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#21 Josh

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Posted 14 May 2007 - 11:06 AM

please could you explain question 10 testB. ive been trying it for the past 20 minutes and can't get it to work sad.gif I'm scared for tomorrow now unsure.gif

#22 Josh

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Posted 14 May 2007 - 11:19 AM

Oh, and could someone help with question6 from Test B as well? Thanks

As you can tell, im not the brightest button on the calculator


#23 mikey4020

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Posted 14 May 2007 - 11:22 AM

This is how I done it but I don't know if it's right. Using the point (4,18) you substitute 18 for y and 4 for x. 18=3e^4^k then you take log_{e} of both sides to give you, log_{e}18=4klog_{e}3 Then rearange to give 4k=log_e18-log_e3 After subtracting the left hand side you get log_e15 take the 4 from the 4k over and divide to give k=\frac{log_e15}{4} which is the same as k=\frac{1}{4} log_e15 and hence answer C

Hope that helps, took me ages to do using that LaTex stuff laugh.gif

IGNORE ABOVE: Where I said you take it over and subtract you don't you take it over and divide and this means that you take 3 away from 18 to give you log_e15 the rest shoudl still be valid! rolleyes.gif

#24 George

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Posted 14 May 2007 - 11:36 AM

QUOTE(Josh @ May 14 2007, 12:06 PM) View Post
please could you explain question 10 testB. ive been trying it for the past 20 minutes and can't get it to work sad.gif I'm scared for tomorrow now unsure.gif

Don't be too scared - logs is only a small bit of the course, and it just takes work to get on top of it.

As I explained above, after subbing in the point you get 18=3e^{4k}. Here's how to get the answer:



\begin{align*}
3e^{4k} &=18 \\
e^{4k} &= \frac{18}3 = 6 \\
4k &= \log_e 6 \;\; \textrm{(taking }\log_e\textrm{ on both sides)}\\
k &= \frac14 \log_e 6
\end{align*}

So the answer is B.

(mikey, it went wrong for you when you said "take log on both sides", because you still had the 3 on the RHS, so \log_e (3e^{4k}) = \log_e 3 + \log_e e^{4k} = \log_e 3 + 4k )


Question 6 is a (fairly difficult) completing the square:



Here's my working:



\begin{align*}
-x^2+6x-4 &= -( x^2-6x+4 ) \\
&= -\bigl( (x-3)^2 - 9 + 4 \bigr) \\
&= -\bigl( (x-3)^3 - 5\bigr) \\
&= -(x-3)^2 + 5
\end{align*}

So the "p" is 5; answer C.

Notice that I took out the negative first of all, so that I could do the usual completing the square with a +x^2, since that's easier.

Going onto the second line is really the "completing the square" step - to get x^2-6x we use (x-3)^2 (the -3 is "half the x-coefficient"). But that bracket is (x-3)^2 = x^2 - 6x + 9 - so it's giving us an extra 9, which we then take off in our working.

So all I've done going from line 1 to line 2 is replace x^2-6x with (x-3)^2-9 since those are the same. After that, it's just a matter of tidying up.

#25 mikey4020

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Posted 14 May 2007 - 11:51 AM

Woops that was rather embarrassing!! Least I've learned something lol. I agree that whilst logs are only a small part of the course I didn't feel they were covered that well in my school they kind of left it to the last minute.

#26 Josh

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Posted 14 May 2007 - 11:54 AM

Thanks for that. We do completing the square differently to that (multiply out and equate) but, having looked at your answer, I've managed to get it to work.

Also, I see where I went wrong with the Logs, I hadnt taken the 3 down when I should have. I'm going to concentrate v. hard on Logs for the next while and then attempt some pp questions.

Thanks again!


#27 BuckminsterFullerene

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Posted 14 May 2007 - 09:01 PM

I did the tests, got 9/10 in the first one, It won't let me see the test again, so I can't tell where I went wrong! AAAAAAAAAAAAAAAAAAAAAAAAARGH, MUST FIND OUT!
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#28 Hassan

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Posted 14 May 2007 - 09:11 PM

I got 9/10 but really 8 cos I had to look for help for the vectors question. And in Q8 I integrated and differentiated, whoops.
Thanks again for these!

#29 George

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Posted 14 May 2007 - 09:12 PM

It was this question in Test 2:



The answer is C - remember that you get -sin x when you differentiate cos x.



Also this one from Test 1:



The answer is B - just differentiate, and see where f'(x)>0 (this tells you where f is increasing).

#30 BuckminsterFullerene

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Posted 14 May 2007 - 09:18 PM

QUOTE(George @ May 14 2007, 10:12 PM) View Post
It was this question in Test 2:



The answer is C - remember that you get -sin x when you differentiate cos x.



Also this one from Test 1:



The answer is B - just differentiate, and see where f'(x)>0 (this tells you where f is increasing).


Thank you very very very very very very very much, Yeah I know what I did wrong, didn't differentiate, just drew the graph and looked at it!


George could you possibly help me with this question, powerx.gif power2.gif + y power2.gif + 4kx - 2ky - k - 2 = 0

For what range of values of k does the equation represent a circle. The answer is, for all k.
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#31 George

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Posted 14 May 2007 - 10:25 PM

QUOTE(BuckminsterFullerene @ May 14 2007, 10:18 PM) View Post
George could you possibly help me with this question, x[^2] + y power2.gif + 4kx - 2ky - k - 2 = 0

For what range of values of k does the equation represent a circle. The answer is, for all k.

This just comes from the formula for the radius of a circle in that form. It's the \sqrt{g^2+f^2-c} bit that causes trouble - since we can't square root a negative number. So just work out g^2+f^2-c and see if it's ever negative (and going by the answer, it must be positive for any k in this case).





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