Q16 (1 Viewer)

[Jashua_Long]

Re: radians/degrees?=/

Q16 (b) was so easy. Just did it in less than 5 minutes.

 

[Rawf]

This is for 16b)

gradient OT = tan theta, therefore gradient tp = -1/tan theta...

point T is (sin theta, cos theta) and use equation of a line formula

oh my frik, i was so close.. i had a bunch of cos, sin and tan crap but then I didn't know what to do

 

[gr_111]

Point T is (cos theta, sin theta) or the other way around ?

yes you're correct sorry, typo. Definitely got it in the exam though!

 

[Jashua_Long]

Hi! so i can do worked solutions if people want but basically here was how to do 16c i. took me a solid 30min to work out how to do it. Anyways.

equate the two graphs. so you get

y + (y-c)^2 = R^2.
y + y^2 - 2yc + c^2 - R^2 = 0
y^2 + y(1-2c) + (c^2 - R^2) = 0

now there are 2 equal and identical solutions to the above equation (because they intersect twice at the same location, just on oposite sides of the y axis)

hence the discriminate = 0

(1-2c)^2 - 4(c^2 - R^2) = 0 will then simplify down into the required answer, 4c = 1-4C^2

No idea how to do part ii though...

Don't be surprised if you lose a mark.

That is incorrect reasoning.

The graph has two intersection points. The correct reasoning is that the quartic equation should have a discriminant of zero to ensure two values of x.. If the quartic has zero discriminant, x^2 has only one solution. Therefore, x has two solutions.

They should penalise the people that did not include the reasoning (similiar to what I said above). I know tons of people would have just blindly set discriminant to zero (as the answer involved an equality, not a inequality) without actually thinking about what they were doing.

 

[jnigga07]

This. Working up to question 16 I was just like "make sure you don't make silly mistakes and should get a solid mark", nek minute hit Question 16, an hour later, fuck it I'll just check for silly mistakes. But if they hadn't of put in a fucked up question 16 then they would not have got a very good spread of marks imo.

 

[Jashua_Long]

This. Working up to question 16 I was just like "make sure you don't make silly mistakes and should get a solid mark", nek minute hit Question 16, an hour later, fuck it I'll just check for silly mistakes. But if they hadn't of put in a fucked up question 16 then they would not have got a very good spread of marks imo.

Q16 was easy, what u talking about

 

[TheOptimist]

For those who still don't get part b), here is my solution that I used in the exam:
View attachment Q16 b).pdf
Sorry if there are some minor mistakes, I did this in like a couple of minutes.

 

[Jashua_Long]

For those who still don't get part b), here is my solution that I used in the exam:
View attachment 26773
Sorry if there are some minor mistakes, I did this in like a couple of minutes.

Thats the newb way of finding the equation of the line.

 

[TheOptimist]

Thats the newb way of finding the equation of the line.

Its the safest way to ensure you get the marks imo

 

[captainplanet7]

it was to easy... like i think i wasted time doing 11 past papers. kinda pissed tbh good have achieved the same resurlt with 4-5 papers.

 

[P.T.F.E]

Nailed most of it apart from 16 bi) the rest of b was quite simple

 

[RealiseNothing]

How I got the equation of the line:

Gradient of OT is

Therefore gradient of TP is

T has co-ordinates

Therefore:



Then you just re-arrange to get:

 

[RishBonjour]

How I got the equation of the line:

Gradient of OT is

Therefore gradient of TP is

T has co-ordinates

Therefore:



Then you just re-arrange to get:

yep, easiest and most efficient way of doing that question.

 

[RishBonjour]

Don't be surprised if you lose a mark.

That is incorrect reasoning.

The graph has two intersection points. The correct reasoning is that the quartic equation should have a discriminant of zero to ensure two values of x.. If the quartic has zero discriminant, x^2 has only one solution. Therefore, x has two solutions.

They should penalise the people that did not include the reasoning (similiar to what I said above). I know tons of people would have just blindly set discriminant to zero (as the answer involved an equality, not a inequality) without actually thinking about what they were doing.

I think they will penalise. It wasn't a question 16 for nothing. 1/2 without reasoning prolly.

 

[Jashua_Long]

Just looked at Q16 (a). So easy.

Easy similiarity proof with parallel lines and angles and then a simple ratio of corresponding sides.

Was a joke.

Done in 3 minutes.

 

[RealiseNothing]

I think they will penalise. It wasn't a question 16 for nothing. 1/2 without reasoning prolly.

You can just as easily solve for 'y' and avoid any reasoning.

 

[cookeemonstah]

For those who still don't get part b), here is my solution that I used in the exam:
Q16 b).pdf
Sorry if there are some minor mistakes, I did this in like a couple of minutes.

shit...im an idiot.

 

[gr_111]

Q16 was easy, what u talking about

Thats the newb way of finding the equation of the line.

Just looked at Q16 (a). So easy.

Easy similiarity proof with parallel lines and angles and then a simple ratio of corresponding sides.

Was a joke.

Done in 3 minutes.

Big deal mate. What are you trying to prove by coming on here and dissing everyone?

 

[Stevm]

The real solution to Q16. (B) (ii)

Since there's many BS answers flying everywhere, here's (one of) the real answers:

 

[BKM1]

part (a) was alright, (b) just took a lot of working but i was completely lost on (c)