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Free Multiple-Choice Test - HSN forum

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Free Multiple-Choice Test


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#1 George

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Posted 11 May 2007 - 09:20 PM

I hope everyone's revision is going well smile.gif

If you can spare about half an hour, you might like to try this quick multiple-choice test that we've put together:

http://www.hsn.uk.net/resources/Higher-Maths/mc/

There are 10 questions, each of which would be worth 2 or 3 marks in the exam. Hopefully you won't find them too hard tongue.gif

Update: There are now two different tests available on that page; Test A and Test B. Each one has 10 questions, and you can do both of them if you like.


Please use this thread to ask questions about the questions blink.gif - i.e. if you don't know how to do something, or wonder why you got one wrong, etc.

Good luck!


P.S. - you need to be logged in to the forum to take the test, so if you haven't got an account yet, please register!

#2 Amazoph

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Posted 12 May 2007 - 12:50 PM

Thanks - it was really useful. Helped to jog my memory on some of the bits of the course I'd almost forgotten like polynominal division smile.gif

Got 10/10 so looks like my revision is going well biggrin.gif

#3 celticfcz

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Posted 12 May 2007 - 12:53 PM

This looks good have just tried it - didnt do as well as i thought i would!! - so many daft mistakes!! - will defo be revising all day today!

Thanks for putting the questions up!

#4 Hassan

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Posted 12 May 2007 - 02:00 PM

Thanks, this was helpful!

Could anyone please explain question 8 about the maximum values? Thanks.

#5 celticfcz

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Posted 12 May 2007 - 03:12 PM

QUOTE(Hassan @ May 12 2007, 03:00 PM) View Post
Thanks, this was helpful!

Could anyone please explain question 8 about the maximum values? Thanks.


yep that was one of the questions i was stuck at.

#6 babeasc

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Posted 12 May 2007 - 03:15 PM

Very useful it's flagged up some stuff!

#9 im struggling with! I know u do synthetic division, but does anyone have the working because I think my mistake comes once I've set the whole thing = 0

It would be awesome if you put another one up!

Ta much!
Beyond it, and us, shines a greater hope and a brighter future
Can't wait 'til the exams are over!

#7 George

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Posted 12 May 2007 - 03:20 PM

I'm glad this has been useful smile.gif

Question 8 is about the maximum and minimum values of the sine function. Remember that, for all x, -1 \leq \sin x \leq 1 - you should be able to see this from a sketch of the sine graph.

Also notice that \sin( \frac{\pi}2 )=1 and \sin( \frac{3\pi}2 )=-1 (that's 90, 270 degrees).

Now, the expression 3-\sin( x+ \frac{\pi}2 ) is like 3-\textrm{something}, which is as big as possible when the "something" is as small as possible. Since our "something" is a sine function, the smallest we can make it is -1. This gives a final answer of 3-(-1)=4, so it's between C and D.

All that remains is to see what "x" we need. Well, we want \sin (x + \frac{\pi}2) = -1, so this means x+\frac{\pi}2=\frac{3\pi}2 (remember we said above that -1 comes from an angle of 270 degress, or \frac{3\pi}2 radians).

So, solving that gives x=\pi, i.e. option D.

I hope that's made it a bit clearer for you smile.gif

#8 celticfcz

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Posted 12 May 2007 - 03:22 PM

QUOTE(George @ May 12 2007, 04:20 PM) View Post
I'm glad this has been useful smile.gif

Question 8 is about the maximum and minimum values of the sine function. Remember that, for all x, -1 \leq \sin x \leq 1 - you should be able to see this from a sketch of the sine graph.

Also notice that \sin( \frac{\pi}2 )=1 and \sin( \frac{3\pi}2 )=-1 (that's 90, 270 degrees).

Now, the expression 3-\sin( x+ \frac{\pi}2 ) is like 3-\textrm{something}, which is as big as possible when the "something" is as small as possible. Since our "something" is a sine function, the smallest we can make it is -1. This gives a final answer of 3-(-1)=4, so it's between C and D.

All that remains is to see what "x" we need. Well, we want \sin (x + \frac{\pi}2) = -1, so this means x+\frac{\pi}2=\frac{3\pi}2 (remember we said above that -1 comes from an angle of 270 degress, or \frac{3\pi}2 radians).

So, solving that gives x=\pi, i.e. option D.

I hope that's made it a bit clearer for you smile.gif


Thanks I get it now!

Another test later would be great also!!

Thanks again!


#9 George

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Posted 12 May 2007 - 03:29 PM

QUOTE(babeasc @ May 12 2007, 04:15 PM) View Post
#9 im struggling with! I know u do synthetic division, but does anyone have the working because I think my mistake comes once I've set the whole thing = 0

I don't really want to try typing out the synthetic division here biggrin.gif

Do you get -8-18a in the last cell? (That's what I got)

Then, the question tells you the remainder is 2. So you have -8-18a=2, i.e. 18a=-10.

So a=-\frac{10}{18} = -\frac59, option C.

Now, if you set -8-18a = 0 that would lead you to a wrong answer (since you're effectively saying the remainder is 0, which contradicts the question).

Just say if you don't get that from the synthetic division, and I'll try to get the table typed up dry.gif

#10 babeasc

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Posted 12 May 2007 - 03:38 PM

QUOTE(George @ May 12 2007, 04:29 PM) View Post
QUOTE(babeasc @ May 12 2007, 04:15 PM) View Post
#9 im struggling with! I know u do synthetic division, but does anyone have the working because I think my mistake comes once I've set the whole thing = 0

I don't really want to try typing out the synthetic division here biggrin.gif

Do you get -8-18a in the last cell? (That's what I got)

Then, the question tells you the remainder is 2. So you have -8-18a=2, i.e. 18a=-10.

So a=-\frac{10}{18} = -\frac59, option C.

Now, if you set -8-18a = 0 that would lead you to a wrong answer (since you're effectively saying the remainder is 0, which contradicts the question).

Just say if you don't get that from the synthetic division, and I'll try to get the table typed up dry.gif


Ta for that...
I checked my working against what u had and i had just made a stupid mistake!
I used (x+2) in the division instead of just '2' which gives the answer you have for the last cell.
Thanks for the help, invaluable resource!
x
Beyond it, and us, shines a greater hope and a brighter future
Can't wait 'til the exams are over!

#11 weerydo

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Posted 12 May 2007 - 07:45 PM

too easy! 10/10 ma revision must be goin well. lol. giv us a harder 1 on sunday plzzzzzzzzz! cheers n i found it v helpful, much appreciated

#12 Hassan

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Posted 12 May 2007 - 10:36 PM

QUOTE(George @ May 12 2007, 04:20 PM) View Post
I hope that's made it a bit clearer for you smile.gif

Thank you, I get it now. But in an exam I would never think to do that huh.gif

#13 Allan_r_123

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Posted 13 May 2007 - 04:57 PM

That was great - really helpful and it's always useful to be doing some new questions.

Please post more! smile.gif

#14 George

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Posted 13 May 2007 - 05:10 PM

I'm really glad people are finding this useful smile.gif

Apart from the ones I've gone over above, the only other one which seems to be causing problems is this one:



You need to watch out for the direction of the arrows - the rule \mbox{\boldmath$a$}.\mbox{\boldmath$b$}=|\mbox{\boldmath$a$}||\mbox{\boldmath$b$}|\cos\theta only makes sense when the vectors a and b both point away from the angle (or both point into the angle).

So there, you need to use the angle of 120 degrees - this is the angle between a and b if you slide a down to point away from the same point as b.


I'm working on more questions for another test, and if all goes well it will go up this evening smile.gif

#15 mikey4020

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Posted 13 May 2007 - 08:40 PM

I tried to do this like a couple of questions at a time and now it looks like I've got a really rubbish score lol! crying.gif blush.gif

#16 william

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Posted 13 May 2007 - 08:55 PM

damm i got 5/10 sad.gif mainly cuase i lost interest half way through
"millis times ten to the power of minus 3 yeah?"

#17 George

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Posted 13 May 2007 - 09:42 PM

I've just put up the second test now:

http://www.hsn.uk.net/resources/Hmath/mc/testB

If you're reading this on Sunday night, it would probably be best to wait and try it tomorrow, after a good night's sleep biggrin.gif

Again, if you have any problems, just reply in this thread and I'll try to explain.

#18 mikey4020

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Posted 14 May 2007 - 09:41 AM

These were really useful, highlighted a few isuues to correct. I got Q5 wrong in Test B and I don't know how to get it any chance you could explain it for me. Also could you explain Q10 in test B because I got the answer and I don't know how.

#19 George

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Posted 14 May 2007 - 10:37 AM

Sure, here's question 5:



We have to find \dls{XE}, which just means "how to get from X to E". If you have a copy of the diagram, it might help to mark on the things we know - \dls{AC} and \dls{CE}.

From that, it looks like X to C to E is a good route, and we have enough information to work it out.

So \dls{XE} = \dls{XC}+\dls{CE}. The only problem now is finding \dls{XC}, but this is just half of \dls{AC} since X is the centre of the base.

Can you get the answer from that?


Now question 10:



The best way to do this is to just use the point (4,18). Subbing in to the equation of the curve gives 18=3e^{4k}, and you now just have to solve that exponential equation. Post again if you're not sure about that smile.gif

#20 mikey4020

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Posted 14 May 2007 - 11:06 AM

Yeah that's a bit better I think I've got it now. The answers I got are B & C respectively.





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