Author Topic: Dekoyl's question thread (Read 22036 times) Tweet Share

dekoyl · « **Reply #120 on:** November 06, 2008, 07:03:15 pm »

Thank you guys.
14 hours until the exam and I still can't read a question properly..

onlyfknhuman · « **Reply #121 on:** November 06, 2008, 07:05:18 pm »

Quote from: dekoyl on November 06, 2008, 12:57:18 pm

Quote from: Mao on November 06, 2008, 11:42:26 am
always give domain of inverse functions
Yep but if it asks for the rule, do you give the domain as well? Our teacher said only to give the domain if it asks for a function.

Quote from: DivideBy0 on November 06, 2008, 01:07:08 pm

Quote from: dekoyl on November 06, 2008, 12:57:18 pm
Quote from: Mao on November 06, 2008, 11:42:26 am
always give domain of inverse functions
Yep but if it asks for the rule, do you give the domain as well? Our teacher said only to give the domain if it asks for a function.

No

"When the inverse function is asked for, the domain must be given. Students will be penalised in the future if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given."

I read in one of the methods ass report, saying that despite it not asking for the domain of the inverse you still have to state it... im pretty sure this was the case.

dekoyl · « **Reply #122 on:** November 06, 2008, 07:35:13 pm »

^Yeah the quote /0 posted was from an assessment report. (I think the 2007 one)

/0 · « **Reply #123 on:** November 06, 2008, 07:50:21 pm »

Quote from: dekoyl on November 06, 2008, 07:03:15 pm

Thank you guys.
14 hours until the exam and I still can't read a question properly..

Don't count it by hours, count it by days. Tomorrow is a new day. Get some rest, relax your mind, you don't want to be tight as a coiled spring on exam day!

rowshan · « **Reply #124 on:** November 06, 2008, 07:50:57 pm »

Quote from: Mao on November 06, 2008, 06:52:13 pm

case 1, $b\ge 0$

$\implies -b \le 0$

since A need to be greater than 0, we hence can only use the positive strand.

$A=\frac{-b+\sqrt{b^2-4}}{2}$

$\sqrt{b^2-4}\le \sqrt{b^2} = |b|$

$A = \frac{-b+\sqrt{b^2-4}}{2} \le \frac{-b + |b|}{2} = 0$

here, we arrive at the inequality of $A\le 0$ , which is impossible since $A=e^x >0$ , so b cannot be positive.

case 2: $b<0$

$\implies -b = |b| > 0$

if we take the positive strand, we will have two positive terms, $|b|$ and $\sqrt{b^2-4}$ adding together, hence A is positive, all is well.

if we take the negative strand, remembering that $|b|\ge \sqrt{b^2-4}$ , hence $\frac{|b|-\sqrt{b^2-4}}{2}\ge 0$ , it is also acceptable.

ALAS, $b\in (-\infty,-2]$

but isnt the answer b?
this is so confusing...

Mao · « **Reply #125 on:** November 06, 2008, 07:54:42 pm »

rowshan:

the answer is B, it is the only legal 'subset' of the maximal acceptable 'b' values.

onlyfknhuman · « **Reply #126 on:** November 06, 2008, 08:03:19 pm »

Is this a japanese thread?

rowshan · « **Reply #127 on:** November 06, 2008, 08:23:35 pm »

divided by zero - thanks
mao - ok
onlyfknhuman - japanese algebra!!!

dekoyl · « **Reply #128 on:** November 09, 2008, 03:08:00 pm »

The largest set of real values of a for which $|a^{2}-2a|\geq 1$ is _____?

I'm still quite unfamiliar with modulus stuff.

Thank you.

humph · « **Reply #129 on:** November 09, 2008, 03:16:59 pm »

Easiest way probably is to graph $y = x^2 - 2x$ .

Alternatively, split it up into the two cases: first, if $a^2 - 2a \geq 0$ then $a(a-2) \geq 0$ , which by considering the graph of $y = x(x-2)$ , is when $a \leq 0 , a \geq 2$ . So we can rewrite the original question as the set of numbers $a \in (\-\infty,0] \cup [2,\infty)$ satisfying $a^2 - 2a \geq 1$ , or equivalently, by factorising, that $(a - (\sqrt{2}+1))(a+(\sqrt{2}-1)) \geq 0$ , which by considering the graph of $y = (x - (\sqrt{2}+1))(x+(\sqrt{2}-1))$ , is when $a \leq 1-\sqrt{2} , a \geq 1+\sqrt{2}$ .

For the second case, if $-(a^2 - 2a) \geq 0$ then $a(a-2) \leq 0$ , which by considering the graph of $y = x(x-2)$ , is when $0 \leq a \leq 2$ . So we can rewrite the original question as the set of numbers $a \in [0,2]$ satisfying $-(a^2 - 2a) \geq 1$ , or equivalently, by factorising, that $(a - 1)^2 \leq 0$ , which by considering the graph of $y = (x-1)^2$ , is when $a=1$ .

Thus the set of real values of $a$ such that $|a^2 - 2a| \geq 1$ is $(-\infty,1-\sqrt{2}] \cup \{1\} \cup [1+\sqrt{2},\infty)$ .

dekoyl · « **Reply #130 on:** November 09, 2008, 03:30:17 pm »

Thanks humph but I'm trying to understand it algebraically. The solutions don't make much sense to me

Edit: Thanks to above and below.

ReVeL · « **Reply #131 on:** November 09, 2008, 03:37:31 pm »

Quote from: dekoyl on November 09, 2008, 03:08:00 pm

The largest set of real values of a for which $|a^{2}-2a|\geq 1$ is _____?

I'm still quite unfamiliar with modulus stuff.

Thank you.

Just take the positive and negative of whats inside the modulus.

If $|a^{2}-2a|\geq 1$

Then $a^2-2a\geq 1$

so $a^2-2a-1\geq 0$

$a=\frac {2\pm \sqrt{4-4\times 1\times -1}}{2}$

$a=\frac {2\pm \sqrt{8}}{2}=1\pm \sqrt {2}$

Hence, first two solutions are $1+\sqrt {2}$ and $1-\sqrt {2}$

However seeing as its the modulus we must consider the negative.

Therefore, the same but $2a-a^2\geq 1$ (just switched signs)

$-a^2+2a-1\geq 0$

$a=\frac {-2\pm \sqrt {4-4\times -1\times -1}}{-2}$

$a=\frac {-2\pm \sqrt {0}}{-2}$

$a=1$

Therefore, $|a^{2}-2a|\geq 1$ for when $a\leq 1-\sqrt {2}$ and when $a\geq 1+\sqrt {2}$ or when $a=1$

Hope that makes sense.

dekoyl · « **Reply #132 on:** November 10, 2008, 05:00:09 pm »

This signals the end of my time in methods.

LOCK.

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