memeeee
Badges: 13
Rep:
?
#1
Report Thread starter 3 years ago
#1
paper is here! anyone got any answers?

i got A, B, D, A, C, C, D, C, B, A for the multiple choice but it's very likely those are wrong.
0
reply
Blazy
Badges: 8
Rep:
?
#2
Report 3 years ago
#2
(Original post by memeeee)
paper is here! anyone got any answers?

i got A, B, D, A, C, C, D, C, B, A for the multiple choice but it's very likely those are wrong.
At a quick glance, you may want to take another look at Question A

Also, http://www.wolframalpha.com/ is pretty useful for double checking some answers.
0
reply
memeeee
Badges: 13
Rep:
?
#3
Report Thread starter 3 years ago
#3
(Original post by Blazy)
At a quick glance, you may want to take another look at Question A

Also, http://www.wolframalpha.com/ is pretty useful for double checking some answers.
i did question A for 2sinx, not sin2x, oops! thought it was too easy. thank you
0
reply
mathphysics123
Badges: 5
Rep:
?
#4
Report 3 years ago
#4
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
0
reply
DFranklin
Badges: 18
Rep:
?
#5
Report 3 years ago
#5
(Original post by mathphysics123)
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
It's going to be hard to provide a solution without diagrams.

Sketch solution:

Spoiler:
Show



(a) area ABC = \frac{1}{2} bc \sin \alpha is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write \Delta(ABC) for the area of ABC, then we can deduce the sine rule fom the fact that \Delta(ABC) = \frac{1}{2} bc \sin \alpha = \frac{1}{2} ca \sin \beta = \frac{1}{2} ab \sin \gamma, then dividing by abc and taking the reciprocal of each fraction.

(b) BR = BC \cos \beta = a \cos \beta,  BP = AB \cos \beta = c \cos \beta, so area BRP = \frac{1}{2}ac \sin \beta \cos^2 \beta = \Delta(ABC) \cos^2 \beta, Similarly area CQP = \Delta(ABC) \cos^2 \gamma and area ARQ = \Delta(ABC) \cos^2 \alpha. Subtracting all 3 regions from the total area gives the desired result.


0
reply
Math1791
Badges: 9
Rep:
?
#6
Report 3 years ago
#6
(Original post by DFranklin)
It's going to be hard to provide a solution without diagrams.

Sketch solution:

Spoiler:
Show






(a) area ABC = \frac{1}{2} bc \sin \alpha is shown using 1/2 base x height, with the "height" being calculated using sin \alpha. Write \Delta(ABC) for the area of ABC, then we can deduce the sine rule fom the fact that \Delta(ABC) = \frac{1}{2} bc \sin \alpha = \frac{1}{2} ca \sin \beta = \frac{1}{2} ab \sin \gamma, then dividing by abc and taking the reciprocal of each fraction.

(b) BR = BC \cos \beta = a \cos \beta,  BP = AB \cos \beta = c \cos \beta, so area BRP = \frac{1}{2}ac \sin \beta \cos^2 \beta = \Delta(ABC) \cos^2 \beta, Similarly area CQP = \Delta(ABC) \cos^2 \gamma and area ARQ = \Delta(ABC) \cos^2 \alpha. Subtracting all 3 regions from the total area gives the desired result.






The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
0
reply
DFranklin
Badges: 18
Rep:
?
#7
Report 3 years ago
#7
(Original post by Math1791)
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
I'll give you 10:1 it's the expected solution, however.
0
reply
RichE
Badges: 15
Rep:
?
#8
Report 3 years ago
#8
(Original post by mathphysics123)
Does anybody have a quick solution for question 4, I have a solution but it is fairly longwinded
This question appears in the specimen papers, and they have solutions.
0
reply
RichE
Badges: 15
Rep:
?
#9
Report 3 years ago
#9
(Original post by DFranklin)
I'll give you 10:1 it's the expected solution, however.
Yes I recall this being noted after the fact.

PS I haven't thought about this in detail, but I imagine the question still makes sense for an obtuse triangle if the result is appreciated in terms of signed areas.
0
reply
DFranklin
Badges: 18
Rep:
?
#10
Report 3 years ago
#10
(Original post by Math1791)
The solution to part b only makes sense if none of the angles are obtuse, and the question does not mention that alpha,beta,delta are less than equal to 90 degrees.
(Original post by RichE)
Yes I recall this being noted after the fact.
@Math1791: Your ability to spot corner cases in MAT questions is really impressive (I'm not being sarcastic), but I really hope you're not going to let it stop you finding the expected solutions when it comes to the actual exam. When sitting an exam, there are times when you have to go with "there's a possible complication, but given the flow of the question and how different a solution dealing with the complication would be, I'm just going to ignore it" (possibly mention it if you're concerned).
0
reply
Math1791
Badges: 9
Rep:
?
#11
Report 3 years ago
#11
(Original post by DFranklin)
@Math1791: Your ability to spot corner cases in MAT questions is really impressive (I'm not being sarcastic), but I really hope you're not going to let it stop you finding the expected solutions when it comes to the actual exam. When sitting an exam, there are times when you have to go with "there's a possible complication, but given the flow of the question and how different a solution dealing with the complication would be, I'm just going to ignore it" (possibly mention it if you're concerned).
Thanks a lot for your advice and the praise I will keep it in mind and try my best to follow it.
0
reply
mrandrews
Badges: 6
Rep:
?
#12
Report 1 year ago
#12
Can someone help me with question 3 on this paper.

Part A is fine, but from part B on, I am unsure.

I had to form 4 equations using various values of T and then create a matrix. Using the inverse matrix I found the 4 coefficients and found the cubic.

But how are you actually meant to do this question.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (357)
56.13%
I don't have everything I need (279)
43.87%

Watched Threads

View All