Author Topic: 4U Maths Question Thread (Read 665369 times) Tweet Share

RustyWasTaken · « **Reply #1665 on:** January 25, 2018, 12:31:40 pm »

Cheeers

RuiAce · « **Reply #1666 on:** January 25, 2018, 01:05:51 pm »

Quote from: RustyWasTaken on January 25, 2018, 12:31:40 pm

Cheeers

This question lacks information - it is not stated whether $a$ and $b$ are real or not.

philgee · « **Reply #1667 on:** January 25, 2018, 01:23:40 pm »

Hello how would i go about finding the integral of the last question?

RustyWasTaken · « **Reply #1668 on:** January 25, 2018, 01:29:16 pm »

Quote from: RuiAce on January 25, 2018, 01:05:51 pm

This question lacks information - it is not stated whether $a$ and $b$ are real or not.

Oh. The answer from the textbook for a and b is real.
a = -2 and b =-2
I cant get the ans though

RuiAce · « **Reply #1669 on:** January 25, 2018, 02:02:09 pm »

Quote from: philgee on January 25, 2018, 01:23:40 pm

Hello how would i go about finding the integral of the last question?

This technically is not 4U material as improper integrals are not in the course. But having said that, many bits and pieces will be provided as it's quite easily doable regardless.

$\text{The problem is an area-between-curves problem}\\ \text{with the upper curve being }y=1\\ \text{and the lower curve being }y=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$
$\text{Since our area extends forever (i.e. as }x\to \infty),\\ \text{our area integral is }A=\int_0^\infty \left(1-\frac{e^x-e^{-x}}{e^x+e^{-x}} \right) \,dx\\ \text{i.e. }\boxed{A=\lim_{R\to \infty} \int_0^R \frac{2e^{-x}}{e^x +e^{-x}}\,dx}$
$\text{To evaluate the integral, first let }u=e^{-x}\implies du = -e^{-x}\,dx.\\ \text{The rest then follows relatively nicely.}\\ \begin{align*}A&=\lim_{R\to \infty}\int_1^{e^{-R}} \frac{-2\, du}{u^{-1}+u}\\ &= \int_0^{1}\frac{2u}{1+u^2}\, du\\ &= \left[\ln (1+u^2) \right]_0^1\\ &= \ln 2\end{align*}$
Extra: The function provided is actually the hyperbolic tan function $ f(x) = \tanh x$

Quote from: RustyWasTaken on January 25, 2018, 01:29:16 pm

Oh. The answer from the textbook for a and b is real.
a = -2 and b =-2
I cant get the ans though

$\text{Assuming that }a\text{ and }b\text{ are real,}\\ \text{all the coefficients of }P(z)\text{ are real and}\\ \text{thus since }i\text{ is a zero of }P(z),\text{ so is }-i.$
$\text{Call the last root }\alpha\text{ (noting }P(z)\text{ has multiplicity 4)}.\\ \text{Via the product of roots, }\\ i (-i)(1+\sqrt2)\alpha = -1\implies \alpha = \frac{-1}{1+\sqrt2} = 1-\sqrt2.$
$\text{Hence our roots are }\pm i\text{ and }1\pm \sqrt{2}.$
$\text{The sum of one root at a time immediately gives }2 = -a \implies a=-2.$
$\text{From the sum of three roots at a time,}\\ i(-i)(1-\sqrt2) + i(-i)(1+\sqrt2) + i(1+\sqrt2)(1-\sqrt2) - i(1+\sqrt2)(1-\sqrt2) = -b \\\implies b=-2.$

philgee · « **Reply #1670 on:** January 25, 2018, 02:18:15 pm »

Dear Rui, i dont get why why it is 1- the original curve inside the integral.

RuiAce · « **Reply #1671 on:** January 25, 2018, 02:23:32 pm »

Quote from: philgee on January 25, 2018, 02:18:15 pm

Dear Rui, i dont get why why it is 1- the original curve inside the integral.

Like mentioned, area between two curves.

If we just considered $\int_0^\infty \frac{e^x+e^{-x}}{e^{x}+e^{-x}}\,dx $, we would've been considering the region that's merely below the curve. Not the region sandwiched between the curve and its asymptote.

(In fact, that integral would be infinite, because that area would just blow up.)

philgee · « **Reply #1672 on:** January 25, 2018, 02:26:52 pm »

ohhhh okay thank you i understand now!

clovvy · « **Reply #1673 on:** January 25, 2018, 11:19:01 pm »

This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.

RuiAce · « **Reply #1674 on:** January 25, 2018, 11:23:27 pm »

Quote from: clovvy on January 25, 2018, 11:19:01 pm

This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.

I’m at the gym right now but here’s some thoughts upon first glance.

Step 1. Divide both sides by t^2.
Step 2. Make the substitution u = t - 1/t. If that doesn’t work, try t + 1/t

clovvy · « **Reply #1675 on:** January 25, 2018, 11:31:57 pm »

Quote from: clovvy on January 25, 2018, 11:19:01 pm

This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.

God that's a typo, it is t^4+8t^3-6t^2-8t+1=0

RuiAce · « **Reply #1676 on:** January 26, 2018, 12:27:30 am »

Quote from: clovvy on January 25, 2018, 11:31:57 pm

God that's a typo, it is t^4+8t^3-6t^2-8t+1=0

$\text{In that case, the substitution }u = t-\frac{1}{t}\\ \text{is an excellent idea.}$
______________________
$\text{Clearly }t=0\text{ is not a solution, so consider}\\ t^2 + 8t - 6 - \frac{8}{t} + \frac{1}{t^2} = 0.$
$\text{Let }u = t - \frac{1}{t}.\text{ Then, }u^2 = t^2 - 2 + \frac{1}{t^2}\\ \text{This reduces the equation to }u^2 + 8u - 4 = 0$
$\text{Of course, this is just a quadratic.}\\ \text{From completing the square, it is easy to see that the solutions}\\ \text{are }u =-4\pm \sqrt{20}$
$\text{From here, sub back in to have }t - \frac{1}{t} = -4\pm \sqrt{20}\\ \text{i.e. }t^2 + (4+\sqrt{20})t - 1 = 0\text{ or }t^2+(4-\sqrt{20})t - 1 =0.\\ \text{The remainder is just brute force quadratic formula.}$
Note that the intuition behind the substitution $ u = t - \frac{1}{t} $ was essentially dependent on the symmetric nature of the coefficients (of the original equation). This kind of intuition isn't expected for a 4U student.

clovvy · « **Reply #1677 on:** January 26, 2018, 09:10:23 pm »

Quote from: RuiAce on January 26, 2018, 12:27:30 am

$\text{In that case, the substitution }u = t-\frac{1}{t}\\ \text{is an excellent idea.}$
______________________
$\text{Clearly }t=0\text{ is not a solution, so consider}\\ t^2 + 8t - 6 - \frac{8}{t} + \frac{1}{t^2} = 0.$
$\text{Let }u = t - \frac{1}{t}.\text{ Then, }u^2 = t^2 - 2 + \frac{1}{t^2}\\ \text{This reduces the equation to }u^2 + 8u - 4 = 0$
$\text{Of course, this is just a quadratic.}\\ \text{From completing the square, it is easy to see that the solutions}\\ \text{are }u =-4\pm \sqrt{20}$
$\text{From here, sub back in to have }t - \frac{1}{t} = -4\pm \sqrt{20}\\ \text{i.e. }t^2 + (4+\sqrt{20})t - 1 = 0\text{ or }t^2+(4-\sqrt{20})t - 1 =0.\\ \text{The remainder is just brute force quadratic formula.}$
Note that the intuition behind the substitution $ u = t - \frac{1}{t} $ was essentially dependent on the symmetric nature of the coefficients (of the original equation). This kind of intuition isn't expected for a 4U student.

I tried doing this again on paper and for some reason it didn't work..., the u sub I tried on the equation, I get the u^2 but I did not get 8u...

RuiAce · « **Reply #1678 on:** January 26, 2018, 09:18:19 pm »

Quote from: clovvy on January 26, 2018, 09:10:23 pm

I tried doing this again on paper and for some reason it didn't work..., the u sub I tried on the equation, I get the u^2 but I did not get 8u...

$t^2 + \boxed{8t} - 6 - \boxed{\frac{8}{t}} + \frac{1}{t^2}.$
$\text{Observe how }8t - \frac{8}{t} = 8\left(t-\frac1t\right) = 8u.$
Edit: Typo

Dragomistress · « **Reply #1679 on:** January 27, 2018, 09:46:36 pm »

How should I do this?

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