Author Topic: 3U Maths Question Thread (Read 1495040 times) Tweet Share

Dragomistress · « **Reply #3255 on:** February 11, 2018, 03:54:22 pm »

I don't quite understand these questions.

RuiAce · « **Reply #3256 on:** February 11, 2018, 04:17:23 pm »

Quote from: Dragomistress on February 11, 2018, 03:54:22 pm

I don't quite understand these questions.

$\text{Using the formula }\boxed{n! = n(n-1)!}\\ \begin{align*}RHS &= (k+1)! - k!\\ &= (k+1)k! - k!\\ &= (k+1) \times k! - 1 \times k!\\ &= \left[(k+1)-1\right]\times k!\\ &= k\times k!\\ &= LHS\end{align*}$
$\text{And hence applying this formula for all }k=1,2,3,\dots,n:\\ \begin{align*}1\times 1! + 2\times 2! + 3\times 3! + \dots + k \times k!&= (2! - 1!) + (3! - 2!) + (4! -3!) + \dots + [(k+1)! - k!]\\ &= (k+1)! - 1\end{align*}\\ \text{since we have a whole bunch of terms cancelling out.}$

fireives1967 · « **Reply #3257 on:** February 11, 2018, 08:53:01 pm »

For this question, would b be needed to be found algebraically? (from part a)

Or would it need to be found graphically from a? Not quite sure how to do it algebraically :/, hoping it isn't in a test

RuiAce · « **Reply #3258 on:** February 11, 2018, 09:01:31 pm »

Quote from: fireives1967 on February 11, 2018, 08:53:01 pm

For this question, would b be needed to be found algebraically? (from part a)

Or would it need to be found graphically from a? Not quite sure how to do it algebraically :/, hoping it isn't in a test

There is no way of doing it algebraically. You must rely on the graphs.

Quote from: Dragomistress on February 11, 2018, 03:54:22 pm

I don't quite understand these questions.

$\text{Note that }\boxed{10! = 10\times 9\times8\times7\times6\times5\times4\times3\times2\times 1}$
$\text{10 is a factor of 10! and the highest power of 2 that divides into it is }2^1.\\ \text{8 is a factor of 10! and the highest power of 2 that divides into it is }2^3.\\ \text{6 is a factor of 10! and the highest power of 2 that divides into it is }2^1.\\ \text{4 is a factor of 10! and the highest power of 2 that divides into it is }2^2.\\ \text{2 is a factor of 10! and the highest power of 2 that divides into it is }2^1.$
$\text{So therefore, the number }2^1 \times 2^3 \times 2^1 \times 2^2 \times 2^1\\ \text{must therefore be the }\textbf{highest}\text{ power of 2}\\ \text{that is also a factor of }10!.$
$\text{Using our index laws, this is }2^8\\ \text{and hence your answer is }8.$
Following a similar process, $5^2$ is the greatest power of 5 that is a factor of $10!$, and hence your answer for that part is just 2.

You will do the same thing for $100!$ however it will be tedious. There are clever ways of speeding up the computations a bit but I will not go over them unless you ask for them, since they are more advanced.

Edit: Alternate method for $10!$

$\text{We can, alternatively, find a prime factorisation for 10!.}$
\begin{align*}&\quad 10!\\ &= 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\\ &= (5\times2)\times (3^2)\times (2^3)\times 7 \times (3\times 2) \times 5 \times (2^2)\times 3\times 2\\ &= 2^8\times 3^4 \times 5^2 \times 7\end{align*}
So just reading off that, the answer to the first one is 8 and the answer to the second one is 2.

brooksykait · « **Reply #3259 on:** February 12, 2018, 09:29:16 am »

Hi, I'm really struggling with 3u curve sketching. Here's a question that I'm struggling with:
Sketch, showing all stationary points, asymptotes and points of inflextion
y = 2/(x²+3)

(From Magaret Grove Textbook)
Thank you

RuiAce · « **Reply #3260 on:** February 12, 2018, 09:40:15 am »

Quote from: brooksykait on February 12, 2018, 09:29:16 am

Hi, I'm really struggling with 3u curve sketching. Here's a question that I'm struggling with:
Sketch, showing all stationary points, asymptotes and points of inflextion
y = 2/(x²+3)

(From Magaret Grove Textbook)
Thank you

$\text{Subbing }y=0\text{ gives }0 = 2\\ \text{so this curve has no }x\text{-intercepts}$
$\text{Also, setting the denominator to 0, }x^2+3=0\text{ has no solution}\\ \text{so this curve has no vertical asymptotes.}$
_________________________________________________________________
$\text{The }y\text{-intercept is at }\left(0, \frac23\right).\\ \text{Doing the classic 'divide by highest power on }x\\ \text{our horizontal asymptote can be shown to be }y = 0.$
_________________________________________________________________
$\frac{dy}{dx} = \frac{-4x}{(x^2+3)^2}\\ \text{Setting }\frac{dy}{dx} = 0\text{ shows that we have a stationary point at }x=0.$
$\text{Following an extremely messy quotient rule computation}\\ \frac{d^2y}{dx^2} = \frac{12(x^2-1)}{(x^2+3)^3}\\ \text{First note that at }x=0, \frac{d^2y}{dx^2} = -\frac49 < 0\\ \text{ and hence }\left(0,\frac23\right)\text{ is a local maximum.}$
The possible points of inflexion occur when $ \frac{d^2y}{dx^2}=0 $ which will be when $x = \pm 1$. You can then do a check to determine that indeed, we do have stationary points there.
$\text{So ultimately the graph looks something like this.}$

These types of questions are fairly straightforward so you are expected to know how to do them. The difference is that they're more computationally demanding so if there are any particular bits you're unsure about, clearly state what they are.

Never.Give.Up · « **Reply #3261 on:** February 12, 2018, 06:55:20 pm »

hello,
im having some difficulties with induction...

i would truly appreciate it if someone could help me with this question.
-4+8-16+...-4(-2)^n-1= 4((-2)-1) all over 3, for all n>=1
I have proved n=1 and assumed n=k, but i am just stuck on proving n=k+1

....this seems to be the trend with most questions i do

...if there are any tricks or something to help it would be great to know

thanks

Dragomistress · « **Reply #3262 on:** February 12, 2018, 07:34:46 pm »

Quote from: RuiAce on February 12, 2018, 09:40:15 am

$\text{Subbing }y=0\text{ gives }0 = 2\\ \text{so this curve has no }x\text{-intercepts}$
$\text{Also, setting the denominator to 0, }x^2+3=0\text{ has no solution}\\ \text{so this curve has no vertical asymptotes.}$
_________________________________________________________________
$\text{The }y\text{-intercept is at }\left(0, \frac23\right).\\ \text{Doing the classic 'divide by highest power on }x\\ \text{our horizontal asymptote can be shown to be }y = 0.$
_________________________________________________________________
$\frac{dy}{dx} = \frac{-4x}{(x^2+3)^2}\\ \text{Setting }\frac{dy}{dx} = 0\text{ shows that we have a stationary point at }x=0.$
$\text{Following an extremely messy quotient rule computation}\\ \frac{d^2y}{dx^2} = \frac{12(x^2-1)}{(x^2+3)^3}\\ \text{First note that at }x=0, \frac{d^2y}{dx^2} = -\frac49 < 0\\ \text{ and hence }\left(0,\frac23\right)\text{ is a local maximum.}$
The possible points of inflexion occur when $ \frac{d^2y}{dx^2}=0 $ which will be when $x = \pm 1$. You can then do a check to determine that indeed, we do have stationary points there.
$\text{So ultimately the graph looks something like this.}$
(Image removed from quote.)
These types of questions are fairly straightforward so you are expected to know how to do them. The difference is that they're more computationally demanding so if there are any particular bits you're unsure about, clearly state what they are.

I am actually interested in this more clever method.

Also, how do I do this?

RuiAce · « **Reply #3263 on:** February 12, 2018, 08:18:53 pm »

Quote from: Dragomistress on February 12, 2018, 07:34:46 pm

I am actually interested in this more clever method.

Also, how do I do this?

Alright, here's the more clever method. Let me know if there's some bits that confuse you.

Setting the whole idea up using the 10! example

$\text{Consider the question for 10! from another perspective.}\\ \text{We will still do part a) where we're interested in multiples of 2.}$
$\text{The first step is to note that the powers of 2 are}\\ 1, 2, 4, 8, 16, 32, 64, \dots\\ \text{and that of these, 8 is the largest number that is still less than 10.}$
________________________________________________________
$\text{We recall, again, that}\\ 10! = 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1.\\ \text{This following approach focuses on the frequency of 2's appearing}\\ \textbf{however }\text{by analysing the powers of 2, up to 8.}$
$\text{In the expansion of 10!}\\ \text{every number that is a multiple of 2 will obviously spit out a factor of 2.}$
E.g. 6 is a multiple of 2, and it will spit out a factor of 2 since $6 = 2\times 3$.
$\text{Every time a multiple of 2 appears, the amount of times 2 divides into 10!}\\ \text{will always go up by one.}$
What that says is consider this. The multiples of 2 that appear are 2, 4, 6, 8 and 10. Since 2 appears, the amount of times 10! is divisible by 2 is at least one. Since 4 also appears, the amount of times 10! is divisible by 2 will also go up again by one (in fact, two, but we temporarily ignore that and say one). Since 6 also appears, 10! is divisible by 2 yet again.
_______________________________________________________
$\text{Now, this is nice, as }2 = 2\times 1, 6 = 2\times 3\text{ and }10 = 2\times 5.\\ \text{But of course,} 4 = 2^2.\\ \text{So the idea is that we still aren't done.}$
$\text{If you think about it, every time }\textbf{multiples of 4}\\ \text{are }\textbf{also}\text{ present in the expansion of 10!}\\ \text{the amount of times 10! is divisible by 2}\\ \textbf{must go up yet again.}$
That is to say, since 10! is divisible by the following multiples of 2: (2,4,6,8,10), 10! is divisible by $2^5$ already. But since 10! is also divisible by the following multiples of 4: (4,8), 10! is also divisible by yet another $2^2$. And hence, 10! is now divisible by $2^7$.
$\text{Essentially, we first consider numbers as multiples of 2,}\\ \textbf{before}\text{ addressing those again, that are }\textbf{also}\text{ multiples of 4.}$
_______________________________________________________
$\text{And lastly, we must consider the numbers that are also multiples of 8.}$
Note that the whole point of noting that 8 was the largest number below powers of 2 was so that we knew when to stop. We do not have to consider numbers that are also multiples of 16, since 16 doesn't even appear in the expansion of 10!.
$\text{Of course, the only number that's a multiple of 8 here is just 8 itself.}\\ \text{And thus we only increase the amount of times 10! is divisible by 2, once.}$
Giving us our final result of $2^8$, and hence the answer is 8.

$\text{And now I'll take all of that, and do it for the 100! problem.}\\ \text{For 100!, the largest power of 2 that's less than 100 is 64.}$
$\text{The number of multiples of 2 that are less than or equal to 100}\\ \text{will simply be }\frac{100}{2} = 50$
So $100!$ must be divisible by $2^{50}$
$\text{The number of multiples of 4 that are less than or equal to 100}\\ \text{will also simply be }\frac{100}{4} = 25$
So $100!$ must be divisible by a further $2^{25} $
$\text{Now, as for the number of multiples of 8 less than or equal to 100,}\\ \text{we realise that by doing }\frac{100}{8}\text{ we get }12.5\\ \text{How do we resolve this issue?}$
$\text{The correct answer is that when we have some decimal left over, we will always round down.}\\ \text{This is because the first 12 multiples of 8 will be }8, 16, 24, \dots, 96\\ \text{but then you realise that the thirteenth multiple of 8 is }104.\\ \text{This is obviously too large.}$
Hence, $100!$ must be divisible by a further $2^{12}$ as well.
$\text{And then we rinse and repeat for 16, 32 and 64.}$
After considering multiples of 16, $100!$ is divisible by another $2^6 $
After considering multiples of 32, $100!$ is divisible by another $2^3$
After considering multiples of 64, $100!$ is divisible by another $2^1$
$\text{And thus, the greatest power of 2 that 100! is divisible by will be}\\ 2^{50}\times 2^{25}\times 2^{12}\times 2^6\times 2^3\times 2^1 =\boxed{2^{97}}$
...giving us a final answer of 97.

Going a bit beyond HSC and into uni level maths...

$\text{This can then be generalised.}$
$\text{Suppose that we replace 100 with just any arbitrary number }n\\ \text{and 2 with another arbitrary number }k\text{, where }k < n.$
$\text{It can then be shown that the number of times }k\text{ divides into }n!\text{ will be}\\ \sum_{j=1}^{\lfloor \log_k n\rfloor} \left\lfloor \frac{n}{k^j}\right\rfloor$
where $ \lfloor x \rfloor$ denotes the floor function of $x$, which gives you the value of $x$ rounded down to the nearest integer

Dragomistress · « **Reply #3264 on:** February 13, 2018, 06:33:07 am »

Ohhh, I get it

So if you were to do something like find the factors of 3 in 120!

You would do,

120/3 = 40
120/9 ≈ 13
120/27 ≈ 4
120/81 ≈ 1
Therefore, the greatest power of 3 is 3^58 in 120!

RuiAce · « **Reply #3265 on:** February 13, 2018, 08:41:01 am »

Quote from: Dragomistress on February 13, 2018, 06:33:07 am

Ohhh, I get it

So if you were to do something like find the factors of 3 in 120!

You would do,

120/3 = 40
120/9 ≈ 13
120/27 ≈ 4
120/81 ≈ 1
Therefore, the greatest power of 3 is 3^58 in 120!

Yep, looks good to me

________________________________________________________________

Your other question:
$\text{In the case }k \le n,\\ \text{If }f(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_k x^k +\dots + a_1 x + a_0\\ \text{then differentiating }k\text{ times gives}\\ f^{(k)}(x) = b_{n}x^{n-k} + b_{n-1}x^{n-k-1} + \dots + b_k$
$\text{where}\\ \begin{align*}b_n &= a_n \times n(n-1)\dots (n-k+1)\\ b_{n-1}&=a_{n-1}\times (n-1)(n-2)\dots(n-k)\\ &\vdots\\ b_k &= a_k \times k(k-1)\dots (1)\end{align*}$
I'm fairly sure you'd be safe doing something like this in 3U. However if you feel this method is a bit dodgy you can always prove it by induction.
$\text{Upon plugging in 0, we see that a whole bunch of terms cancel out.}\\ \text{All we are left with is }\boxed{f^{(k)}(0) = a_k \times k(k-1)\dots(1)}\\ \text{which implies }f^{(k)}(0) = a_k k!$
________________________________________________
$\text{And when }k > n\text{, we can properly prove it the same way as above}\\ \text{or we can use common sense.}$
$\text{Just consider }x^n.\\ \text{Whenever you differentiate this }\textbf{exactly }n\text{ times, you get }n!.\\ \text{So if you differentiate it more than }n\text{ times, you differentiate a constant which gets you back to 0.}$
$\text{Of course, in our case this doesn't just happen to }x^n,\\ \text{but also to }x^{n-1}\text{ and everything that follows it.}\\ \text{Therefore, for }k > n\text{ we have }f^{(k)}(x) = 0\\ \text{and then just subbing }x=0\text{ gives }\boxed{f^{(k)}(0)=0}$
________________________________________________
$\text{Before concluding, we note that when }k > n\\ \text{if }f^{(k)}(0)=0\\\text{then multiplying by stuff not equal to 0 will keep things that way. i.e.}\\ \frac{f^{(k)}(0)}{k!} x^k = 0$
$\text{From here, we just use what we know and do some clever rearranging.}$
We basically plug in values for $k = n, n-1, \dots, 1, 0$ first, and then also plug in for $k = n+1, n+2, \dots$
$\begin{align*}f(x) &= a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\\ &= \left( \frac{a_n n!}{n!} x^n + \frac{a_{n-1} (n-1)!}{(n-1)!} x^{n-1} + \dots + \frac{a_1 1!}{1} x + \frac{a_0 0!}{0!}\right) + \left( 0 + 0 + \dots\right )\\ &= \left(\frac{f^{(n)}(0)}{n!} x^n + \frac{f^{(n-1)}(0)}{(n-1)!} x^{n-1} + \dots + \frac{f^\prime (0)}{1!}x + \frac{f(0)}{0!}\right) +\left( \frac{f^{(n+1)(0)}x^{n+1}}{(n+1)!}x^{n+1} + \frac{f^{(n+2)(0)}x^{n+2}}{(n+2)!}x^{n+2} + \dots\right) \\ &= \sum_{k=0}^n \frac{f^{(k)}(0) \times x^k}{k!} + \sum_{k=n+1}^{\infty} \frac{f^{(k)}(0) \times x^k}{k!}\\ &= \sum_{k=0}^\infty \frac{f^{(k)}(0) \times x^k}{k!}\end{align*}$
Aside: This is actually a sneak peek into Taylor series. It goes on to eventually proving that the Taylor series for any polynomial is actually itself.

Quote from: Never.Give.Up on February 12, 2018, 06:55:20 pm

hello,
im having some difficulties with induction...

i would truly appreciate it if someone could help me with this question.
-4+8-16+...-4(-2)^n-1= 4((-2)-1) all over 3, for all n>=1
I have proved n=1 and assumed n=k, but i am just stuck on proving n=k+1

....this seems to be the trend with most questions i do ...if there are any tricks or something to help it would be great to know

thanks

By the way, you most likely mistyped your question. Your RHS doesn't make sense since you don't have $n$ anywhere.

mirakhiralla · « **Reply #3266 on:** February 13, 2018, 03:29:37 pm »

I just started trigonometric functions,
can someone help me with this question:

ABC has a perimeter of 12m, show that the area is given by:
A= 72x/(x+2)^2 (x is the angle theta, I was unable to write it)

Thank you.

RuiAce · « **Reply #3267 on:** February 13, 2018, 06:20:47 pm »

Quote from: mirakhiralla on February 13, 2018, 03:29:37 pm

I just started trigonometric functions,
can someone help me with this question:

ABC has a perimeter of 12m, show that the area is given by:
A= 72x/(x+2)^2 (x is the angle theta, I was unable to write it)

Thank you.

I presume you want the area of ABC but what is ABC? Is it a triangle, sector, ...?

And which angle is theta? Is it angle ABC, or angle BAC, ...

Never.Give.Up · « **Reply #3268 on:** February 13, 2018, 06:29:31 pm »

Quote from: RuiAce on February 13, 2018, 08:41:01 am

By the way, you most likely mistyped your question. Your RHS doesn't make sense since you don't have $n$ anywhere.

Ohhh my bad!! Sorry...
Well- I am stuck on Step 3 of both of these

9+14+19+...+(5n+4)= 5n^2/2+13n/2 for all n>=1

and

-4+8-16...-4(-2)^(n-1) = 4((-2)^n - 1)/3 for all n>= 1

dont know if that works....
thanks

RuiAce · « **Reply #3269 on:** February 13, 2018, 10:20:47 pm »

Quote from: Never.Give.Up on February 13, 2018, 06:29:31 pm

Ohhh my bad!! Sorry...
Well- I am stuck on Step 3 of both of these

9+14+19+...+(5n+4)= 5n^2/2+13n/2 for all n>=1

and

-4+8-16...-4(-2)^(n-1) = 4((-2)^n - 1)/3 for all n>= 1

dont know if that works....
thanks

$\text{If we assume that}\\ \boxed{9+14+19+\dots + (5k+4) = \frac{5k^2}{2} + \frac{13k}{2}}$
$\text{Then we can consider how}\\ \begin{align*}9+14+19+\dots + (5k+4) + (5k+9)&= \frac{5k^2}{2} + \frac{13k}{2} + 5k+9\\ &= \frac{5}{2}(k^2+2k+1) + \frac{13}{2}(k+1)\\ &= \frac{5(k+1)^2}{2} + \frac{13k}{2}.\end{align*}$

Making the transition from line 2 to line 3 a bit clearer..

\begin{align*}\frac{5k^2}{2} + \frac{13k}{2} + 5k + 9 &= \frac{5k^2}{2} + 5k + \frac{5}{2} + \frac{13k}{2} + \frac{13}{2}\\ &= \frac{5}{2}(k^2+2k+1) + \frac{13}{2}(k+1)\end{align*}

________________________________________________________________________
$\text{If we assume that}\\ \boxed{-4 + 8 - 16 + \dots -4(-2)^{k-1} = \frac{4\left((-2)^k-1 \right)}{3}}$
$\text{Then we can consider how}\\ \begin{align*}-4+8-16+\dots -4(-2)^{k-1} - 4(-2)^k &=\frac{4\left((-2)^k-1 \right)}{3} - 4(-2)^k\\ &= \frac{4(-2)^k - 4 - 12(-2)^k}{3} \\ &= \frac{-8(-2)^k - 4}{3}\\ &= \frac{4 \times (-2)(-2)^k - 4}{3}\\ &= \frac{4(-2)^{k+1}-4}{3}.\end{align*}$

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