Author Topic: 3U Maths Question Thread (Read 1237680 times) Tweet Share

cxmplete · « **Reply #2940 on:** October 22, 2017, 04:05:49 pm »

Hi! How would you do part i?

jamonwindeyer · « **Reply #2941 on:** October 22, 2017, 05:36:18 pm »

Quote from: cxmplete on October 22, 2017, 04:05:49 pm

Hi! How would you do part i?

Alrighty back into it! So essentially, if this formula works, we should be able to integrate the RHS and get the same thing, since that is what would happen to the LHS:

$\int\left(1+x+\frac{x^2}{2!}...\frac{x^n}{n!}...\right)dx=x+\frac{x^2}{2}+\frac{x^3}{3\times2!}...+\frac{x^{n+1}}{n!\times(n+1)}...$

See how the series is actually the same? Remember $2!\times3=3!$, and even though we've shown a different power in the middle of those two ellipses, it is still the same form, same series

that's actually all we need to do! Since we can integrate both sides and the formula still holds, that has verified it

Edit: Shit, forgot the constant. So there is a +C on the RHS, if we set x=0, we can easily evaluate C=1!

RuiAce · « **Reply #2942 on:** October 22, 2017, 05:43:07 pm »

Quote from: jamonwindeyer on October 22, 2017, 05:36:18 pm

Alrighty back into it! So essentially, if this formula works, we should be able to integrate the RHS and get the same thing, since that is what would happen to the LHS:

$\int\left(1+x+\frac{x^2}{2!}...\frac{x^n}{n!}...\right)dx=x+\frac{x^2}{2}+\frac{x^3}{3\times2!}...+\frac{x^{n+1}}{n!\times(n+1)}...$

See how the series is actually the same? Remember $2!\times3=3!$, and even though we've shown a different power in the middle of those two ellipses, it is still the same form, same series that's actually all we need to do! Since we can integrate both sides and the formula still holds, that has verified it

...+C...+1...
Why did you close Facebook!!

justwannawish · « **Reply #2943 on:** October 23, 2017, 07:47:47 pm »

Hey guys, how do you convert:
tan theta= (sqrt 3)/2 into radians?

K9810 · « **Reply #2944 on:** October 23, 2017, 09:27:08 pm »

Hey! How do I do part d?

Thanks

jamonwindeyer · « **Reply #2945 on:** October 23, 2017, 11:28:31 pm »

Quote from: justwannawish on October 23, 2017, 07:47:47 pm

Hey guys, how do you convert:
tan theta= (sqrt 3)/2 into radians?

Hey! You can't really convert those into radians, they are just expressions, but in general if you have something in degrees, and you want it in radians, you multiply by:

$\frac{\pi}{180}$

So for example, 1 degree is equal to $\frac{\pi}{180}$ radians

Quote from: K9810 on October 23, 2017, 09:27:08 pm

Hey! How do I do part d?

Thanks

In that domain, the function $f$ is invertible - Meaning if we apply the function and the inverse, they cancel each other out.

$f^{-1}\left(f(a)\right)=a$

winstondarmawan · « **Reply #2946 on:** October 25, 2017, 12:04:47 am »

Would appreciate help with this question from the 2001 HSC:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22831344_1359447470847439_790338668_o.png?oh=c9fb0ccdb28e9526e17a90387c5d52cb&oe=59F17EE7
Part c (ii).
The answer I got for (b) was:
If n is odd, 2p^n.
If n is even, 2 nC1 q p^(n-1)
TIA.

itssona · « **Reply #2947 on:** October 25, 2017, 09:29:05 am »

pls help how do I factorise
3x^3-7x^2+4

RuiAce · « **Reply #2948 on:** October 25, 2017, 11:34:16 am »

Quote from: winstondarmawan on October 25, 2017, 12:04:47 am

Would appreciate help with this question from the 2001 HSC:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22831344_1359447470847439_790338668_o.png?oh=c9fb0ccdb28e9526e17a90387c5d52cb&oe=59F17EE7
Part c (ii).
The answer I got for (b) was:
If n is odd, 2p^n.
If n is even, 2 nC1 q p^(n-1)
TIA.

I'm doing all of what's there mainly because I want to add it to the compilation.
$\text{In the expansion of }(q-p)^n\\ \text{The last term involves a factor of }(-1)^n\\ \text{When }n\text{ is odd, this equals to }-1\\ \text{When }n\text{ is even, this equals to }1.$
$\text{So it's possible to compare which will be the last term.}\\ \begin{align*}(q-p)^n &=\binom{n}{0}q^n -\binom{n}1 q^{n-1}p + \binom{n}2 q^{n-2}p^2 - \binom{n}3 q^{n-3}p^3 + \dots + \binom{n}{n-1}qp^{n-1} - \binom{n}{n} p^n \tag{n odd}\\ \tag{n even} (q-p)^n &=\binom{n}{0}q^n -\binom{n}1 q^{n-1}p + \binom{n}2 q^{n-2}p^2 - \binom{n}3 q^{n-3}p^3 + \dots - \binom{n}{n-1}qp^{n-1} + \binom{n}{n} p^n\end{align*}\\ \text{and of course,}\\ (q+p)^n =\binom{n}{0}q^n +\binom{n}1 q^{n-1}p + \binom{n}2 q^{n-2}p^2 + \binom{n}3 q^{n-3}p^3 + \dots + \binom{n}{n-1}qp^{n-1} + \binom{n}{n}p^n$
$\text{So in the case when }n\text{ is odd}\\ (q+p)^n - (q-p)^n = 2\binom{n}{1}q^{n-1}p + 2\binom{n}{3} q^{n-3}p^3 + \dots + \boxed{2\binom{n}{n}p^n}$
$\text{And in the case when }n\text{ is even}\\ (q+p)^n - (q-p)^n = 2\binom{n}1q^{n-1}p+2\binom{n}3q^{n-3}p^3 + \dots + \boxed{2\binom{n}{n-1}qp^{n-1}}$
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$\text{The probability of a 6 in one particular dice roll is }\frac16\\ \text{so since we need }r\text{ 6's in }n\text{ rolls}\\ \text{the binomial probability formula gives }\binom{n}{r}\left(\frac56\right)^{n-r}\left(\frac16\right)^r$
__________________
$\text{Essentially, we put }\boxed{p=\frac16}\text{ and }\boxed{q=\frac56}\\ \text{into the formula in part b).}$
$\text{Whilst we can technically break the cases into }n\text{ odd and }n\text{ even}\\ \text{we really don't need to. This is because the formula}\\ \boxed{ (q+p)^n - (q-p)^n = 2\binom{n}1q^{n-1}p+2\binom{n}3q^{n-3}p^3 + \dots }\\ \text{holds }\textit{regardless}\text{ of the parity of }n.$
All that really changes is the last term, but we'll find that this isn't much of a problem at all. This is what tends to throw people off, and it can be resolved by just considering intuitively what goes on; you don't really care about how MANY terms there are if r is odd, you just care that r is odd and nothing more.
$\text{Now, using our formula from part c) i), we want to sum up all the}\\ \text{probabilities for when }r=1,3,5,\dots\\ \text{This will be}\\ \boxed{\binom{n}{1}\left(\frac56\right)^{n-1}\left(\frac16\right) + \binom{n}{3}\left(\frac56\right)^{n-3}\left(\frac16\right)^3 + \dots}\\ \text{so using the substitutions and our formula in part i), this equals}\\ \boxed{\frac12\left[\left(\frac56+\frac16\right)^n - \left(\frac56-\frac16\right)^n\right]}$
Remember, we had extra 2's floating around in the formula in part b)
$\text{Which is just }\frac12\left[1-\left(\frac23\right)^n\right]$

RuiAce · « **Reply #2949 on:** October 25, 2017, 11:39:09 am »

Quote from: sssona09 on October 25, 2017, 09:29:05 am

pls help how do I factorise
3x^3-7x^2+4

$\text{Testing }x=1\\ 3(1)^3 - 7(1)^2 + 4 = 0\\ \text{so }(x-1)\text{ is a factor.}$
$\text{At this point, we may use polynomial long division.}\\ \text{Alternatively, for a 'sly' way of factorising}\\ \begin{align*}&\quad3x^3-7x^2+4\\ &= 3x^3-3x^2 - 4x^2 + 4\\ &= 3x^2(x-1) - 4x^2 + 4x - 4x +4\\ &= 3x^2(x-1)-4x(x-1)-4(x-1)\\ &= \left(3x^2-4x-4\right)(x-1)\end{align*}$
$\text{Which factorises to}\\ (3x+2)(x-2)(x-1)$

K9810 · « **Reply #2950 on:** October 25, 2017, 05:16:42 pm »

Hi, I need help with this question: Prove by mathematical induction that 2^n>=n^2

Thanks

RuiAce · « **Reply #2951 on:** October 25, 2017, 05:50:16 pm »

Quote from: K9810 on October 25, 2017, 05:16:42 pm

Hi, I need help with this question: Prove by mathematical induction that 2^n>=n^2

Thanks

We're not starting at n=1 are we? Because it fails when n=3.

K9810 · « **Reply #2952 on:** October 25, 2017, 05:54:36 pm »

Quote from: RuiAce on October 25, 2017, 05:50:16 pm

We're not starting at n=1 are we? Because it fails when n=3.

I forgot to put it down, values for n=4,5,6,...

RuiAce · « **Reply #2953 on:** October 25, 2017, 05:58:21 pm »

$\text{Under the condition that }n\ge 4\\ \text{we can prove that }n^2 \ge 2n+1$
This happens mostly because the equation solves to give $ n \ge 1+\sqrt{2} $, so if $ n \ge 4$ this is certainly satisfied.
$\text{When }n=4, LHS = 16\text{ and }RHS=16\\ \text{so certainly }LHS\ge RHS$
$\text{Assume that }\boxed{2^k \ge k^2}\\ \text{We need to prove that }2^{k+1}\ge (k+1)^2$
\begin{align*}2^{k+1}&= 2\left(2^k\right)\\ &\ge 2k^2\tag{assumption}\\ &= k^2 + k^2\\ &\ge k^2+2k+1 \tag{from above}\\ &= (k+1)^2\end{align*}

K9810 · « **Reply #2954 on:** October 25, 2017, 07:36:40 pm »

Hey, how can I use the tangent condition in this question: Find the equation of each of the following parabola; vertex at (3,−1), axis parallel to the y-axis and the line 4x + y−7 = 0 is a tangent.

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