How did Maths 12 go? No discussion

[Ram97]

No

 

[farhan141]

Correct Answers (Joint Answers By Me and Aly).

Note: Question order may be wrong.

Q1) C = 8 And then arrange equation according to that.

2) area of shaded region= r^2(pie/2 -theta + sin2theta/2)

3) x^2= 240 x^3= -160 then 560x^3

4) u=6

5) theta= 71.4 and 63.7 (around these values)

6)i) 120m ii) pie/15 iii) 10<t<20

7) 2y=x-13 (idk the exactly value in replacement of 13) ii) C (9,7)

8)i) 459 ii)k=6 iii)S=54

9)i) 31.8 ii) 1/6 * (4i-2j+4k) iii) same magnitude for isosceles of OC and OA

10)i) 16pi/3 ii) i guess c was 2.5 and -3.5

11)i) p>0.5 ii) 2(x-1.5)^2 +0.5 iii) g(x)>=1/2 (iv) k=1.5 v)h inverse= 3/2 + underroot((2x-1)/4)

 

[Ram97]

wasnt p less than 0.5

 

[SonamKhan]

wasnt p less than 0.5

I think so too

 

[Sifat Bin Quadery]

Can someone explain how 10<t<20??

 

[<><> Ice <><>]

when i pluged in t=10 i got my answer as 90, thats why i gave my answer as inequality like yours, i said h>90 then solved for what t equals to. not sure though.

Well after leaving the exam, I thought of it in another way
what if the question actually meant
for how long will the thing be above 90m?

 

[Sifat Bin Quadery]

well... i thought....
h>90
so, 60(1-coskt)>90
or, 1-coskt>(90/60)
or, -coskt>1.5-1
or, -coskt>0.5
or, coskt<-0.5
or, kt<cos-1(-.5)
hence, t<cos-1(-.5)/k
this gives t<10.
so my answer is also correct ryt....? my result shows that for about less than 10 mins the wheel was above 90.
not completely sure whether it's correct or not.

 

[farhan141]

well... i thought....
h>90
so, 60(1-coskt)>90
or, 1-coskt>(90/60)
or, -coskt>1.5-1
or, -coskt>0.5
or, coskt<-0.5
or, kt<cos-1(-.5)
hence, t<cos-1(-.5)/k
this gives t<10.
so my answer is also correct ryt....? my result shows that for about less than 10 mins the wheel was above 90.
not completely sure whether it's correct or not.

Majority have the answer that I mentioned. Don't worry its only 2 marks

 

[asadalam]

well... i thought....
h>90
so, 60(1-coskt)>90
or, 1-coskt>(90/60)
or, -coskt>1.5-1
or, -coskt>0.5
or, coskt<-0.5
or, kt<cos-1(-.5)
hence, t<cos-1(-.5)/k
this gives t<10.
so my answer is also correct ryt....? my result shows that for about less than 10 mins the wheel was above 90.
not completely sure whether it's correct or not.

That is illogical. Think about it .0 is also less than ten.does that mean the ride instantaneously went to 90m even before time was measured. You must have made an error somewhere.

Correct Answers (Joint Answers By Me and Aly).

Note: Question order may be wrong.

Q1) C = 8 And then arrange equation according to that.

2) area of shaded region= r^2(pie/2 -theta + sin2theta/2)

3) x^2= 240 x^3= 160 then 560x^3

4) u=6

5) theta= 71.4 and 63.7 (around these values)

6)i) 120m ii) pie/15 iii) 10<t<20

7) 2y=x-13 (idk the exactly value in replacement of 13) ii) C (9,7)

8)i) 459 ii)k=6 iii)S=54

9)i) 31.8 ii) 1/6 * (4i-2j+4k) iii) same magnitude for isosceles of OC and OA

10)i) 16pi/3 ii) i guess c was 2.5 and -3.5

11)i) p>0.5 ii) 2(x-1.5)^2 +0.5 iii) g(x)>=1/2 (iv) k=1.5 v)h inverse= 3/2 + underroot((2x-1)/4)

X^3 was -160 not 160
And we had to prove the trig identity using tancos in place of sin.

 

[farhan141]

That is illogical. Think about it .0 is also less than ten.does that mean the ride instantaneously went to 90m even before time was measured. You must have made an error somewhere.

X^3 was -160 not 160
And we had to prove the trig identity using tancos in place of sin.

Where have I mentioned identity of sin? I have corrected the -160.

 

[asadalam]

Where have I mentioned identity of sin? I have corrected the -160.

I was just mentioning as that parts ans was not included,not correct it

 

[Sifat Bin Quadery]

What working did u show for the value of t thing?

 

[DeadlYxDemon]

Forget it folks! Nothing can be done now.. better prepare for upcoming exams!

 

[Master Minds]

well... i thought....
h>90
so, 60(1-coskt)>90
or, 1-coskt>(90/60)
or, -coskt>1.5-1
or, -coskt>0.5
or, coskt<-0.5
or, kt<cos-1(-.5)
hence, t<cos-1(-.5)/k
this gives t<10.
so my answer is also correct ryt....? my result shows that for about less than 10 mins the wheel was above 90.
not completely sure whether it's correct or not.

The maximum height of the wheel was 120m. To calculate the time it will stay above 90m, we need to calculate the time taken to reach from 90m to 120m and then return back to 90m. Just imagine a huge wheel completing its revolution. I think that was the correct solution and my ans was 10 minutes.

 

[Master Minds]

Forget it folks! Nothing can be done now.. better prepare for upcoming exams!

Agreed
btw there was some kind of security breach in this paper in pakistan. CIE posted it on their fb page. Investigation is still under progress, can't say if we will get another chance to improve our math grade

 

[Cs Gunslinger]

Hello! I've got screen shots of the maths paper p1!
Sneak peek!

 

[DeadlYxDemon]

Hello! I've got screen shots of the maths paper p1!
Sneak peek!

View attachment 53257

OMG! .. how??

 

[BhaiArshad]

OMG! .. how??

 

[DeadlYxDemon]

Woah!

 

[farhan141]

Paper was leaked after exam last year too so nothing to worry about