Author Topic: 4U Maths Question Thread (Read 665257 times) Tweet Share

RuiAce · « **Reply #990 on:** March 16, 2017, 01:51:55 pm »

Quote from: bluecookie on March 16, 2017, 01:14:17 pm

Suppose that z^7=1, z is not equal to 1.
i) deduce that z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.
ii) by letting x=z+1/z, reduce the equation in i to a cubic equation in x
iii) Hence deduce that cospi/7cos2pi/7cos3pi/7=1/8

First two parts addressed in post #885 albeit with different pronumerals
$\text{Now, solving }z^7=1\text{ gives the solutions:}\\ z=\text{cis }\frac{-6\pi}{7}, \text{cis }\frac{-4\pi}{7}, \text{cis }\frac{-2\pi}{7},1, \text{cis }\frac{2\pi}{7}, \text{cis }\frac{4\pi}{7}, \text{cis }\frac{6\pi}{7}$
$\text{Hence, the solutions to }z^6+z^5+z^4+z^3+z^2+z+1=0\\ \text{i.e. the solutions to }z^3+z^2+z+1+\frac1z+\frac1{z^2}+\frac1{z^3}=0\\ \text{are just the same as above, without the }z=1\\ \text{i.e. } z=\text{cis }\frac{-6\pi}{7}, \text{cis }\frac{-4\pi}{7}, \text{cis }\frac{-2\pi}{7}, \text{cis }\frac{2\pi}{7}, \text{cis }\frac{4\pi}{7}, \text{cis }\frac{6\pi}{7}$
$\text{The point of part ii was to make it easier.}\\ \text{Note, for example, that }\left(\text{cis }\frac{4\pi}{7}\right)^{-1}=\text{cis }\frac{-4\pi}{7}\\ \text{The point we're trying to get at here is that}\\ \boxed{\text{If }z\text{ is a root, then }\frac{1}{z}\text{ is a root }\textbf{as well.}}$
_________
$\text{Note that:}\\ \text{cis }\frac{2\pi}{7}+\text{cis }\frac{-2\pi}{7}=2\cos \frac{2\pi}{7} \\\text{cis }\frac{4\pi}{7}+\text{cis }\frac{-4\pi}{7}=2\cos \frac{4\pi}{7}\\ \text{cis }\frac{6\pi}{7}+\text{cis }\frac{-6\pi}{7}=2\cos \frac{6\pi}{7}\\ \text{This I will leave as your exercise to prove}\\ \text{The idea is that }\overline{\text{cis }\frac{4\pi}{7}}=\text{cis }\frac{-4\pi}{7}\text{ as well}\\ \text{and }z+\overline{z}=2\text{ Re }(z)$
$\boxed{\text{Indeed, here we actually have }\overline{z}=\frac1z}$
_________
$\text{Since we know that }\text{cis }\frac{-6\pi}{7}, \text{cis }\frac{-4\pi}{7}, \text{cis }\frac{-2\pi}{7}, \text{cis }\frac{2\pi}{7}, \text{cis }\frac{4\pi}{7}, \text{cis }\frac{6\pi}{7}\\ \text{are the roots to the ugly one}\\ \text{Noting that }\boxed{x=z+\frac{1}{z}}\\ \text{and the above three equations}$
$\textbf{We see that the roots of }x^3+x^2-2x-1=0\\ \text{end up being }2\cos \frac{2\pi}{7}, 2\cos \frac{4\pi}{7}, 2\cos \frac{6\pi}{7}$
$\text{But note that }\cos \frac{4\pi}{7} = -\cos \frac{3\pi}{7}\\ \text{and }\cos \frac{6\pi}{7}=-\cos \frac{\pi}{7}\\ \text{This is because }\cos (\pi-x)=-\cos x$
$\text{Hence, it can be deduced that the roots are in fact}\\ -\cos \frac{3\pi}{7}, -\cos \frac{\pi}{7}, \cos \frac{2\pi}{7}\\ \text{Product of roots finishes everything off}$

chelseam · « **Reply #991 on:** March 16, 2017, 09:20:05 pm »

Thank you so much!

Quote from: RuiAce on March 16, 2017, 07:27:27 am

Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).

There are many other solutions better than the BoS ones which you should consider reviewing. However if those don't make sense, post up your interpretation of whichever solutions you looked at and we will fill in the remainder of the puzzle.

hanaacdr · « **Reply #992 on:** March 17, 2017, 06:32:16 am »

Hi
i am not sure how to find the asymptote
do you just do lim x approaches 0?
thank you

ellipse · « **Reply #993 on:** March 17, 2017, 07:22:41 am »

Quote from: hanaacdr on March 17, 2017, 06:32:16 am

Hi
i am not sure how to find the asymptote
do you just do lim x approaches 0?
thank you

This is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
Alternatively, you can try factorising the LHS
so (x-2y)(x+2y)=4. As the RHS is a non-zero constant, the LHS must not be equal to 0.
Therefore x-2y =/= 0 and x+2y =/= 0
So the asymptotes are y=1/2x and y=-1/2x

hanaacdr · « **Reply #994 on:** March 17, 2017, 05:12:39 pm »

Quote from: ellipse on March 17, 2017, 07:22:41 am

This is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
Alternatively, you can try factorising the LHS
so (x-2y)(x+2y)=4. As the RHS is a non-zero constant, the LHS must not be equal to 0.
Therefore x-2y =/= 0 and x+2y =/= 0
So the asymptotes are y=1/2x and y=-1/2x

thank you!

RuiAce · « **Reply #995 on:** March 17, 2017, 11:04:18 pm »

Quote from: bluecookie on March 16, 2017, 12:20:02 pm

a) Show that the roots of y^4+y^3+y^3+y+1=0 are y=cos(2kpi/5)+1sin(2kpi/5), k=1,2,3,4.
b) Hence deduce that cos36 deg = 1/2 + cos 72 deg.
^ I need help with part b.

Your expression is wrong. cos 36deg = -1/2 - cos 72deg.

cutiepie30 · « **Reply #996 on:** March 19, 2017, 02:56:47 am »

How would you go about doing these questions ? (attached)

RuiAce · « **Reply #997 on:** March 19, 2017, 07:37:53 am »

Quote from: cutiepie30 on March 19, 2017, 02:56:47 am

How would you go about doing these questions ? (attached)

I'm just gonna give this warning now.

Hyperbolic functions are NOT a part of the 4U course. In the previous case it was okay because they defined the functions for you and just made you do algebraic manipulation. I will not do any more questions that are beyond the scope of the course.

First image - Use similar identities to the trig ones.
Second image - Consider sinh for the first and cosh for the others

VydekiE · « **Reply #998 on:** March 19, 2017, 08:52:09 am »

Hi, it would be great if I could get some help on this question
1) Prove that if T (x₀, y₀) lies on a directrix on the hyperbola x²/a² - y²/b²=1 then the chord of contact will be a focal chord
Thank you!!

RuiAce · « **Reply #999 on:** March 19, 2017, 09:03:26 am »

Quote from: VydekiE on March 19, 2017, 08:52:09 am

Hi, it would be great if I could get some help on this question
1) Prove that if T (x₀, y₀) lies on a directrix on the hyperbola x²/a² - y²/b²=1 then the chord of contact will be a focal chord
Thank you!!

$\text{The chord of contact has equation}\\ \frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1\\ \text{You should be able to prove this.}$
$\text{Assume w.l.o.g. we are talking about the right-side directrix}\\ \text{If }T\text{ lies on the directrix then }x_0=\frac{a}{e}$
$\text{So the equation of the chord of contact becomes }\frac{x}{ae}-\frac{yy_0}{b^2}=1$
$\text{Suppose we wish to find the }x\text{-intercept of the chord of contact}\\ \text{This is done by setting }y=0\\ \text{Immediately, we have }\frac{x}{ae}-0=1 \implies x=ae\\ \text{Hence, }(ae,0)\text{ lies on the chord of contact}\\ \text{Which is of course, none other than the focus}$

VydekiE · « **Reply #1000 on:** March 19, 2017, 09:25:33 am »

Quote from: RuiAce on March 19, 2017, 09:03:26 am

$\text{The chord of contact has equation}\\ \frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1\\ \text{You should be able to prove this.}$
$\text{Assume w.l.o.g. we are talking about the right-side directrix}\\ \text{If }T\text{ lies on the directrix then }x_0=\frac{a}{e}$
$\text{So the equation of the chord of contact becomes }\frac{x}{ae}-\frac{yy_0}{b^2}=1$
$\text{Suppose we wish to find the }x\text{-intercept of the chord of contact}\\ \text{This is done by setting }y=0\\ \text{Immediately, we have }\frac{x}{ae}-0=1 \implies x=ae\\ \text{Hence, }(ae,0)\text{ lies on the chord of contact}\\ \text{Which is of course, none other than the focus}$

Thank you!!

cutiepie30 · « **Reply #1001 on:** March 19, 2017, 12:38:37 pm »

Quote from: RuiAce on March 19, 2017, 07:37:53 am

I'm just gonna give this warning now.

Hyperbolic functions are NOT a part of the 4U course. In the previous case it was okay because they defined the functions for you and just made you do algebraic manipulation. I will not do any more questions that are beyond the scope of the course.

First image - Use similar identities to the trig ones.
Second image - Consider sinh for the first and cosh for the others

yea i know they are not rui

, but i think that maths is a very beautiful subject and so it doesn't hurt here and there to do questions outside of the scope of the 4U course

next time i will post a disclaimer before posting these questions.

Just wondering rui for the second image, what sinh(u) substituition would i have to make, i have tryed x=sinh(u) for the 1st question of the 2nd image , it didnt work ?

RuiAce · « **Reply #1002 on:** March 19, 2017, 02:00:28 pm »

Quote from: cutiepie30 on March 19, 2017, 12:38:37 pm

yea i know they are not rui , but i think that maths is a very beautiful subject and so it doesn't hurt here and there to do questions outside of the scope of the 4U course

next time i will post a disclaimer before posting these questions.

Just wondering rui for the second image, what sinh(u) substituition would i have to make, i have tryed x=sinh(u) for the 1st question of the 2nd image , it didnt work ?

x=3sinh(u)

Treat it like a trigonometric substitution. You need a coefficient in front of the function. √9=3

Kle123 · « **Reply #1003 on:** March 20, 2017, 03:58:38 pm »

Hey there is no solution let alone answer to this question within the past paper. Idk how to get an expression such as the ones in the choices, but i tried subbing each of the MC and i got C (working out may be wrong though). This method seems too long to me for a MC question. Could someone show me a more efficient way? THank you.

RuiAce · « **Reply #1004 on:** March 20, 2017, 06:58:15 pm »

Quote from: Kle123 on March 20, 2017, 03:58:38 pm

Hey there is no solution let alone answer to this question within the past paper. Idk how to get an expression such as the ones in the choices, but i tried subbing each of the MC and i got C (working out may be wrong though). This method seems too long to me for a MC question. Could someone show me a more efficient way? THank you.

$\text{Since }\left(cq, \frac{c}q\right)\text{ lies on the normal}\\ \text{Upon substitution of the coordinates we have}$
$\begin{align*}cp^3q-\frac{cp}{q}&=c(p^4-1)\\ p^3q-p^4+1-\frac{p}{q}&=0\\ p^3(q-p)-\frac{1}{q}(q-p)&=0\\ \left(p^3-\frac{1}{q}\right)(q-p)&=0\end{align*}$
$\text{Obviously }p\neq q\text{ as they're distinct points}\\ \text{So }p^3-\frac{1}{q}=0 \implies p^3q=1$

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