Math sl paper 1 TZ2

[IBN95]

So what do you guys think?

I found it harder than usual and very tricky mostly, but overall not as good as expected.

Last part of last question was pretty diffecult although i got the gradient 1/3.

[Pepe]

TZ2: I didnt find it very difficult but I was too tired after 2.5 h of history P3... I didnt finish the last question (you had to use the product rule to find the equation of the normal) and in some way couldnt count the cosine from the previous question.

Do you remember the cumulative frequency question and two last results in it? I put 30 (number of runners) and 70% (do not remember what it was) but a friend of mine said they were wrong :<

Edited May 10, 2013 by Pepe

[Megan Dimmick]

Well, just wondering everyone's thoughts on yesterday's paper one in tz2. Personally the last question was horrible I got myself confused with numbers and so I failed that bit

What about everyone else

[Claudiosee]

Since the older topic (from 2012) is still blocked, while the other Math levels are accepted, I wanted to create this topic in order to initiate the discussion of Paper 1 SL in Math (TZ2).

Generally, I felt that the paper was rather simple.

Was I the only who felt this way?

Also: On Q7 part a)

The question that asked the candidate to identify the Integration of the part of the function below the x axis. As the function was not given to us, I assume most of you used the area of a circle in order to identify this value.

However, the thing that worries me, is that they did not specifically mention "area" but only gave the command for Integration.

This worries me me since area is the absolute value of the integration, and hence would make the result "-pi" instead of just "pi".

The following question b) was however not impacted on by this.

Best Regards.

[Pepe]

In the integration question the total area was 3pi. The area of the circle was pi. Therefore the area under the second part of the graph was 2pi

[Megan Dimmick]

In the integration question the total area was 3pi. The area of the circle was pi. Therefore the area under the second part of the graph was 2pi

That's what I got

[Claudiosee]

In the integration question the total area was 3pi. The area of the circle was pi. Therefore the area under the second part of the graph was 2pi

Thank you for your response, but that was not my question. My question was if the part a) was "-pi".

I do understand that part b) was 3pi, but since area is absolute integration that does not make any difference because it would cause the "-pi" to be turned into "pi". My question was, and remains, if the answer to a) was -pi.

As i said, the question asked for INTEGRATION, which is negative when below the x axis. Therefore if one were to find the area, but look for the integration, one would have to consider the negative result.

Nevertheless I am not sure about the answer, hence why I am asking. Therefore if someone were to explain to why i was incorrect (if that is the case) id easily accept it.

Thank you and best regards.

[maereth]

I found it really easy, except the question 7b which I think was not properly asked ;/ but the paper was surprisingly nice, I actually expect to get 7 from it.

[Pepe]

You should have found the area of a part of graph on the left using the formula: A=pi * r^2 * 0.25
The result was pi, no minus here.

EDIT: I will ask once more because I really want to know the answer for that:
Do you remember the cumulative frequency question and two last results in it? I put 30 (number of runners) and 70% (do not remember what it was) but a friend of mine said they were wrong :<

Edited May 10, 2013 by Pepe

[maereth]

yeah, I meant part b ;d (unless you were talking to this German guy with annoyingly formal English, then sorry)

Because everyone says that it was 3pi-1pi, because 3pi is the whole area, but for me the question said 3pi was the area enclosed by f(x) and x and y axis which means it should only be this part of the graph that was like in the first quarter, if you know what I mean.

But that's probably too simple for 4 points, so I'll keep telling that I didn't like the way it was asked ;p

Edited May 10, 2013 by maereth

[Hussam]

Hi guys ,

for the second question the matricies , what values did u get for p and q ?

thanks

[Claudiosee]

You should have found the area of a part of graph on the left using the formula: A=pi * r^2 * 0.25

The result was pi, no minus here.

Okay one last time.

∫f(x)dx refers to the "integration of f(x)".

Now, as you correctly identified area=pi.

However since area = absolute of inverse, one has to consider the value of the integration.

http://www.intmath.com/applications-integration/2-area-under-curve.php

I quote (from the above source) :

"In this case, we have to sum the individual parts, taking the absolute value for the section where the curve is below the x-axis" and

"If we don't do it like this, the "negative" area (the part below the

x

x-axis) will be subtracted from the "positive" part, and our total area will not be correct."

Therefore if we were to find the area and were asked for the integration, the integration=+/-area. However, since this part of the function was below the x-axis, this should make the integration "-pi"-

[Pepe]

You should have found the area of a part of graph on the left using the formula: A=pi * r^2 * 0.25

The result was pi, no minus here.

Okay one last time.

∫f(x)dx refers to the "integration of f(x)".

Now, as you correctly identified area=pi.

However since area = absolute of inverse, one has to consider the value of the integration.

http://www.intmath.com/applications-integration/2-area-under-curve.php

I quote (from the above source) :

"In this case, we have to sum the individual parts, taking the absolute value for the section where the curve is below the x-axis" and

"If we don't do it like this, the "negative" area (the part below the

x

x-axis) will be subtracted from the "positive" part, and our total area will not be correct."

Therefore if we were to find the area and were asked for the integration, the integration=+/-area. However, since this part of the function was below the x-axis, this should make the integration "-pi"-

But you didnt do any integration in this question. The only think you were supposed to do is to recognise that the part of the graph under x-axis is simply 1/4 of the area of the circle (with radius 2 as I remember well) and then subtract this area from the given value of 3pi. Nothing more to do here.

The sign would change if you were integrating the function with limits, but there was no function to integrate

[rach_h]

I hadn't thought about the -pi thing actually - I suppose it could be given that they did ask for the intergration of the function and not the area (if I remember correctly), but the minus can't be worth any more than 1 mark, most of it would probably be for working out to use the area of a circle.

Also I got 30 and 70% too! I hope your friend's wrong, what did they think it was?

And I can't remember the matrices one exactly but I think I got something like p=4 and q=-1... maybe...

Glad you all found the last question weird too, did anyone get an answer for the equation of the normal?

[Pepe]

I hadn't thought about the -pi thing actually - I suppose it could be given that they did ask for the intergration of the function and not the area (if I remember correctly), but the minus can't be worth any more than 1 mark, most of it would probably be for working out to use the area of a circle.

Also I got 30 and 70% too! I hope your friend's wrong, what did they think it was?

Glad you all found the last question weird too, did anyone get an answer for the equation of the normal?

His result was 69.5%. In the question it was said that 'girls who run faster than 20 minutes are qualified' or something like that. It would mean that we shouldnt have taken the value for 20 min from the cum freq (that was 20 girls) but the value beneath - 19 girls. Simply, girls whose results was equal to 20 min didnt qualify according to him, therefore we should count 19 girls in our percentage (who for sure run faster than 20 minutes) what would give 69.5%. Ugh, complicated.

The gradient of the normal was -1/27 but I didnt have time to count the equation :<

[Megan Dimmick]

Hi guys ,

for the second question the matricies , what values did u get for p and q ?

thanks

I think everyone in my class got 4 as p, but for a we had either -1 or 0

[rach_h]

I hadn't thought about the -pi thing actually - I suppose it could be given that they did ask for the intergration of the function and not the area (if I remember correctly), but the minus can't be worth any more than 1 mark, most of it would probably be for working out to use the area of a circle.

Also I got 30 and 70% too! I hope your friend's wrong, what did they think it was?

Glad you all found the last question weird too, did anyone get an answer for the equation of the normal?

His result was 69.5%. In the question it was said that 'girls who run faster than 20 minutes are qualified' or something like that. It would mean that we shouldnt have taken the value for 20 min from the cum freq (that was 20 girls) but the value beneath - 19 girls. Simply, girls whose results was equal to 20 min didnt qualify according to him, therefore we should count 19 girls in our percentage (who for sure run faster than 20 minutes) what would give 69.5%. Ugh, complicated.

The gradient of the normal was -1/27 but I didnt have time to count the equation :<

Oh ok, I see his logic, but I think because the classes were organised as a</=x<b, when plotting a cumulative frequency diagram you would still plot the point for that class at b, here being 20, but that value wouldn't actually include 20. So the point at 20 should represent the number of girls who ran it in <20 minutes. I think anyway, but to be honest statistics has never been my strength.

And at least you should get marks for working out the gradient, I barely wrote anything for that question!

[jinye]

When I left the examination room, I was thrilled because I thought I had aced paper 1.

But now, I see how many mistakes I've made, and my desired 7 is seriously compromised...

Damn, I wish the boundaries could lower this year

[maereth]

I hadn't thought about the -pi thing actually - I suppose it could be given that they did ask for the intergration of the function and not the area (if I remember correctly), but the minus can't be worth any more than 1 mark, most of it would probably be for working out to use the area of a circle.

Also I got 30 and 70% too! I hope your friend's wrong, what did they think it was?

Glad you all found the last question weird too, did anyone get an answer for the equation of the normal?

His result was 69.5%. In the question it was said that 'girls who run faster than 20 minutes are qualified' or something like that. It would mean that we shouldnt have taken the value for 20 min from the cum freq (that was 20 girls) but the value beneath - 19 girls. Simply, girls whose results was equal to 20 min didnt qualify according to him, therefore we should count 19 girls in our percentage (who for sure run faster than 20 minutes) what would give 69.5%. Ugh, complicated.

The gradient of the normal was -1/27 but I didnt have time to count the equation :<

yeah, I did the same with the percentages I think (I mean like your friend). And I somehow managed to do the normal equation question but I don't remember what I got...

Edited May 10, 2013 by maereth

[Matsu]

Paper 1 was pretty easy. I messed up the normal question though and didn't get the negative reciprocal of the slope I got. Whoopsy. I also mucked up the number with the "find g(4)" question a lil bit