Author Topic: 3U Maths Question Thread (Read 1465267 times) Tweet Share

queenie09 · « **Reply #1485 on:** February 13, 2017, 10:14:51 pm »

Quote from: RuiAce on February 13, 2017, 09:26:00 pm

If the answer given was 32 (like your presumably deleted post said) then that's definitely wrong, because 32 is larger than the answer for part a), which is 24.

There should never be more outcomes with a restriction than without.

I'm sticking with my answer, so if you got the same thing then yes the answer is wrong.

Ok thanks for letting me know!

legorgo18 · « **Reply #1486 on:** February 14, 2017, 07:14:18 pm »

Hey guys need some help with the following inequality proving questions

If a,b and c are positive real numbers, prove that:
a) a^2 + b^2 + c^2 greater than or equal to ab + ac + bc (i tried (a-b-c)^2>0 but it didnt seem legit)
b) a^3 + b^3 + c^3 greater than or equal to 3abc (no clue)
c) (a+5b)(a+2b) greater than or equal to 9b(a+b) [i think it has something to do with a+b greater than or equal to 2 root ab but i couldnt get the answer]

Thanks for helping/ trying

RuiAce · « **Reply #1487 on:** February 14, 2017, 07:42:37 pm »

Quote from: legorgo18 on February 14, 2017, 07:14:18 pm

Hey guys need some help with the following inequality proving questions

If a,b and c are positive real numbers, prove that:
a) a^2 + b^2 + c^2 greater than or equal to ab + ac + bc (i tried (a-b-c)^2>0 but it didnt seem legit)
b) a^3 + b^3 + c^3 greater than or equal to 3abc (no clue)
c) (a+5b)(a+2b) greater than or equal to 9b(a+b) [i think it has something to do with a+b greater than or equal to 2 root ab but i couldnt get the answer]

Thanks for helping/ trying

These questions are 4U difficulty. I am not sure why they are being asked of a 3U student. What is the source of these questions?
$\text{Given }x,y\ge 0\\ \begin{align*}(x-y)^2&\ge 0\\x^2+2xy+y^2&\ge 4xy\\ (x+y)^2 &\ge 4xy\\ x+y&\ge 2\sqrt{xy}\end{align*}$
$\text{Let }x=a^2\text{ and }y=b^2\text{ to get}\\ a^2+b^2\ge 2ab\\ \text{Rinse and repeat with other combinations to get}\\ b^2+c^2\ge 2bc\\ a^2+c^2\ge 2ac$
$\text{Add:}\\ \begin{align*}a^2+b^2+b^2+c^2+a^2+c^2&\ge 2ab+2ac+2bc\\ a^2+b^2+c^2&\ge ab+ac+bc\end{align*}$
____________________________________________________________________________
$\text{From earlier, }\frac{x+y}{2}\ge \sqrt{xy}$
Tagged: AM-GM inequality
$\text{Commence by letting }x=a, y=b\\ \text{then }x=c, y=d\text{ before summing and halving:}\\ \begin{align*}\frac{a+b+c+d}4&=\frac{\frac{a+b}2+\frac{c+d}2}2\\&\ge \frac{\sqrt{ab}+\sqrt{cd}}{2}\\ &\ge \sqrt{\sqrt{ab}\sqrt{cd}}\tag{*}\\ &\ge {\sqrt[4]{abcd}}\end{align*}\\ (*)-\text{ apply the above result again with }x=\sqrt{ab}\text{ and }y=\sqrt{cd}$
$\text{Let }d=(abc)^{\frac13}\text{, then}\\ \begin{align*}\frac{1}{4}(a+b+c+d)&\ge \left(abc(abc)^\frac13\right)^\frac14\\ &= \left((abc)^\frac43\right)^\frac14\\ &\ge (abc)^\frac13\\ &= d\end{align*}$
$\text{So:}\\ \begin{align*}\frac{1}{4}(a+b+c+d)&\ge d\\\implies \frac{1}{4}(a+b+c)+\frac{d}4&\ge d\\ \implies \frac{1}{4}(a+b+c)&\ge \frac{3}{4}d\\ \implies \frac{1}{3}(a+b+c)&\ge d\\ \implies a+b+c&\ge \sqrt[3]{abc}\end{align*}$
$\text{Let }a=A^3, \, b=B^3, \, c=C^3\text{ to get the result}$
Clearly a bit inaccurate, but the flow of the proof is all there
____________________________________________________________________________
$\text{Work backwards on scrap paper, then write the}\\ \text{correct working out down as your answer:}$
$\begin{align*}(a-b)^2&\ge 0\\ \implies a^2-2ab+b^2&\ge 0\\ \implies a^2+7ab+10b^2&\ge 9ab+9b^2\\ \implies (a+5b)(a+2b)&\ge 9b(a+b)\end{align*}$

legorgo18 · « **Reply #1488 on:** February 14, 2017, 07:52:02 pm »

Hi rui, all these questions are in my dr du (math coaching in hurstville) homework and we finished projectiles last lesson and now this harder 3u stuff, thanks for helping anyway

RuiAce · « **Reply #1489 on:** February 14, 2017, 07:58:18 pm »

Quote from: legorgo18 on February 14, 2017, 07:52:02 pm

Hi rui, all these questions are in my dr du (math coaching in hurstville) homework and we finished projectiles last lesson and now this harder 3u stuff, thanks for helping anyway

That explains a lot.

Just an extra thing to leave: Another proof of the 3-variable AM-GM inequality given the 4-variable version. It splits off at the "let d=..." part

jamesq · « **Reply #1490 on:** February 15, 2017, 07:12:07 pm »

Hey guys, just wondering how you'll approach induction questions with a </> 0. This is from the 2005 paper.

RuiAce · « **Reply #1491 on:** February 15, 2017, 07:21:42 pm »

Quote from: jamesq on February 15, 2017, 07:12:07 pm

(Image removed from quote.)
Hey guys, just wondering how you'll approach induction questions with a </> 0. This is from the 2005 paper.

$\text{The expression is clearly true when }n=2\text{ as}\\ 4^2-1-7(2)=1>0\\ \text{Assuming the truth when }n=k\text{ gives}\\ 4^k-1-7k > 0$
$\text{The assumption can be rearranged (ala divisibility style) to}\\ \boxed{4^k > 1+7k}\\ \text{We must use this assumption to prove the truth of the statement when }n=k+1\\ \text{i.e.} \quad 4^{k+1}-1-7(k+1)>0$
$\begin{align*}LHS&=4^{k+1}-1-7(k+1)\\ &= 4(4^k) - 7k - 8\\ &> 4(1+7k) - 7k + 8 \tag{assumption used by substitution}\\ &= 12+21k\\ &> 0 \tag{*}\\ &= RHS\\ \therefore LHS &> RHS\end{align*}\\ \text{Hence true for }n=k+1\text{ if true for }n=k\\ \text{so true by mathematical induction.}$
$(*) -\text{ Note that since }k\ge 2\text{ we have}\\ 12+21k \ge 12+21(2) = 54 > 0$

bsdfjnlkasn · « **Reply #1492 on:** February 16, 2017, 08:12:29 pm »

Hey there!

When graphing trig functions, do we need to find the x-intercepts of the graph if it's been translated horizontally?
If so, could I please get some help with finding the x intercepts of y=2sinx+1 ? Thank you!!

RuiAce · « **Reply #1493 on:** February 16, 2017, 08:16:28 pm »

Quote from: bsdfjn;lkasn on February 16, 2017, 08:12:29 pm

Hey there!

When graphing trig functions, do we need to find the x-intercepts of the graph if it's been translated horizontally?
If so, could I please get some help with finding the x intercepts of y=2sinx+1 ? Thank you!!

$\text{For this one, it's possible by considering the general solution}\\ \begin{align*}2\sin x + 1 &=0\\ \sin x &= -\frac12\\ x&= n\pi +(-1)^n\left(-\frac{\pi}6\right)\end{align*}$
$\therefore x=\dots,-\frac{5\pi}{6},-\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}6, \dots$
$\text{Though in general, this is usually not necessary}\\ \text{As a rule of thumb, go for it when the answer is neat.}$

bsdfjnlkasn · « **Reply #1494 on:** February 16, 2017, 08:21:08 pm »

Quote from: RuiAce on February 16, 2017, 08:16:28 pm

$\text{For this one, it's possible by considering the general solution}\\ \begin{align*}2\sin x + 1 &=0\\ \sin x &= -\frac12\\ x&= n\pi +(-1)^n\left(-\frac{\pi}6\right)\end{align*}$
$\therefore x=\dots,-\frac{5\pi}{6},-\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}6, \dots$
$\text{Though in general, this is usually not necessary}\\ \text{As a rule of thumb, go for it when the answer is neat.}$

Thank you - so quick!
Can I get some help with:

lim 5xcos2x
x-->0 sinx

Thanks again

RuiAce · « **Reply #1495 on:** February 16, 2017, 08:23:59 pm »

Quote from: bsdfjn;lkasn on February 16, 2017, 08:21:08 pm

Thank you - so quick!
Can I get some help with:

lim 5xcos2x
x-->0 sinx

Thanks again

$\text{Split the limit}\\ \begin{align*}\lim_{x\to 0}\frac{5x\cos 2x}{\sin x}&= 5\left(\lim_{x\to 0}\frac{x}{\sin x}\right)\left(\lim_{x\to 0}\cos 2x\right)\\ &= 5\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^{-1}(\cos 0)\\ &= 5\times 1^{-1}\times 1\\ &= 5\end{align*}$

bsdfjnlkasn · « **Reply #1496 on:** February 16, 2017, 08:29:15 pm »

Quote from: RuiAce on February 16, 2017, 08:23:59 pm

$\text{Split the limit}\\ \begin{align*}\lim_{x\to 0}\frac{5x\cos 2x}{\sin x}&= 5\left(\lim_{x\to 0}\frac{x}{\sin x}\right)\left(\lim_{x\to 0}\cos 2x\right)\\ &= 5\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^{-1}(\cos 0)\\ &= 5\times 1^{-1}\times 1\\ &= 5\end{align*}$

Hey I'm still struggling with these limit questions...
Could i get some help with

lim x²
x->0 2-2cos2x

RuiAce · « **Reply #1497 on:** February 16, 2017, 08:35:34 pm »

Quote from: bsdfjn;lkasn on February 16, 2017, 08:29:15 pm

Hey I'm still struggling with these limit questions...
Could i get some help with

lim [u ]x² [/u]
x->0 2-2cos2x

$\text{That one requires opening up the double angle because there's nothing to split}\\ \begin{align*}\lim_{x\to 0}\frac{x^2}{2-2\cos 2x}&= \frac12\lim_{x\to 0}\frac{x^2}{1-\cos 2x}\\ &= \frac12\lim_{x\to 0}\frac{x^2}{2\sin^2 x}\\ &= \frac14 \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^{-2}\\ &= \frac14\times 1^{-2}\\ &= \frac14\end{align*}$

legorgo18 · « **Reply #1498 on:** February 16, 2017, 10:54:18 pm »

Hey need some help with the following and want to check answers for some:

1) given f(x) = lnx/x

a) find the stationary pt and determine its nature (i did, max x=e)
b) Show that f(e) > f(pi) errrr
c) show that e^pi > pi^e wtf

2) Show that sinx < x for x>0
I attempted this and moved sinx over and differentiated and got hpoi, can someone double check and tell me the correct method if i am completely wrong

3) Show that tan^-1 x > x - 1/3 x^3

RuiAce · « **Reply #1499 on:** February 16, 2017, 11:14:39 pm »

Quote from: legorgo18 on February 16, 2017, 10:54:18 pm

Hey need some help with the following and want to check answers for some:

1) given f(x) = lnx/x

a) find the stationary pt and determine its nature (i did, max x=e)
b) Show that f(e) > f(pi) errrr
c) show that e^pi > pi^e wtf

2) Show that sinx < x for x>0
I attempted this and moved sinx over and differentiated and got hpoi, can someone double check and tell me the correct method if i am completely wrong

3) Show that tan^-1 x > x - 1/3 x^3
Similarly, i moved to other side and diff and also got hpoi. can someone double check and tell me the correct method if i am completely wrong

ty

Once again - Warning to everyone else, these are 4U difficulty. Except potentially the first one because it's guided.
$\text{From a), the stationary point is at }x=e\\ \text{and its nature is a local maximum}$
$\text{Since the stationary point at }x=e\text{ is a local max}\\ \text{If }0 < x < e\text{ then the curve is increasing}\\ \text{If }x > e\text{ then the curve is decreasing.}$
$\text{If the curve is decreasing, then the greater the }x\text{ value}\\ \text{The less the }y\text{ value}\\ \textbf{Because that's just what it means to decrease}\\ \textbf{Make sure you revise the basic graphical definition}$
$\text{So since }e < \pi\\ \text{we have }f(e) > f(\pi)$
$\text{Now, given }f(e)>f(\pi)\\ \text{Upon substituting back }f(x)=\frac{\ln x}{x}\text{ we obtain}\\ \begin{align*}\frac{\ln e}{e}&>\frac{\ln \pi}{\pi}\\ \implies \frac{\pi}{e} &> \ln \pi\\ \implies e^{\frac{\pi}e} &> \pi\\ \implies \left(e^{\frac{\pi}e}\right)^e& > \pi^e \\ \implies e^\pi &> \pi^e\end{align*}$
_________________

Only going to do Q3 because Q2 is just an easier replica.
$\text{The problem is equivalent to proving}\\ \tan^{-1}x - x + \frac13x^3\ge 0\\ \text{which you correctly spotted}$
Important: I will presume that x > 0 again for this question. It was not stated (even if it's repetition) and thus, is not normally
$\text{Let }f(x)=\tan^{-1}x - x + \frac13x^3 > 0\\ \text{Then }f^\prime (x)=\frac{1}{1+x^2}-1+x^2$
$\text{This simplifies to}\\ f^\prime(x) = \frac{x^4}{1+x^2}$
$\text{Now, clearly }f^\prime (x) \ge 0\text{ for all }x\\ \text{Here's why:}\\ \text{Note that }1+x^2\text{ is always positive}\\ \text{and then }x^4\text{ is always positive or equal to 0}$
$\text{If you divide a positive number by a positive number}\\ \text{you still get a positive number}\\ \text{and if you divide 0 by anything (except 0)}\\ \text{you get }0\text{ again}\\ \text{So in every case, }f^\prime(x)\text{ is positive or equals 0}\\ \text{i.e. }f^\prime(x) \ge 0$
$\text{Since }f^\prime(x) \ge 0\\ \text{We know that }f(x)\text{ is monotonic increasing}$
$\text{So since }f(x)\text{ is monotonic increasing}\\ \text{we know that as the }x\text{ values get larger}\\ \text{the }y\text{ values also get larger}$
$\text{Now, note that }f(0) = 0\\ \text{From the above statement, if }f(x)\text{ is monotonic increasing}\\ \text{Then for }x > 0\text{, we must have }f(x) > 0\\ \text{This is because the curve will keep increasing away from 0}$
$\text{So finally, by substitution}\\ f(x) = 0 \implies \tan^{-1}x - x + \frac13x^3\ge 0 > 0\\ \text{which rearranges to what we needed to prove.}$

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Author Topic: 3U Maths Question Thread (Read 1465267 times) Tweet Share

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