Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

arii · « **Reply #1740 on:** March 01, 2018, 07:32:20 pm »

If u_n = 2ⁿ⁺² + 3²ⁿ⁺¹, show that u_n+1 = 2u_n + 7*3²ⁿ⁺¹. Hence, show that u_n is divisible by 7 for n≥1.

jazzycab · « **Reply #1741 on:** March 01, 2018, 09:01:46 pm »

Quote from: arii on March 01, 2018, 07:32:20 pm

If u_n = 2ⁿ⁺² + 3²ⁿ⁺¹, show that u_n+1 = 2u_n + 7*3²ⁿ⁺¹. Hence, show that u_n is divisible by 7 for n≥1.

If $u_n=2^{n+2}+3^{2n+1}$ then:
$\begin{align*}u_{n+1}&=2^{\left(n+1\right)+2}+3^{2\left(n+1\right)+1}\\&=2^{\left(n+2\right)+1}+3^{\left(2n+1\right)+2}\\&=2\times2^{n+2}+9\times3^{2n+1}\end{align*}$ . We can split the second term into $2\times3^{2n+1}+7\times3^{2n+1}$:
$\begin{align*}2\times2^{n+2}+9\times3^{2n+1}&=2\left(2^{n+2}+3^{2n+1}\right)+7\times3^{2n+1}\\&=2u_n+7\times3^{2n+1}\end{align*}$
For the second part, we can prove this using induction:
$\begin{align*}\text{Initial case, }n&=1:u_1=2^{1+2}+3^{2\times1+1}\\&=2^3+3^3\\&=8+27\\&=35\\&=7\times5\\\text{Therefore }&u_1\text{ is divisible by 7}\\\text{Assume }u_k&=7\alpha\\u_{k+1}&=2u_k+7\times3^{2k+1}\\&=2\times7\alpha+7\times3^{2k+1}\\&=7\left(2\alpha+3^{2k+1}\right)\\&=7\beta\\\text{Therefore, if }u_k&\text{ is divisible by 7, then so is }u_{k+1}\\\text{Therefore, by mathematical induction, we}&\text{ have proven that }u_n\text{ is divisible by 7 for }n\in\mathbb{N}\end{align*}$

Lefkiiii6 · « **Reply #1742 on:** March 02, 2018, 08:41:28 pm »

How do you graph y=e^f(x) when given the graph of y=f(x). I think it is simply dealing what is below x-axis and then everything else is kind of higher up than original f(x) but thats just by looking at desmos and looking at some graphs

RuiAce · « **Reply #1743 on:** March 02, 2018, 08:46:20 pm »

Quote from: Lefkiiii6 on March 02, 2018, 08:41:28 pm

How do you graph y=e^f(x) when given the graph of y=f(x). I think it is simply dealing what is below x-axis and then everything else is kind of higher up than original f(x) but thats just by looking at desmos and looking at some graphs

$\text{Consider the key characteristics of }y=e^x.\\ e^0 = 1\\ \text{As }x \to -\infty, e^x\to 0\\ \text{As }x\to +\infty, e^x \to +\infty.$
$\text{So any }x\text{-intercepts of }y=f(x)\text{ will result in}\\ y=e^{f(x)}\text{ intersecting the line }y=1$
$\text{If }f(x)\text{ decreases as }x\to \infty\text{ (or to }-\infty)\\ \text{then }e^{f(x)}\text{ will asymptotically approach }y=0$
$\text{If }f(x)\text{ shoots all the way down to a vertical asymptote}\\ e^{f(x)}\text{ will decrease towards the line }y=0\\ \text{and eventually we'd have a hole in the graph on the line }y=0.$
$\text{Wherever }f(x)\text{ blows up, }e^{f(x)}\text{ blows up as well.}$
$\text{Since }e^x\text{ is monotonic increasing,}\\ \text{if }f(x)\text{ increases then so does }e^{f(x)}\\ \text{if }f(x)\text{ decreases then so does }e^{f(x)}$
These are just some examples of heaps of things you can consider.

kaustubh.patel · « **Reply #1744 on:** March 03, 2018, 09:05:11 pm »

A bit of help with complex polynomial division please.

RuiAce · « **Reply #1745 on:** March 03, 2018, 09:11:17 pm »

Quote from: kaustubh.patel on March 03, 2018, 09:05:11 pm

A bit of help with complex polynomial division please.

$\text{Noting that the coefficients are real, another root is }x=1+i\\ \text{and hence a quadratic factor of the polynomial is}\\ \big( x- (1-i) \big) \big( x - (1+i)\big) = x^2 -2x + 2$

(For now, I'll presume you know how to finish off the remainder of the question.)

(Note that LaTeX "might've" done polynomial long division a bit differently to what you're used to, with the formatting.) Alternatively, if you view my lecture slides for the half yearly lectures, there's a different method I introduce that avoids polynomial long division altogether.

kaustubh.patel · « **Reply #1746 on:** March 04, 2018, 01:39:52 pm »

Ohh damn Rui i completely overlooked the fact that if the coefficients are real than the conjugate is also a root. I actually attempted long division with complex divisor (my teacher did something similar in class) but it was horrendously long.
Yikes i do feel a bit dumb now for forgetting a good amount of what I'd noted in the slides, thanks heaps for sharing em I realised the equalities & equating coefficients method is hella lot cleaner.

justwannawish · « **Reply #1747 on:** March 04, 2018, 08:36:41 pm »

Hi, could you plese help me out with this question?

Given that arg(z+1)=pi/6 and arg(z-1)=2pi/3, write z in the form x+iy where x and y are real numbers

Thank you!

RuiAce · « **Reply #1748 on:** March 04, 2018, 08:52:18 pm »

Quote from: justwannawish on March 04, 2018, 08:36:41 pm

Hi, could you plese help me out with this question?

Given that arg(z+1)=pi/6 and arg(z-1)=2pi/3, write z in the form x+iy where x and y are real numbers

Thank you!

$\text{Let }P\text{ represent }z.$

$\text{Note that from the angle sum of a triangle, }\angle APB = \frac\pi2.\\ \text{Hence, from circle geometry, }AB\text{ is a diameter of the circle through the points }A,B,P.$
$\text{But }AB\text{ is literally the diameter of the circle }|z| = 1.\\ \text{This means that }P\text{ also lies on the circle }\boxed{|z|=1}.$
$\text{Also note that from circle geometry, since the angle subtended}\\ \text{to the centre of the circle is twice the angle subtended to the circumference,}\\ \angle POB = 2\angle PAB \implies \boxed{\angle POB = \frac\pi3} \implies \boxed{\arg(z) = \frac\pi3}$
\begin{align*}\therefore z &= 1 \left( \cos \frac\pi3 + i \sin \frac\pi3\right)\\ &= \frac12 + \frac{\sqrt3}{2}i\end{align*}

Lefkiiii6 · « **Reply #1749 on:** March 05, 2018, 05:53:59 pm »

Hi I came across this in the 2001 HSC paper and I don't know what to do and can't find solutions. Can anyone assist?

Hi, I'm also unsure about this question. Any help appreciated. Thankyou!

Mod edit: Merged. Please use the "Modify" button at times like this to refrain from chain posting.

RuiAce · « **Reply #1750 on:** March 05, 2018, 06:58:53 pm »

Quote from: Lefkiiii6 on March 05, 2018, 05:53:59 pm

Hi I came across this in the 2001 HSC paper and I don't know what to do and can't find solutions. Can anyone assist?

Hi, I'm also unsure about this question. Any help appreciated. Thankyou!

Mod edit: Merged. Please use the "Modify" button at times like this to refrain from chain posting.

Please find the solution for the latter linked in the compilation. (The first one will be added shortly.)
$\text{Assuming that }x = \frac{p}{q}\text{ is a root, upon subbing in we have}\\ \begin{align*}\frac{ap^3}{q^3} - \frac{3p}{q} + b & = 0\\ ap^3 - 3pq^2 + bq^3 &= 0\\ ap^3 - 3pq^2&= bq^3\\ p(ap^2 - 3q^2) &=- bq^3\end{align*}$
$\text{Now, carefully note that }\textbf{if}\, p\text{ times "something", is equal to }-bq^3\\ \text{then }p\text{ must be a }\textbf{factor}\text{ of }bq^3.$
$\text{But if }b\text{ is a factor of }bq^3,\\ \text{then it must either be a factor of }b\text{, or it is a factor of }q.$
Carefully note that the minus sign here can be safely ignored, because -1 is a factor of every integer and thus is not really relevant.
$\text{However the latter is a contradiction. This is because we were }\textbf{told}\\ \text{that }p\text{ and }q\text{ have no common integer divisors}\\\text{except for }\pm 1\text{, which means nothing here.}$
$\text{Therefore, the }\textbf{only}\text{ plausible conclusion here is that}\\ \boxed{p\text{ is a factor of }b.}$
$\text{Alternatively, we could've rearranged our equation from earlier}\\ \text{to yield }ap^3 = 3pq^2-bq^3 \implies \boxed{q(3pq - bq^2) = ap^3}.\\ \text{A similar argument to the above shows that }\boxed{q\text{ is a factor of }a.}$
________________________
$\text{To now show that (*) has no rational roots, observe that}\\ (*)\text{ is basically the case when }a=1\text{ and }b=-1.$
$\text{Since }b\text{ is divisible by }p,\text{ this means }-1\text{ is divisible by }b.\\ \text{The only possible values here are }p=\pm 1.\\ \text{Similarly, }q=\pm 1.$
$\text{Grouping these cases, the only candidate values for }x\\ \text{are therefore }x=\pm 1.\\ \text{But when }x=-1, \quad x^3-3x-1 = 1\neq 0\\ \text{and when }x=1, \quad x^3-3x-1 = -3\neq 0.$
$\text{Hence the only candidate values don't work.}\\ \text{So it must be the case that }x^3+3x-1=0\text{ has no rational roots.}$
____________________________________________________________
$\text{Upon subbing the root }x=r+s\sqrt{d}\text{ into }(*)\\ \begin{align*}(r+s\sqrt{d})^3 - 3(r+s\sqrt{d}) - 1 &= 0\\ r^3 + 3r^2s\sqrt{d} + 3rs^2d + s^3d\sqrt{d} - 3r - 3s\sqrt{d} - 1 &= 0\\ \underbrace{\left(r^3+3rs^2d-3r-1 \right)}_{\text{this is }\textbf{rational}} + \underbrace{\sqrt{d}\left(3r^2s+s^3d-3s\right)}_{\text{this is }\textbf{irrational}}&=0\end{align*}$
$\text{The underbraces are important.}\\ \text{The second term is irrational, because thanks to }r,s\text{ and }d\text{ being rational,}\\ \text{so must }3r^2s+s^3d-3s\text{ be rational.}$
$\text{Yet }\sqrt{d}\text{ is irrational.}\\ \text{Whenever you multiply a rational and irrational number,}\\ \text{the result is guaranteed to be irrational.}$
________________________
$\text{So what we're saying is that the rational number }\boxed{r^3+3rs^2d-3r-1}\\ \text{plus the IRrational number }\boxed{\sqrt{d}\left(3r^2s+s^3d-3s\right)}\\ \text{is equal to 0.}$
$\text{Which normally cannot happen.}\\ \text{The only way this can happen is if both the rational 'part'}\\ \text{and the irrational 'part', are }\textbf{both equal to 0 themselves.}$
$\text{Setting the irrational part equal to 0 gives}\\ \sqrt{d}\left(3r^2s+s^3d-3s\right)=0\\ \text{Note that }d\text{ cannot be 0, or else }\sqrt{d}\text{ is rational.}\\ \text{So dividing by }\sqrt{d}\text{ produces the desired result }\boxed{3r^2s+s^3d-3s=0}$
________________________
$\text{Furthermore, setting the rational part equal to 0 gives}\\ \boxed{r^3+3rs^2d - 3r - 1 = 0}$
$\text{Now, when }x=r - s\sqrt{d}\\ x^3-3x-1 = \left(r^3+3rs^2d-3r-1 \right) -\sqrt{d}\left(3r^2s+s^3d-3s\right)\\ \text{But thanks to the above, this is just equal to }0 + \sqrt{d} (0) \text{, which is equal to 0.}\\ \text{Therefore }r-s\sqrt{d}\text{ is also a root of }(*)$
____________________________________________________________
$\text{Now for the demanding part.}\\ \text{For the remainder of the question, }\sqrt{d}\textit{ can }\text{be rational.}\\ \text{So we have two cases.}$
(Note that this deduce part is actually not linked, notation wise, to the above. We just happen to use the same letter.)
$\text{Case 1: }\sqrt{d}\text{ is rational.}\\ \text{Then }r + s\sqrt{d}\text{ is also rational, so we may write it in the form }x = \frac{p}{q}\\ \text{as in part i).}$
Essentially, case 1 is equivalent to saying (*) actually has a rational root.
$\text{Using the results of part i), }p \text{ and }q\text{ must both divide 1.}\\ \text{Hence, the only candidates are }x = \pm 1.\\ \text{But we can try subbing }x = \pm 1\text{ in, and clearly both of them are not roots of }(*).\\ \text{Therefore this case yields nor roots of the form }r+s\sqrt{d}$
________________________
$\text{Case 2: }\sqrt{d}\text{ is irrational.}\\ \text{From earlier in part ii), this means }r + s\sqrt{d}\textbf{ and }r - s\sqrt{d}\text{ are roots of }(*).$
$\text{Because things cancel out easily, it makes sense to consider the sum of roots.}\\ \text{Let the third root be }\alpha\text{. Then, }\\ (r + s\sqrt{d}) + (r - s\sqrt{d}) + \alpha = 0 \implies \boxed{\alpha = -2r}$
$\text{Since }r\text{ is rational, so must }\alpha.\\ \text{But we literally just contradicted what we deduced in case 1.}\\ \text{Therefore this case is impossible as well,}$
$\textbf{Hence, the overall effect is that NO root}\\ \text{takes the form }x = r + s\sqrt{d}\\ \text{for any }r, s, d\text{ rational.}$
____________________________________________________________
$\text{The last part is easy.}\\ \text{Upon subbing }x = 2\cos \frac\pi9\text{, we have}\\ \begin{align*}x^3-3x^2-1&=8\cos^3\frac\pi9 - 6\cos \frac\pi9 - 1\\ &= 2\left(4\cos^3\frac\pi9 - 3\cos \frac\pi9 \right)-1\\ &= 2\cos \frac\pi3 - 1 \tag{using the identity}\\ &= 2 \times \frac12 - 1 \\ &= 0\end{align*}\\ \text{so }2\cos \frac\pi9 \text{ is a root.}$

Dragomistress · « **Reply #1751 on:** March 06, 2018, 05:54:46 pm »

How do I do this question?
There are 11 people and 2 circular tables, one has 5 seats and the other has 6 seats. How many arrangements are possible?

roygbivmagic · « **Reply #1752 on:** March 06, 2018, 06:17:25 pm »

Hi, can you please explain how to do this question?
Prove that (x + y)^2 >= 4xy
Hence prove that 1/x^2 + 1/y^2 >= 4/(x^2 + y^2)

RuiAce · « **Reply #1753 on:** March 06, 2018, 06:24:22 pm »

Quote from: Dragomistress on March 06, 2018, 05:54:46 pm

How do I do this question?
There are 11 people and 2 circular tables, one has 5 seats and the other has 6 seats. How many arrangements are possible?

$\text{First choose 5 out of 11 people for the first table: }\binom{11}{5}\text{ choices.}\\ \text{Then arrange the 5 people: }4!\text{ ways.}\\ \text{Now choose 6 out of 6 people for the second table: }\binom66\text{ choices}\\ \text{and lastly arrange the 6 people: }5!$
$\therefore \text{Ans = }\binom{11}5 4! \binom{6}{6}5!$
Alternatively, the answer is also $ \binom{11}{6}5! \binom55 4! $ if we choose the people for the second table first.

RuiAce · « **Reply #1754 on:** March 06, 2018, 06:32:31 pm »

Quote from: roygbivmagic on March 06, 2018, 06:17:25 pm

Hi, can you please explain how to do this question?
Prove that (x + y)^2 >= 4xy
Hence prove that 1/x^2 + 1/y^2 >= 4/(x^2 + y^2)

$\text{The first part is standard and you're expected to know how to do it.}\\ \text{Using the fact that the square of any real number is positive,}\\ \begin{align*}(x-y)^2&\ge 0\\ x^2-2xy+y^2&\ge 0\\ x^2+2xy+y^2 &\ge 4xy\\ (x+y)^2 &\ge 4xy\end{align*}$
_____________________________________________________________________________________
$\text{Note that the given inequality rearranges to}\\ \boxed{x + y \ge 2\sqrt{xy}}$
$\text{First, replace }x \text{ with }\frac{1}{x^2}\text{ and }y\text{ with }\frac{1}{y^2}.\\ \text{We have }\boxed{\frac{1}{x^2}+\frac{1}{y^2} \ge \frac{2}{xy}}.\\ \text{It remains to check whether }\frac{2}{xy} \ge \frac{4}{x^2+y^2}$
$\text{Note from above, that}\\ \begin{align*}x^2-2xy+y^2&\ge 0\\ x^2+y^2&\ge 2xy\\ 2(x^2+y^2)&\ge 4xy\\ \frac{2}{xy}\ge \frac{4}{x^2+y^2}\end{align*}\\ \text{as required, noting that }x, y > 0\text{ for our purposes.}$
$\text{Hence combining the two inequalities, we have}\\ \frac{1}{x^2}+\frac{1}{y^2} \ge \frac{2}{xy} \ge \frac{4}{x^2+y^2} \implies \boxed{\frac{1}{x^2}+\frac{1}{y^2} \ge \frac{4}{x^2+y^2}}$

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