Author Topic: 4U Maths Question Thread (Read 663242 times) Tweet Share

RuiAce · « **Reply #1380 on:** August 03, 2017, 11:37:01 pm »

Quote from: beau77bro on August 03, 2017, 09:22:28 pm

ok so it's kind of a weird comparison/equating of co-efficients for different expansions, i can see what you did but I'm struggling with the reasoning. is it just re-denoting the question so that you can show a new relationship between a with the power? - seems liek they just gave u a question where u just use the answer to find something - i can't see a straightforward process other than changing the powers.

Isn't that what you're meant to do in a proof? Look at the answer and find something?

I did mention in my lecture that one of the key ingredients to proving is to actually know what you're trying to prove.
____________

Although, to be fair, I might've made my response sounded like I was really just overly using what I was trying to prove, so the explanation looked dodge.

The problem is that the standard method of actually writing out the coefficient of $x^r$ and trying to invoke the identity $ \binom{N}{K}=\binom{N}{N-K}$ means we have to deal with the expansion of $ ((1+x)+x^2)^n $ and then the presence of the $x^2$ makes things awkward. If you want an example of a question where they actually did that, look at the binomial theorem question of the HSC 3U 2013 paper.

johnk21 · « **Reply #1381 on:** August 04, 2017, 11:00:04 am »

Could i please get some help with part i.
Thanks

RuiAce · « **Reply #1382 on:** August 04, 2017, 11:21:03 am »

Quote from: johnk21 on August 04, 2017, 11:00:04 am

Could i please get some help with part i.
Thanks

$\text{Using the standard gradient formula}\\ \text{the gradient of }OP\text{ is }\boxed{m=\frac{y_1}{x_1}}$
$\text{Also, the point }P\text{ lies on the curve }y=(\ln x)^2\\ \text{Hence }\boxed{y_1 = (\ln x_1)^2}\\ \text{Substituting into the above gives }\boxed{m=\frac{(\ln x_1)^2}{x_1}}$
$f(x) = (\ln x)^2 \implies f^\prime(x) = \frac{2\ln x}{x}\\ \text{So we also have }\boxed{m = f^\prime(x_1) = \frac{2\ln x_1}{x_1}}$
$\text{Equating the boxed expressions give}\\ \begin{align*}\frac{(\ln x_1)^2}{x_1}&=\frac{2\ln x_1}{x_1}\\ \ln x(\ln x - 2)&=0\\ \ln x-2&=0\\ x &= e^2\end{align*}\\ \text{noting of course we're not interested the point }(1,0)$
$\text{Hence the coordinates are }(e^2,4)$

johnk21 · « **Reply #1383 on:** August 04, 2017, 11:25:33 am »

Quote from: RuiAce on August 04, 2017, 11:21:03 am

$\text{Using the standard gradient formula}\\ \text{the gradient of }OP\text{ is }\boxed{m=\frac{y_1}{x_1}}$
$\text{Also, the point }P\text{ lies on the curve }y=(\ln x)^2\\ \text{Hence }\boxed{y_1 = (\ln x_1)^2}\\ \text{Substituting into the above gives }\boxed{m=\frac{(\ln x_1)^2}{x_1}}$
$f(x) = (\ln x)^2 \implies f^\prime(x) = \frac{2\ln x}{x}\\ \text{So we also have }\boxed{m = f^\prime(x_1) = \frac{2\ln x_1}{x_1}}$
$\text{Equating the boxed expressions give}\\ \begin{align*}\frac{(\ln x_1)^2}{x_1}&=\frac{2\ln x_1}{x_1}\\ \ln x(\ln x - 2)&=0\\ \ln x-2&=0\\ x &= e^2\end{align*}\\ \text{noting of course we're not interested the point }(1,0)$
$\text{Hence the coordinates are }(e^2,4)$

OOOH... obviously. I used the same method but failed in isolating for x1 and y1.
Thanks so much

chelseam · « **Reply #1384 on:** August 13, 2017, 05:09:39 pm »

Hi! Could someone please help me with proving this statement? I'm stuck on how to get the left hand side, but I managed to use (x-y)^2≥0 to get the right hand side (not sure if I'm meant to do that though)! Thank you

RuiAce · « **Reply #1385 on:** August 13, 2017, 05:17:42 pm »

Quote from: chelseam on August 13, 2017, 05:09:39 pm

Hi! Could someone please help me with proving this statement? I'm stuck on how to get the left hand side, but I managed to use (x-y)^2≥0 to get the right hand side (not sure if I'm meant to do that though)! Thank you

That question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?

chelseam · « **Reply #1386 on:** August 13, 2017, 05:20:09 pm »

Quote from: RuiAce on August 13, 2017, 05:17:42 pm

That question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?

Yes it is! I'm so sorry, I drew it wrong in the picture!

RuiAce · « **Reply #1387 on:** August 13, 2017, 05:26:12 pm »

$\text{In that case working backwards can help a lot}$
$\begin{align*}0&\le (x-y)^2\\ 0&\le x^2-2xy+y^2\\ x^2+2xy+y^2&\le 2x^2+2y^2\\ \frac{(x+y)^2}{4}&\le \frac{x^2+y^2}{2}\\ \frac{x+y}{2}&\le \sqrt{\frac{x^2+y^2}2}\end{align*}$
I know you were at my lecture so i know you know what I'm talking about

I wrote the bottom line first and worked my way up.

chelseam · « **Reply #1388 on:** August 13, 2017, 05:34:31 pm »

Quote

I know you were at my lecture so i know you know what I'm talking about I wrote the bottom line first and worked my way up.

Thanks so much Rui!

The tips at your lecture have helped me like inequalities questions a lot more now knowing that there's different ways to approach them!

itssona · « **Reply #1389 on:** August 18, 2017, 06:33:00 pm »

pls help
(x+iy)^2=7-24i
find x and y

RuiAce · « **Reply #1390 on:** August 18, 2017, 06:37:56 pm »

Quote from: sssona09 on August 18, 2017, 06:33:00 pm

pls help
(x+iy)^2=7-24i
find x and y

$\text{Follow the usual process}$
$\text{Expanding gives }x^2-y^2 + 2ixy = 7-24i\\ \text{Hence equating real/imag parts,}\\ \begin{align*}x^2-y^2&=7 \tag{1}\\ 2xy&=24\\ xy&=12\tag{2}\end{align*}$
$\text{Then apply the trick. In the original equation,}\\ \text{taking modulus of both sides, }|x+iy|^2 = |7-24i|^2\\ \begin{align*}x^2+y^2=25\tag{3}\end{align*}$
$(1)+(3)\text{ gives }2x^2 = 32\implies x = \pm 4\\ \text{You should be able to finish it off from here.}$

itssona · « **Reply #1391 on:** August 18, 2017, 06:54:28 pm »

Quote from: RuiAce on August 18, 2017, 06:37:56 pm

$\text{Follow the usual process}$
$\text{Expanding gives }x^2-y^2 + 2ixy = 7-24i\\ \text{Hence equating real/imag parts,}\\ \begin{align*}x^2-y^2&=7 \tag{1}\\ 2xy&=24\\ xy&=12\tag{2}\end{align*}$
$\text{Then apply the trick. In the original equation,}\\ \text{taking modulus of both sides, }|x+iy|^2 = |7-24i|^2\\ \begin{align*}x^2+y^2=25\tag{3}\end{align*}$
$(1)+(3)\text{ gives }2x^2 = 32\implies x = \pm 4\\ \text{You should be able to finish it off from here.}$

ohhh!
thank you Rui!!

bluecookie · « **Reply #1392 on:** August 26, 2017, 06:54:34 pm »

If P(acosθ , bsinθ ) and Q(acos(-θ ), bsin(-θ )] are the eremities of the latus rectum x=ae of the ellipse x^2/a^2y^2/b^2=1.

Show that PQ has length 2b^2/a
____________________________

P(asecθ , btanθ ) lies on the hyperbola x^2/a^2-y^2/b^2=1 with foci S(ae,0) and S'(-ae,0).
a) Show that PS=a(esecθ -1) and PS'a(esecθ +1)
b) Deduce that |PS-PS'|=2a
____________________________

P(asecθ , btanθ ) and Q(asecphi, btanphi) lie on the hyperbola x^2/a^2-y^2/b^2=1. Use the result that the chord PQ has the equation (x/a)*cos[(θ -pi)/2]-(y/b)*sin[(θ +phi)/2]=cos[(θ +phi)/2] to show that if PQ is a focal chord, then tanθ /2tanphi/2 takes one of the values of (1-e)/(1+e) or (1+e)/(1-e).

P(2rt(3), 3rt(3)) is one extremity of a focal chord on the hyperbola x^2/3 - y^2/9=1. Find the coordinates of the other extremity Q.
____________________________

Show that cos4θ =8(cosθ )^4-8(cosθ )^2+1.
a) Solve the equation 8x^4-8x^2+1=0 and deduce the exact values of cospi/8 and cos5pi/8.
b) Solve the equation 16x^4-16x^2+1=0 and deduce the exact v alues of cospi/12 and cos5pi/12.

Moderator action: Posts merged. At times like these, please resort to the modify at the top right corner of a post, to refrain from multi-posting.

RuiAce · « **Reply #1393 on:** August 26, 2017, 07:01:07 pm »

Quote from: bluecookie on August 26, 2017, 06:54:34 pm

If P(acosθ , bsinθ ) and Q(acos(-θ ), bsin(-θ )] are the eremities of the latus rectum x=ae of the ellipse x^2/a^2y^2/b^2=1.

Show that PQ has length 2b^2/a

$\text{Note that }P(a\cos \theta,b\sin \theta)\text{ lies on }x=ae\\ \text{Hence equating coordinates, we must have }a\cos\theta = ae \implies \boxed{\cos \theta = e}$
$\text{Noting }\cos(-\theta)=\cos\theta, \sin(-\theta)=-\sin\theta\\ \text{we find}\\ \begin{align*}PQ &= \sqrt{(a\cos \theta - a \cos \theta)^2 + (b\sin \theta + b\sin \theta)^2}\\ &= \sqrt{4b^2\sin \theta}\\ &= 2 |b\sin\theta|\end{align*}$
$\text{Now, given }\cos \theta = e\\ \begin{align*}\sin^2\theta &= 1-\cos^2\theta\\ &= 1-e^2\\ \implies |\sin\theta |&= \sqrt{1-e^2}\end{align*}$
$\text{So subbing back in, }PQ = 2b \sqrt{1-e^2}\\ \text{but we know that }b^2 = a^2 (1-e^2) \iff \frac{b}{a} = \sqrt{1-e^2}\\ \text{Hence }\boxed{PQ = \frac{2b^2}{a}}$

RuiAce · « **Reply #1394 on:** August 26, 2017, 07:22:12 pm »

Quote from: bluecookie on August 26, 2017, 06:54:34 pm

P(asecθ , btanθ ) lies on the hyperbola x^2/a^2-y^2/b^2=1 with foci S(ae,0) and S'(-ae,0).
a) Show that PS=a(esecθ -1) and PS'a(esecθ +1)
b) Deduce that |PS-PS'|=2a
____________________________

P(asecθ , btanθ ) and Q(asecphi, btanphi) lie on the hyperbola x^2/a^2-y^2/b^2=1. Use the result that the chord PQ has the equation (x/a)*cos[(θ -pi)/2]-(y/b)*sin[(θ +phi)/2]=cos[(θ +phi)/2] to show that if PQ is a focal chord, then tanθ /2tanphi/2 takes one of the values of (1-e)/(1+e) or (1+e)/(1-e).

P(2rt(3), 3rt(3)) is one extremity of a focal chord on the hyperbola x^2/3 - y^2/9=1. Find the coordinates of the other extremity Q.

$\text{Recalling relevant identities}\\ \begin{align*}PS&=\sqrt{(a\sec \theta - ae)^2 + (b\tan\theta - 0)^2}\\ &= \sqrt{a^2(\sec \theta -e)^2 + b^2\tan^2\theta}\\ &= \sqrt{a^2(\sec^2\theta -2e\sec\theta + e)+a^2(e^2-1)(\sec^2\theta-1)}\\ &= a\sqrt{e^2\sec^2\theta - 2e\sec \theta + 1}\tag{from simplifying}\\ &= a(e\sec \theta - 1)\end{align*}\\ \text{under a potentially implied assumption that }e\sec\theta - 1 \ge 0$
$\text{Repeat the exact same process for }PS^\prime.$
$\text{The second question is a no-brainer. Literally just substitute straight in.}$
______________________________________________________
$\text{We require the compound angle identities}$
$\text{First, suppose that }PQ\text{ passes through }S(ae,0)\\ \text{Then, upon substituting, doing battle with algebra we have}\\ \begin{align*}\frac{ae}{a}\cos \left(\frac{\theta-\phi}{2}\right)-0 &= \cos \left(\frac{\theta+\phi}{2}\right)\\ e\cos\frac{\theta}{2}\cos \frac{\phi}{2} + e\sin \frac{\theta}{2}\sin \frac{\phi}{2} &=\cos\frac{\theta}{2}\cos \frac{\phi}{2} - e\sin \frac{\theta}{2}\sin \frac{\phi}{2} \\ \sin \frac{\theta}{2}\sin \frac{\phi}{2}+e\sin \frac{\theta}{2}\sin \frac{\phi}{2}&= \cos\frac{\theta}{2}\cos \frac{\phi}{2}-e\cos\frac{\theta}{2}\cos \frac{\phi}{2}\\ (1+e) \sin \frac{\theta}{2}\sin \frac{\phi}{2}&=(1-e)\cos\frac{\theta}{2}\cos \frac{\phi}{2}\\ \frac{1-e}{1+e}&=\frac{\sin \frac{\theta}{2}\sin \frac{\phi}{2}}{\cos\frac{\theta}{2}\cos \frac{\phi}{2}}\\ &= \tan \frac{\theta}{2} \tan \frac{\phi}{2}\end{align*}$
$\text{If instead it passed through }S^\prime(-ae,0)\\ \text{using a similar approach, we arrive at the other answer}$

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