Author Topic: 4U Maths Question Thread (Read 668003 times) Tweet Share

RuiAce · « **Reply #1695 on:** February 02, 2018, 02:43:45 pm »

Quote from: stesoo on February 01, 2018, 09:20:53 pm

Hi i need help with this q,
for a real number r, the polynomial 8x^3-4x^2-42x+45 is divisible by (x-r)^2. Find the value of r.

thanks

$\text{This is a classical scenario of a double root problem}\\ \text{where that root is }x=r.$
$\text{Let }P(x) = 8x^3-4x^2-42x+45 \implies P^\prime(x) = 24x^2-8x-42\\ \text{If }(x-r)^2\text{ is a factor of }P(x)\\ \text{then }P(r) = P^\prime(r) = 0.$
$\text{Solving }P^\prime(r)=0\text{ gives}\\ 24r^2-8r-42=0 \implies \boxed{2(6r+7)(2r-3)=0}$
$\text{Hence, either }r=-\frac76\text{ or }r=\frac32.\\ \text{To check which one is correct, sub back into }P(x).$
$P\left( -\frac76 \right) =\frac{2048}{27}\neq 0\\ \text{however }P\left(\frac32\right) = 0\text{ is true}.\\ \therefore \boxed{r=\frac32}$

Lefkiiii6 · « **Reply #1696 on:** February 02, 2018, 07:54:20 pm »

Hi, what are the techniques used to graph y = f(x²)?

Thankyou!!

RuiAce · « **Reply #1697 on:** February 02, 2018, 08:57:21 pm »

Quote from: Lefkiiii6 on February 02, 2018, 07:54:20 pm

Hi, what are the techniques used to graph y = f(x²)?

Thankyou!!

When what's inside the $f$ gets altered there usually isn't any strategy. I just rely on common sense.

Common sense here suggests that I break up the cases for positive $x$ and negative $x$ first. (We can put $x=0$ with the positive ones or just do that one separately.)

I have $y=f(x)$ sitting in front of me. I now do $y=f(x^2)$. So if I put in $x=2$, I don't get $f(2)$ anymore, but rather $f(4)$.
When I put in $x=5$, I don't get $f(5)$ anymore, but rather $f(25)$. Effectively speaking, I'm getting something further away from the $x$-axis closer to it

So if $x > 1$, I'm really just squeezing the graph inwards. But I'm squeezing it at a faster rate as $x$ grows bigger.

Then, I could do a similar analogy with $ 0 < x < 1$. If i'm careful enough, I'll see that the graph expands outwards from the $y$-axis.

And of course, since at $x=0$ we have $f(x^2)=f(0)$ as well, and similarly for $x=1$, nothing changes between $y=f(x)$ and $y=f(x^2)$ for these two specific values.

You won't be able to get a perfectly accurate sketch, because you're just squeezing things. But you should be able to see the idea after you do some simulations on GeoGebra/Desmos.

And lastly, for negative $x$? That is just a matter of reflecting whatever you've done for positive $x$ over into the left of the $y$-axis as well. You should be able to see why that's the case.

RustyWasTaken · « **Reply #1698 on:** February 04, 2018, 03:51:47 pm »

Cheers

RuiAce · « **Reply #1699 on:** February 04, 2018, 04:55:05 pm »

Quote from: RustyWasTaken on February 04, 2018, 03:51:47 pm

Cheers

Hint: Just quote $ PS = ePM$, where in your case $e = \frac35 $

clovvy · « **Reply #1700 on:** February 04, 2018, 06:24:12 pm »

Show that cos(2kpi/7), where k=1,2,3,4,5,6 are roots of the equation 64x^6-64x^5-48x^4-48x^3+8x^2+8x+1=0

RuiAce · « **Reply #1701 on:** February 06, 2018, 11:56:54 am »

Quote from: clovvy on February 04, 2018, 06:24:12 pm

Show that cos(2kpi/7), where k=1,2,3,4,5,6 are roots of the equation 64x^6-64x^5-48x^4-48x^3+8x^2+8x+1=0

I just had a brief at this and they're not. $ \cos \frac{2\pi}7 $ does not go to 0 when I plug it into that thing.

clovvy · « **Reply #1702 on:** February 06, 2018, 01:27:14 pm »

Quote from: RuiAce on February 06, 2018, 11:56:54 am

I just had a brief at this and they're not. $ \cos \frac{2\pi}7 $ does not go to 0 when I plug it into that thing.

there is a k after pi

RuiAce · « **Reply #1703 on:** February 06, 2018, 01:46:38 pm »

Quote from: clovvy on February 06, 2018, 01:27:14 pm

there is a k after pi

Yeah.

I put $k=1$ in, because it was a value of $k$ listed. And it didn't work. (I haven't explicitly tested all of the other values of $k$, but it is very likely that there will be more values of $k$ that won't work either, just because of how these types of questions were designed in the first place.)

beau77bro · « **Reply #1704 on:** February 07, 2018, 11:30:22 pm »

Trying to help a friend. This question(2a) is ridiculous tho. I've broken it up into two integral of negative inverse cot and inverse tan but I can't work out wat to use as the separating boundary. (1 works but I don't know why)

RuiAce · « **Reply #1705 on:** February 08, 2018, 09:18:29 am »

Quote from: beau77bro on February 07, 2018, 11:30:22 pm

Trying to help a friend. This question(2a) is ridiculous tho. I've broken it up into two integral of negative inverse cot and inverse tan but I can't work out wat to use as the separating boundary. (1 works but I don't know why)

$\text{As you have correctly done,}\\ \int_{\tan x}^{\cot x}\frac{dt}{1+t^2} = \tan^{-1}(\cot x) - \tan^{-1}(\tan x)$
$\text{Now, for }0 < x < \frac\pi2, \tan^{-1}(\tan x) = x\\ \text{i.e. the tangent and inverse tangent cancel out.}$
$\text{To handle the cotangent, let}\\ \boxed{y = \tan^{-1}(\cot x)}\text{ where }0 < x < \frac\pi2.\text{Then, }\\\begin{align*}\tan y &= \cot x\\ \tan y &= \tan \left(\frac\pi2 - x\right)\\ y &= \frac\pi2 - x\end{align*}$
Or you could've skipped the "let y" bit and just recognised $\tan^{-1}(\cot x)=\tan^{-1}\left(\tan \left(\frac\pi2-x\right)\right) $ and realised that $0 < x < \frac\pi2\implies 0 < \frac\pi2-x < \frac\pi2 $ as well.
$\text{Hence}\\ \int_{\tan x}^{\cot x}\frac{dt}{1+t^2} = \frac\pi2 - 2x$

radnan11 · « **Reply #1706 on:** February 15, 2018, 01:43:51 pm »

Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)² + y² = 1 about the y axis."

from Cambridge 4u

jazzycab · « **Reply #1707 on:** February 15, 2018, 02:43:23 pm »

Quote from: radnan11 on February 15, 2018, 01:43:51 pm

Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)² + y² = 1 about the y axis."

from Cambridge 4u

If we're rotating about the y-axis, the radius of each cylinder is the x-value.
When finding the volume using the formula for volume of a solid, it is the rotated volume between the curve and an axis. In this particular case, the region being rotated is the region within the circle (not between the curve an the y-axis, due to the translation from the base circle 1 unit right) - giving an annulus (donut-shape). However, the 'hole' of the annulus is technically infinitesimally small due to the fact that there is an x-intercept at the origin.
Thus, to find the volume required, we need to find the volume between the right-hand side of the circle and the y-axis and subtract the volume obtained between the left-hand side of the circle and the y-axis:
$\left(x-1\right)^2+y^2=1\\\left(x-1\right)^2=1-y^2\\x-1=\pm\sqrt{1-y^2}\\x=1\pm\sqrt{1-y^2}$
The left half of the circle has the equation
$x=1-\sqrt{1-y^2}$
Where as the right half of the circle has the equation
$x=1+\sqrt{1-y^2}$
The volume between the right half and the axis is (the terminals are the upper and lower limits of the range of the circle):
$\pi\int\limits_{-1}^{1}{\left(\left(1+\sqrt{1-y^2}\right)^2\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(1+2\sqrt{1-y^2}+1-y^2\right)}dy$
The volume between the left half and the axis is:
$\pi\int\limits_{-1}^{1}{\left(\left(1-\sqrt{1-y^2}\right)^2\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(1-2\sqrt{1-y^2}+1-y^2\right)}dy$
The total volume is therefore:
$\pi\int\limits_{-1}^{1}{\left(1+2\sqrt{1-y^2}+1-y^2\right)}dy-\pi\int\limits_{-1}^{1}{\left(1-2\sqrt{1-y^2}+1-y^2\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(1+2\sqrt{1-y^2}+1-y^2-\left(1-2\sqrt{1-y^2}+1-y^2\right)\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(1+2\sqrt{1-y^2}+1-y^2-1+2\sqrt{1-y^2}-1+y^2\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(1+2\sqrt{1-y^2}+1-y^2-1+2\sqrt{1-y^2}-1+y^2\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(4\sqrt{1-y^2}\right)}dy$
This integrand can be evaluated by using a trig substitution:
$\text{Let }y=\sin{\theta}\\\Rightarrow \frac{dy}{d\theta}=\cos{\theta}\\\Rightarrow\frac{d\theta}{dy}=\frac{1}{\cos{\left(\theta\right)}}\\\therefore 1=\cos{\theta}\frac{d\theta}{dy}\\1=\sin{\theta_1}\Rightarrow \theta_1=\frac{\pi}{2}\\-1=\sin{\theta_2}\Rightarrow \theta_2=\frac{3\pi}{2}\\\therefore\text{Volume}=\pi\int\limits_{-1}^{1}{\left(4\sqrt{1-\sin^2{\left(\theta\right)}}\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(4\sqrt{\cos^2{\left(\theta\right)}}\right)}dy\\=\pi\int\limits_{-1}^{1}{\left(4\cos{\left(\theta\right)}\right)}dy\\=4\pi\int\limits_{-1}^{1}{\left(1\times\cos{\left(\theta\right)}\right)}dy\\=4\pi\int\limits_{-1}^{1}{\left(\cos{\left(\theta\right)}\frac{d\theta}{dy}\times\cos{\left(\theta\right)}\right)}dy\\=4\pi\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\left(\cos^2{\left(\theta\right)}\right)}d\theta\\=2\pi\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\left(2\cos^2{\left(\theta\right)}\right)}d\theta\\=2\pi\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\left(2\cos^2{\left(\theta\right)}-1+1\right)}d\theta\\=2\pi\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{\left(\cos{\left(2\theta\right)}-1\right)}d\theta\\=2\pi\left[\frac{1}{2}\sin{\left(2\theta\right)}-\theta\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\\=2\pi\left(\frac{1}{2}\sin{\left(\pi\right)}-\frac{\pi}{2}-\left(\sin{\left(3\pi\right)}-\frac{3\pi}{2}\right)\right)\\=2\pi\left(\pi\right)\\=2\pi^2\text{ cubic units}$

I'm Victorian, so I'm not familiar with the 4U HSC study design, so not sure how much of this you are expected to know how to do and whether you are taught other techniques or 'standard' integrals to evaluate the integral in this question.

RuiAce · « **Reply #1708 on:** February 15, 2018, 03:03:34 pm »

Quote from: radnan11 on February 15, 2018, 01:43:51 pm

Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)² + y² = 1 about the y axis."

from Cambridge 4u

jazzycab · « **Reply #1709 on:** February 15, 2018, 03:23:58 pm »

Or RuiAce's solution. Much neater and more succinct than mine

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