Preliminary mathematics marathon (1 Viewer)

ohexploitable · Jul 7, 2011

ummm these are pretty much quadratics u do in year 10 lol

$\\2k^4-3k^2-20=0\\(2k^2+5)(k^2-4)=0\\k^2=4,\,k^2=-\frac{5}{2}\\k=\pm2,\,\pm\sqrt{\frac{5}{2}}i$

AAEldar · Jul 7, 2011

ohexploitable said:
ummm these are pretty much quadratics u do in year 10 lol

$\\2k^4-3k^2-20=0\\(2k^2+5)(k^2-4)=0\\k^2=4,\,k^2=-\frac{5}{2}\\k=\pm2,\,\pm\sqrt{\frac{5}{2}}i$

Don't confuse them with the i :<.

SpiralFlex · Jul 7, 2011

spasticaznkid said:
Find the value of k:
The remainder is 25 when:
$2x^{4}-3x^{2}+5$ is divided by $x-k$
I know P(k) = 25 but im unsure what to do after.

$P(k)=2k^4-3k^2+5=25$

$2k^2-3k^2-20=0$

Let $u = k^2$

So now,

$2u^2 -3u-20=0$

$(u-4)(2u+5)=0$

$(k^2-4)(2k^2+5)=0$

$k=2$ , $-2$

Come on boys and girls, stick with my questions.

powlmao · Jul 7, 2011

ohexploitable said:
ummm these are pretty much quadratics u do in year 10 lol

$\\2k^4-3k^2-20=0\\(2k^2+5)(k^2-4)=0\\k^2=4,\,k^2=-\frac{5}{2}\\k=\pm2,\,\pm\sqrt{\frac{5}{2}}i$

Which program/website do you use?

Drongoski · Jul 7, 2011

Supply of free helpers exceeds demand!

Absolutezero · Jul 7, 2011

That's probably LaTex

powlmao · Jul 7, 2011

Absolutezero said:
That's probably LaTex

Thanks

hscishard · Jul 8, 2011

SpiralFlex said:
Yes Sir!

a) Since it is Preliminary, Preliminary student's only answer this!

I will start with a basic question.

Express the following fraction in the partial form,

$\frac{p}{x+2} + \frac{q}{1-x}$ . Where $p$ and $q$ are constants.

The fraction is $\frac{5x+6}{2-x-x^2}$

For Preliminary students,

b) The roots of the equation $ax^2 +bx+c=0$ are in the ratio of $p:q$ . Show that $(p^2+q^2)ac=(b^2-2ac)pq$

For HSC students, [Sorry this question was on the thread already. Didn't know. Attempt it if you want anyway. No cheating. ]

c) Differentiate with respect to $x$ ,

$x^x$ . Hence find the differential coefficient of $x^x^^^x$

[Hint: Don't differentiate normally.]

For HSC students,

d) Show that $\frac{e^x}{e^2^x-1}=\frac{1}{2}{(\frac{e^x}{e^x-1}-\frac{e^x}{e^x+1})}$ and hence evaluate $\int_{1}^{2} \frac{e^x}{e^2^x-1}$

Remarks: HSC students! Don't answer the Preliminary sections! Preliminary students! Attempt HSC questions if you can!

yea...noone answered my questions too here

Make RHS into one fraction then simplify
Use the show part, integrate and use log laws to simplify the answer to
0.5ln[(e+1)^2/e^2+1]

SpiralFlex · Jul 8, 2011

hscishard said:
yea...noone answered my questions too here

Make RHS into one fraction then simplify
Use the show part, integrate and use log laws to simplify the answer to
0.5ln[(e+1)^2/e^2+1]

I see little hscishard on page 1.

hscishard · Jul 8, 2011

For c, I got a pretty long answer...
I'm doubting it's accuracy since I did my working out on a paper that was already full of working out.

I'm pretty this method is correct
let y=x^x^x
Let u=x^x
Take ln on both sides and use implicit differentiation. I doubt this is a mathematics level question

LOL, I thought the question wanted me to differentiate x^x^x...
x^x is so much easier. Same method

hscishard · Jul 8, 2011

SpiralFlex said:
I see little hscishard on page 1.

lmao. 10% of my posts are in this thread

SpiralFlex · Jul 8, 2011

hscishard said:
For c, I got a pretty long answer...
I'm doubting it's accuracy since I did my working out on a paper that was already full of working out.

I'm pretty this method is correct
let y=x^x^x
Let u=x^x
Take ln on both sides and use implicit differentiation. I doubt this is a mathematics level question

It's not that long.

Let $y=x^x$

$\ln y=x \ln x$

$\frac{\mathrm{d} }{\mathrm{d} x}\ln y= \frac{\mathrm{d} }{\mathrm{d} x}(x \ln x)$

$\frac{1}{y} * \frac{dy}{dx} = \ln x + 1$

$\frac{dy}{dx} = x^x (\ln x + 1)$

Let $y=x^x^^^x$

$\ln y=x^x \ln x$

$\frac{\mathrm{d} }{\mathrm{d} x}\ln y= \frac{\mathrm{d} }{\mathrm{d} x}(x^x \ln x)$

$\frac{1}{y} * \frac{dy}{dx} = x^x (\ln x +1) \ln x * x^x *\frac{1}{x}$

$\frac{1}{y} * \frac{dy}{dx} = x^x [(\ln x +1) \ln x + \frac{1}{x}]$

$\frac{dy}{dx} = x^x^^^x * x^x [(\ln x +1) \ln x + \frac{1}{x}]$

hscishard said:
yea...noone answered my questions too here

Make RHS into one fraction then simplify
Use the show part, integrate and use log laws to simplify the answer to
0.5ln[(e+1)^2/e^2+1]

That's not showing, that's telling.

ohexploitable · Jul 8, 2011

^ no need for implicit differentiation, it can be done with 2u methods

$\\\frac{d}{dx}\left[x^x\right]=\frac{d}{dx}\left[e^{x\log x}\right]=(1+\log x)e^{x\log x}=x^x(1+\log x)$

hscishard · Jul 8, 2011

SpiralFlex said:
It's not that long.

Let $y=x^x$

$\ln y=x \ln x$

$\frac{\mathrm{d} }{\mathrm{d} x}\ln y= \frac{\mathrm{d} }{\mathrm{d} x}(x \ln x)$

$\frac{1}{y} * \frac{dy}{dx} = \ln x + 1$

$\frac{dy}{dx} = x^x (\ln x + 1)$

Let $y=x^x^^^x$

$\ln y=x^x \ln x$

$\frac{\mathrm{d} }{\mathrm{d} x}\ln y= \frac{\mathrm{d} }{\mathrm{d} x}(x^x \ln x)$

$\frac{1}{y} * \frac{dy}{dx} = x^x (\ln x +1) \ln x * x^x *\frac{1}{x}$

$\frac{1}{y} * \frac{dy}{dx} = x^x [(\ln x +1) \ln x + \frac{1}{x}]$

$\frac{dy}{dx} = x^x^^^x * x^x [(\ln x +1) \ln x + \frac{1}{x}]$

I had the last line expanded. It's quite long

I'm not bothered showing full working here anymore...too tedious

SpiralFlex · Jul 8, 2011

ohexploitable said:
^ no need for implicit differentiation, it can be done with 2u methods

$\\\frac{d}{dx}\left[x^x\right]=\frac{d}{dx}\left[e^{x\log x}\right]=(1+\log x)e^{x\log x}=x^x(1+\log x)$

Woops, I didn't see that. I am taking your method during exam time!

hscishard said:
I had the last line expanded. It's quite long

I'm not bothered showing full working here anymore...too tedious

What happened to the little hscishard on page 1?

hscishard · Jul 8, 2011

ohexploitable said:
^ no need for implicit differentiation, it can be done with 2u methods

$\\\frac{d}{dx}\left[x^x\right]=\frac{d}{dx}\left[e^{x\log x}\right]=(1+\log x)e^{x\log x}=x^x(1+\log x)$

+1, I remem a question that just asked to differentiate x^x here

ohexploitable · Jul 8, 2011

SpiralFlex said:
I see little Ohexploitable on the first page of this thread!

http://community.boredofstudies.org/showthread.php/219580-What-are-quot-Subject-Lines-quot

i think this was the first thread i ever made, not my first post though

hscishard · Jul 8, 2011

^ lol
Mudkip and "*In a bogan accent* Please explain? " makes you look so immature back then

hscishard · Jul 8, 2011

I thought of polynomials, remainder theorem
Think that's th only way, go learn that yourself with a good book

Alkanes · Jul 8, 2011

^ yep trial and error until it equals to 0 and use polynomials. Can't think of any easier way.

Preliminary mathematics marathon (1 Viewer)

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