Math sl paper 1 TZ2

[Matsu]

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 :<

I got 120 (first try out) + 20 (second try out) runners and 70% from (140/200) too. What's wrong about it?

Edit: nvm, I just saw your other post, I should really read through the whole thread before I respond

Edited May 10, 2013 by Matsu

[jinye]

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 :<

I got 120 (first try out) + 20 (second try out) runners and 70% from (140/200) too. What's wrong about it?

That was my answer too, but I also thought about the answer that the other guy did...

Because it was explicit that the girls were selected only if they ran LESS (not equal) than 20 minutes. In the cumulative frequency graph, you are counting with the girl who did exactly 20 minutes. So the 20th girl did 20 minutes. Since they are only selected if they do less than 20, then 19 girls managed to achieved it, thus giving the 69.5% that the other guy got.

Well, honestly, I am pretty disappointed with myself, I could have gotten great grade, but now I'm just expecting a 6...

[Matsu]

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 :<

I got 120 (first try out) + 20 (second try out) runners and 70% from (140/200) too. What's wrong about it?

That was my answer too, but I also thought about the answer that the other guy did...

Because it was explicit that the girls were selected only if they ran LESS (not equal) than 20 minutes. In the cumulative frequency graph, you are counting with the girl who did exactly 20 minutes. So the 20th girl did 20 minutes. Since they are only selected if they do less than 20, then 19 girls managed to achieved it, thus giving the 69.5% that the other guy got.

Well, honestly, I am pretty disappointed with myself, I could have gotten great grade, but now I'm just expecting a 6...

I think the table's headings were something like time before t (i forgot the value) ≤ t < 20. I'm expecting a 5/6 too depending on how badly I scored on paper 2 haha. What % of our final IB grades are account for by our IAs? I'm really counting on them to pull my grade higher

[Matsu]

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.

Personally, I think it's π because the way I understand integration is that it's the area under the curve of a graph (which as you pointed out is always positive because having negative area is impossible). If it is -π though I really wouldn't stress cause you'd probably only lose 1 mark from that tiny error.

[madame_lexie]

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.

Hey! Yeah I was wondering too, because since it's under the x-axis usually it would be -pi.. However I thought of it after I gave in the paper.

[madame_lexie]

I was expecting paper 1 to be much harder, to be honest. I only had a problem with Q9, because I couldnt find g(4). So I left it out. But then at home I worked it out and got g(4) = 5. Also in Q10 I didnt realise that h' = f'g +g'f (product rule) and simply said h' = f'g'. Such a stupid mistake! Cost me 8 marks

[jinye]

I was expecting paper 1 to be much harder, to be honest. I only had a problem with Q9, because I couldnt find g(4). So I left it out. But then at home I worked it out and got g(4) = 5. Also in Q10 I didnt realise that h' = f'g +g'f (product rule) and simply said h' = f'g'. Such a stupid mistake! Cost me 8 marks

Same thing for Q10. I am so mad at myself for giving in a foolish way 8 marks.... These can be the marks which will determine whether I'll have a 7 or 6...

However I don't think they'll cut the 8 marks, we'll get like 3 or if lucky 4, because we did the normal equation alright, and that also counts for points...

Edited May 10, 2013 by jinye

[Pedro9604]

I didnt understand the question about the intregation of the graph that they didnt gave you the fuction!!!!

The exam was weird. very weird.

[jinye]

I didnt understand the question about the intregation of the graph that they didnt gave you the fuction!!!!

The exam was weird. very weird.

EDIT: ups sorry

Edited May 11, 2013 by jinye

[Kasper]

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

The gradient of the normal could not have been as you suggest, seeing as the tangent at point p was negative, and since they are perpendicular the gradient of normal must be positive. I got the equation of the normal to be 1/22x-59-1/9 and so did another strong student in my class. Anyone who can confirm this?

Also, I'm thinking that this exam was quite a bit easier than the mocks we had from 2012. And seeing as the 7-boundary was a 80/90 last year, it's probably going to be even higher this year. I know of two-three minor mistakes that I did and I am afraid that this could be enough to screw up up my 7 Do people agree that this one was one of the easier paper 1s from the last few years?

[jinye]

I didnt understand the question about the intregation of the graph that they didnt gave you the fuction!!!!

The exam was weird. very weird.

edit

Isn't that from the paper 2? Careful, now...

yes it is, my bad.

can you edit your post, so that my original post quoted there doesn't appear? thanks

again, my bad

And yes kasper, it was one of the easiest, problem is that I made too many silly mistakes, so I can definitely say bye to my 7. And the grade boundary might even go up, worsening my situation...

[iloveindia]

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!

That was by far one of the easiest tests ive ever done.

As for the cum. frequency graph, you were supposed to count from 20 mins and up, getting 20 girls from the second trial + 120 girls form the first trial. This means that 120+20=140, which is 70%..

As for the last questions, the y-coordinate turned out to be -54. The equation of the normal turned out to be y=1/3x-55. Log question was just a gift. The only thing im unsure about is the integration thingy when u had pi and 3pi. I wrote pi on the first one and 2pi on the second one, but i reckon thats wrong. I would have answered -pi and 4pi.. :/

and for quesiton 9, i got the x-coordinate to be pi/2, right?

Edited May 11, 2013 by iloveindia

[kspr]

Was it just me or was paper 1 this year much easier than the previous years?? I mean, exept for that damn statistics question with the runners it was all pretty straight forward stuff.... But paper 2 was just hell for me goodbye seven..

[mardissino]

Hey everyone,

On the running question, i am 100% sure that the group was t<20. Which gives te final answer of 70%, i remember i calculated 69,5% and then went back and changed it! So yeah 70%

[mardissino]

And for the last one i got grad tang = -18-9 = -27

Grad norm therefore 1/27

Substituted into x and y values:

Got a final answer of (1/27)x - 59 and 1/9

= (1/27)x - (487/9)

Or it might have been -1/27, and the +/- were the other way round for tang/norm, i cant really remember haha!

[Kevin Neutrino Lam]

For the last question...

Is it true that if I incorrectly calculated the slope of the normal but used the correct method to find the equation (though it was wrong), I will still get 2-3 marks? I hope they won't take away all 7 marks there...

[Egos]

Was it just me or was paper 1 this year much easier than the previous years?? I mean, exept for that damn statistics question with the runners it was all pretty straight forward stuff.... But paper 2 was just hell for me goodbye seven..

I felt exactly the same. P1 was easy but P2 was terrible. No way I will get a seven.

[jinye]

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!

That was by far one of the easiest tests ive ever done.

As for the cum. frequency graph, you were supposed to count from 20 mins and up, getting 20 girls from the second trial + 120 girls form the first trial. This means that 120+20=140, which is 70%..

As for the last questions, the y-coordinate turned out to be -54. The equation of the normal turned out to be y=1/3x-55. Log question was just a gift. The only thing im unsure about is the integration thingy when u had pi and 3pi. I wrote pi on the first one and 2pi on the second one, but i reckon thats wrong. I would have answered -pi and 4pi.. :/

and for quesiton 9, i got the x-coordinate to be pi/2, right?

The last question, the equation of the normal is not the one you got. It is the same as I got, but I know our method is wrong.

You were given f(x) = g(x) h(x)

Since you needed to differentiate f(x), and seeing that there is a multiplication, the thing you needed to use was the product rule. Thus:

f'(x) = g'(x) f(x) + h'(x) g(x)

I bet you did this:

f'(x) = g'(x) h'(x)

[adhamhussein]

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?

I think I got 0 for q q

Edited May 11, 2013 by adhamhussein

[iloveindia]

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!

That was by far one of the easiest tests ive ever done.

As for the cum. frequency graph, you were supposed to count from 20 mins and up, getting 20 girls from the second trial + 120 girls form the first trial. This means that 120+20=140, which is 70%..

As for the last questions, the y-coordinate turned out to be -54. The equation of the normal turned out to be y=1/3x-55. Log question was just a gift. The only thing im unsure about is the integration thingy when u had pi and 3pi. I wrote pi on the first one and 2pi on the second one, but i reckon thats wrong. I would have answered -pi and 4pi.. :/

and for quesiton 9, i got the x-coordinate to be pi/2, right?

The last question, the equation of the normal is not the one you got. It is the same as I got, but I know our method is wrong.

You were given f(x) = g(x) h(x)

Since you needed to differentiate f(x), and seeing that there is a multiplication, the thing you needed to use was the product rule. Thus:

f'(x) = g'(x) f(x) + h'(x) g(x)

I bet you did this:

f'(x) = g'(x) h'(x)

you dont make any sense bro