* n-1C1 + n* n-1C2 + ... + n* n-1Cn-2

I tried moving n out, then using the identity nCr = nCn-r but then i got lost, ty for helping!

RuiAce · « **Reply #1771 on:** April 05, 2017, 05:44:41 pm »

Quote from: K9810 on April 04, 2017, 02:42:17 pm

Hey!

Can you please help me with these questions?

Thanks

$\text{Let the equations of the lines be of the form}\\ y-1=m(x-7)\\ \text{Which rearranges to}\\ y=mx+(1-7m)$
$\text{Note that we want to find the values of }m\text{ to force the line to be tangential}\\ \text{If a line is tangential, then there is }\textbf{one unique point of intersection}\\ \text{Which is, of course, the point where it is tangential.}$
$\text{Suppose we wish to find the point of intersection}\\ \text{By solving simultaneous equations we have}\\ x^2+(mx+(1-7m))^2=25$
$\text{According to WolframAlpha this should expand to}\\ (m^2+1)x^2+(-14m^2+2m)x+(49m^2-14m-24)=0$
$\text{Recall that this quadratic in }x\text{ only has one solution, as there is one unique point of intersection}\\ \text{Hence, the discriminant of the quadratic should be 0}\\ \text{i.e. }(-14m^2+2m)^2-4(m^2+1)(49m^2-14m-24)=0$
$\text{According to WolframAlpha this expands to}\\ -96m^2+56m+96=0\\ \text{It should be clear that we may use the quadratic formula}\\ \text{This obtains }m=\frac43, -\frac34$

scienceislife · « **Reply #1772 on:** April 05, 2017, 05:45:17 pm »

Just not too great at these simple harmonic motion questions :/ especially when the formula is in the form v^2
"The velocity of a particle is moving in simple harmonic motion in a straight line is given by v^2 = 2 - x - x^2 ms, where x is displacement in metres.
a) find the two point between which the particle is oscillating
b) find the centre of motion
c) find the maximum speed of the particle
d) find the acceleration of the particle in terms of x"
Many thanks!

RuiAce · « **Reply #1773 on:** April 05, 2017, 05:51:09 pm »

Quote from: legorgo18 on April 05, 2017, 05:40:47 pm

I kinda need some help with this bionomial q

Simplify n* n-1C1 + n* n-1C2 + ... + n* n-1Cn-2

I tried moving n out, then using the identity nCr = nCn-r but then i got lost, ty for helping!

$\text{Pick }x=1\text{. Then}\\ \begin{align*}(1+1)^{n-1}&=\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+\binom{n-1}{n-1}\\ 2^n&=1+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+1\end{align*}\\ \text{You should be able to take it from here.}$

legorgo18 · « **Reply #1774 on:** April 05, 2017, 06:02:47 pm »

Quote from: RuiAce on April 05, 2017, 05:51:09 pm

$\text{Pick }x=1\text{. Then}\\ \begin{align*}(1+1)^{n-1}&=\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+\binom{n-1}{n-1}\\ 2^n&=1+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+1\end{align*}\\ \text{You should be able to take it from here.}$

Omg thx rui i got it now... any tricks for binomial identities? i find them the most annoying

RuiAce · « **Reply #1775 on:** April 05, 2017, 06:08:13 pm »

Quote from: scienceislife on April 05, 2017, 05:45:17 pm

Just not too great at these simple harmonic motion questions :/ especially when the formula is in the form v^2
"The velocity of a particle is moving in simple harmonic motion in a straight line is given by v^2 = 2 - x - x^2 ms, where x is displacement in metres.
a) find the two point between which the particle is oscillating
b) find the centre of motion
c) find the maximum speed of the particle
d) find the acceleration of the particle in terms of x"
Many thanks!

$\text{At the two endpoints of motion, the velocity of the particle is 0 specifically for SHM}\\ \therefore \text{We solve }0=2-x-x^2\\ \text{ which gives }x=1, -2$
______________________________________
$\text{The centre of motion is exactly between the two end points}\\ \text{Just averaging }1\text{ and }-2\text{ gives }-\frac12$
______________________________________
$\text{The maximum speed occurs at the centre of motion, which is }x=-\frac12\\ \text{At }x=-\frac12, \, v^2=\frac94\\ \text{Carefully note that this means the max speed is }\frac32\text{ which ignores the }\pm\text{ from square rooting.}$
______________________________________
$\text{To exploit the formula }\frac{d}{dx}\left(\frac12v^2\right)=\ddot{x}\\ \ddot{x}=\frac{d}{dx}\left(1-\frac{x}2-\frac{x^2}2\right)=-\frac12-x$

Quote from: legorgo18 on April 05, 2017, 06:02:47 pm

Omg thx rui i got it now... any tricks for binomial identities? i find them the most annoying

There's quite a ton of tricks. For this one, I started by taking what you said: ALL n's can be factored out. Hence, any coefficient in FRONT of the binomial coefficient is always 1. This implies that I will not need differentiation or integration.

The lower values of the binomial coefficient just go up by 1, and the upper ones are all n-1. This implies I should use $(1+x)^{n-1}$. And I should take x=1 for the same reason as stated in the above paragraph: all the coefficients are 1.

Both these were a tad poorly explained. Come back if confusion arises.

legorgo18 · « **Reply #1776 on:** April 05, 2017, 07:23:15 pm »

Hey rui, sorry for posting another 1. I just cant do this simple question...

its the second part of that simplification and i still cant do it after an hour...

Find the smallest positive integer n such that n* n-1C1 + n* n-1C2 + ... + n* n-1Cn-2 > 20000

so i subbed in the simplification and did some log law stuff and find myself stuck on removing the n... are you not meant to sub in the simplification?

RuiAce · « **Reply #1777 on:** April 05, 2017, 08:20:03 pm »

Quote from: legorgo18 on April 05, 2017, 07:23:15 pm

Hey rui, sorry for posting another 1. I just cant do this simple question... idk if i have autism or something

its the second part of that simplification and i still cant do it after an hour...

Find the smallest positive integer n such that n* n-1C1 + n* n-1C2 + ... + n* n-1Cn-2 > 20000

so i subbed in the simplification and did some log law stuff and find myself stuck on removing the n... are you not meant to sub in the simplification?

$\text{The simplification is }n(2^n-2) = 2n(2^n-1)$
$\text{The inequality }2n(2^n-1) > 0\text{ cannot be solved algebraically}\\ \text{As it involves the product of a linear term and an exponential.}\\ \text{We resort to the dreaded guess-and-check}$
$\text{With the aid of my math nerd memory I claim that }n=10\text{ will be a good choice for an answer.}\\ \text{Check: }2\times 10 \times (2^{10}-1) = 20460 > 20000\\ \text{yet }2\times 9 \times (2^9-1) = 9198 < 20000$
$\text{Hence, assuming I made no silly mistakes, it is the answer.}$

scienceislife · « **Reply #1778 on:** April 06, 2017, 06:20:58 am »

A particle moves in simple harmonic motion with acceleration a = 49x.
a) find the amplitude and period of the motion.
b) find the maximum speed of the particle.
c) find the times when the displacement is 0.5m.

RuiAce · « **Reply #1779 on:** April 06, 2017, 07:00:26 am »

Quote from: scienceislife on April 06, 2017, 06:20:58 am

A particle moves in simple harmonic motion with acceleration a = 49x.
a) find the amplitude and period of the motion.
b) find the maximum speed of the particle.
c) find the times when the displacement is 0.5m.

This question lacks a lot of information. Assuming that there is a minus sign and it should be a = -49x, the only thing that can be said is that the period is 2π/7.

legorgo18 · « **Reply #1780 on:** April 06, 2017, 03:10:19 pm »

Hi, im confused in this q

Using the binomial expansion of (1+x)^2n, prove that n sigma r=0 2nCr = 2^(2n-1) +[(2n)1!/2*(n!)^2]

How do you get n sigma r=0, or how would you start this question... scratching my head (Yr 12 cambridge binomial identity exercise extension q 17)

RuiAce · « **Reply #1781 on:** April 06, 2017, 03:39:22 pm »

Quote from: legorgo18 on April 06, 2017, 03:10:19 pm

Hi, im confused in this q

Using the binomial expansion of (1+x)^2n, prove that n sigma r=0 2nCr = 2^(2n-1) +[(2n)1!/2*(n!)^2]

How do you get n sigma r=0, or how would you start this question... scratching my head (Yr 12 cambridge binomial identity exercise extension q 17)

$\text{Accept the hint they have provided}\\ (1+x)^{2n} =\binom{2n}{0}+\binom{2n}{1}x+\dots+\binom{2n}{n-1}x^{n-1}+\binom{2n}{n}x^n +\binom{2n}{n+1}x^{n+1}+\dots+\binom{2n}{2n}x^{2n}\\ \text{Putting in }x=1\text{ gives }\\ 2^{2n}=\binom{2n}{0}+\binom{2n}{1}+\dots+\binom{2n}{n-1}+\binom{2n}{n}+\binom{2n}{n+1}+\dots+\binom{2n}{2n}$
$\text{As }r\text{ is the constraint on the lower term}\\ \text{To force the lower term to never exceed }n,\text{ apply the symmetry identity}\\ \binom{N}{K}=\binom{N}{N-K}\text{ to the second lot of the terms.}$
$\text{This gives}\\ 2^{2n} = 2\binom{2n}{0}+2\binom{2n}{1}+\dots+2\binom{2n}{n-1}+\binom{2n}{n}$
$\text{The RHS is almost reflective of the sum. We add }\binom{2n}{n}\text{ to both sides to obtain}\\ \begin{align*}2^{2n}+\binom{2n}{n}&=2\binom{2n}{0}+2\binom{2n}{1}+\dots+2\binom{2n}{n-1}+2\binom{2n}{n}\\ &= 2\sum_{r=0}^n\binom{2n}{r}\end{align*}$
$\text{Finally, divide by 2 and open up the binomial coefficient on the LHS to obtain}\\ \frac{1}{2}\left(2^n + \frac{(2n)!}{(n!)(n!)}\right)=\sum_{r=0}^n\binom{2n}{r}$

legorgo18 · « **Reply #1782 on:** April 07, 2017, 09:04:29 am »

Quote from: RuiAce on April 06, 2017, 03:39:22 pm

$\text{Accept the hint they have provided}\\ (1+x)^{2n} =\binom{2n}{0}+\binom{2n}{1}x+\dots+\binom{2n}{n-1}x^{n-1}+\binom{2n}{n}x^n +\binom{2n}{n+1}x^{n+1}+\dots+\binom{2n}{2n}x^{2n}\\ \text{Putting in }x=1\text{ gives }\\ 2^{2n}=\binom{2n}{0}+\binom{2n}{1}+\dots+\binom{2n}{n-1}+\binom{2n}{n}+\binom{2n}{n+1}+\dots+\binom{2n}{2n}$
$\text{As }r\text{ is the constraint on the lower term}\\ \text{To force the lower term to never exceed }n,\text{ apply the symmetry identity}\\ \binom{N}{K}=\binom{N}{N-K}\text{ to the second lot of the terms.}$
$\text{This gives}\\ 2^{2n} = 2\binom{2n}{0}+2\binom{2n}{1}+\dots+2\binom{2n}{n-1}+\binom{2n}{n}$
$\text{The RHS is almost reflective of the sum. We add }\binom{2n}{n}\text{ to both sides to obtain}\\ \begin{align*}2^{2n}+\binom{2n}{n}&=2\binom{2n}{0}+2\binom{2n}{1}+\dots+2\binom{2n}{n-1}+2\binom{2n}{n}\\ &= 2\sum_{r=0}^n\binom{2n}{r}\end{align*}$
$\text{Finally, divide by 2 and open up the binomial coefficient on the LHS to obtain}\\ \frac{1}{2}\left(2^n + \frac{(2n)!}{(n!)(n!)}\right)=\sum_{r=0}^n\binom{2n}{r}$

Hey rui, on the symmetry bit why is there only 1 of the 2nCn? The rest i get ty

RuiAce · « **Reply #1783 on:** April 07, 2017, 09:11:41 am »

Quote from: legorgo18 on April 07, 2017, 09:04:29 am

Hey rui, on the symmetry bit why is there only 1 of the 2nCn? The rest i get ty

There's only one $ \binom{2n}{n} $ in there to begin with in the expansion. You don't get two of them. You only get one of $ 1\, x\, x^2\, \dots$

The second one had to be brought in manually.

legorgo18 · « **Reply #1784 on:** April 07, 2017, 09:16:20 am »

Rui i think i get it, is it because 2n-n = n so it has no doubles?

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