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Author Topic: 4U Maths Question Thread  (Read 665492 times)  Share 

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fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2205 on: March 01, 2019, 07:14:25 pm »
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Sorry, my integral was actually wrong! (but the answer is the same :) )

Going back to fix that right now, but basically it involves a bit of side working considering the integral of the function u^2 e^2u. You do integration by parts twice. Have a shot at it yourself while I fix the code!

You should get down to something similar to 1/2 e^2u(u^2-u+1/2) (in factorised form).

EDIT: Original answer has been fixed.
« Last Edit: March 01, 2019, 07:44:23 pm by fun_jirachi »
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2206 on: March 02, 2019, 10:37:51 am »
+1
That one can be done using only integration by parts if we're cautious with it. (But it definitely needs two applications of it.) Note that according to the rule of LIATE, we wish to differentiate the L, i.e. the logarithm.
\begin{align*}\int_1^e x(\ln x)^2\,dx &= \left[\frac{x^2}{2} (\ln x)^2\right]_1^e - \int_1^e \frac{x^2}{2} \left( \frac{2\ln x}{x} \right)\,dx\\ &= \frac{e^2}{2} - \int_1^e x\ln x\,dx\\ &= \frac{e^2}{2} - \left[ \frac{x^2}{2}\ln x\right]_1^e + \int_1^e \left(\frac{x^2}{2}\right)\left(\frac1x\right)\,dx \\ &= \frac{e^2}{2} - \frac{e^2}{2} + \left[ \frac{x^2}{4} \right]_1^e\\ &= \frac14 (e^2-1)\end{align*}

spnmox

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Re: 4U Maths Question Thread
« Reply #2207 on: March 04, 2019, 10:09:55 pm »
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Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2208 on: March 04, 2019, 10:19:17 pm »
+1
Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!
When you write out your answer, you must start from the beginning, which is either the basic fact that \( (a-b)^2 \geq 0\), or some equation they've given to you. Of course, if there's a part i) or something, you can assume the result of that for part ii).

Working backwards is good to help determine how you would write it up in a forwards direction. But note that your final solution needs to be written the correct way no matter what.
\[ \text{With your problem, }\textbf{further}\text{ working backwards from}\\ (ab+cd)^2 \leq 4abcd\text{ will have provided the path.} \]
\begin{align*}(ab-cd)^2 &\geq 0\\ a^2b^2 - 2abcd + c^2d^2 &\geq 0\\ a^2b^2 + 2abcd + c^2d^2 &\geq 4abcd\\ (ab+cd)^2 &\geq 4abcd \end{align*}
Alternatively, if you've proven \( (x+y)^2 \geq 4xy\) using \( (x-y)^2 \geq 0\) from an earlier part, you can just sub \(x=ab\) and \(y=cd\) without regurgitating the same proof from scratch.

HOWEVER:
Later, when I worked backwards on what you were originally trying to prove, I obtained this.
\begin{align*}ab+cd&\leq \sqrt{(a^2+b^2)(c^2+d^2)}\\ a^2b^2+2abcd+c^2d^2 &\leq a^2b^2+a^2d^2+c^2b^2+c^2d^2\\ a^2d^2-2abcd+c^2b^2 &\geq 0\\ (ad-bc)^2 &\geq 0 \end{align*}
So to prove that, I would've written that proof out backwards.
\begin{align*}(ad-bc)^2 &\geq 0\\ a^2d^2-2abcd + b^2c^2 &\geq 0\\ a^2b^2+a^2d^2+c^2b^2+c^2d^2 &\geq a^2b^2+2abcd+c^2d^2\\ (a^2+c^2)(b^2+d^2) &\geq (ab+cd)^2\\ \therefore ab+cd &\leq \sqrt{(a^2+c^2)(b^2+d^2)} \end{align*}
\[ \text{taking positive roots, since }ab+cd > 0\text{ from the given information.} \]

david.wang28

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Re: 4U Maths Question Thread
« Reply #2209 on: March 06, 2019, 08:20:31 pm »
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Hello,
I haven't got a clue how to do this question in the link below. However, I have attached a separate attachment based on my thoughts of doing this question. Can anyone please help me with this question? Thanks :)
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wlam

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Re: 4U Maths Question Thread
« Reply #2210 on: March 07, 2019, 09:01:57 pm »
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help thanks

RuiAce

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Re: 4U Maths Question Thread
« Reply #2211 on: March 07, 2019, 09:10:09 pm »
+1
Hello,
I haven't got a clue how to do this question in the link below. However, I have attached a separate attachment based on my thoughts of doing this question. Can anyone please help me with this question? Thanks :)
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?
help thanks
Elaborate on what you need help with please?

wlam

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Re: 4U Maths Question Thread
« Reply #2212 on: March 07, 2019, 09:20:16 pm »
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The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?

i cant get the answer for part b

wlam

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Re: 4U Maths Question Thread
« Reply #2213 on: March 07, 2019, 09:21:41 pm »
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heres a better pic

RuiAce

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Re: 4U Maths Question Thread
« Reply #2214 on: March 07, 2019, 10:08:18 pm »
+2
heres a better pic
Would this diagram be helpful?


If you managed to figure out everything on the diagram and it's still not coming out, preferably post up your working. Otherwise give it a try with the diagram drawn. (Also note that the diagram has some right-angles that aren't displayed on it.)

wlam

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Re: 4U Maths Question Thread
« Reply #2215 on: March 07, 2019, 11:19:27 pm »
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thanks! i just misinterpreted the question

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #2216 on: March 08, 2019, 03:30:09 pm »
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Hey rui need some of your maths experties plz

RuiAce

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Re: 4U Maths Question Thread
« Reply #2217 on: March 08, 2019, 06:06:29 pm »
+1
Hey rui need some of your maths experties plz
After tracking down the second question I realise that as it was stated, \(x\) and \(y\) were intended to be real. Of course, this wasn't specified in the question, but apparently it was assumed by the author for some unknown reason.

With that added knowledge, the \(x^2+y^2\) bit hints at the modulus of a complex number. Hence, let \(z=x+iy\).
\begin{align*}x + \frac{3x-y}{x^2+y^2}&=3\tag{1}\\ y-\frac{x+3y}{x^2+y^2}&=0\tag{2} \end{align*}
\[ \text{So upon doing }(1)+i\times (2)\text{ we obtain}\\ \begin{align*} x+iy + \frac{3x-y-ix-3iy}{x^2+y^2}&=3\\ x+iy + \frac{3(x-iy)-i(x-iy)}{x^2+y^2}&=3\\ \therefore z+\frac{(3-i)\overline{z}}{z\overline{z}}&=3\\ z^2-3z+(3-i)&=0\\ \therefore x+iy &= \frac{3\pm \sqrt{9-(12-4i)}}{2} \tag{quadratic formula}\\ &= \frac{3\pm \sqrt{-3+4i}}2\\ &= \frac{3\pm \sqrt{1 + 4i - 4}}2 \tag{completing the square}\\ &= \frac{3\pm \sqrt{1+4i+4i^2}}2 \\ &= \frac{3\pm \sqrt{(1+2i)^2}}2\\ &= \frac{3\pm (1+2i)}2\\ &= 2+i, \, 1-i\end{align*} \]
Note that it is a common MX2 formula that \( z\overline{z} = |z|^2\), and it was used here.
\[ \text{Hence equating real/imaginary parts,}\\ (x,y) = (2,1)\text{ or }(x,y) = (1,-1) \]
Regardless though, the first question still requires Taylor series, and hence should be put in the first year uni maths questions thread.

david.wang28

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Re: 4U Maths Question Thread
« Reply #2218 on: March 08, 2019, 08:37:05 pm »
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The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?
Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2219 on: March 08, 2019, 09:03:19 pm »
+2
Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).
The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.

The wording of the question is also ambiguous, as it could represent the region below. They should have specified that they specifically want the region in the first quadrant.

Nevertheless, the question becomes doable with those assumptions put into place (albeit with some awkwardness). The idea is that they want you to find the green area shown below, but with the aid of the blue area. I will start it off.

\[ \text{Note that when }x=2\sqrt{2}, \, y = 1.\\ \text{The combined area of the blue and green region is therefore just }2\sqrt{2}.\]
\[ \text{Note that the equation given rearranges to }x = \left( y^{2/3}+1\right)^{3/2}.\\ \text{The area of the blue region is therefore equal to}\\ \int_0^1 \left( y^{2/3}+1\right)^{3/2}\,dy\]
\[ \text{Let }y = tan^3\theta\implies dy = 3\tan^2\sec^2\theta.\text{ Then}\\ \begin{align*}\int_0^1 \left( y^{2/3} +1\right)^{3/2}\,dy &= \int_0^{\pi/4} \left( \tan^2\theta + 1 \right)^{3/2} (3\tan^3\theta \sec^2\theta)\,d\theta\\ &= 3\int_0^{\pi/4} \sec^2\theta (\sec^2\theta-1)\sec^2\theta\,d\theta\\ &= 3\left( \int_0^{\pi/4} \sec^6\theta\,d\theta + \int_0^{\pi/4} \sec^4\theta\,d\theta \right) \\ &= 3\left( \left[I_6\right]_0^{\pi/4} + \left[I_4\right]_0^{\pi/4} \right)\\ &= 3\left( \frac{\sec^4\frac\pi4 \tan \frac\pi4}{5} - \frac{\sec^4 0 \tan 0}5 - \frac45 \left[I_4\right]_0^{\pi/4} + \left[ I_4 \right]_0^{\frac\pi4}\right)\\ &= 3\left( \frac45 + \frac15 \left[ I_4\right]_0^{\pi/4}\right)\end{align*}\]
The remaining use of the reduction formulae and the final subtraction at the end is left as your exercise.

Subject to minor computational error.