MAT Prep Thread 2018 Watch

Poll: What do you think you scored in the 2018 MAT exam?
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Quirky Object
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#21
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#21
(Original post by r_gup)
Thanks a lot. I will be more involved with this thread from tomorrow. What is an OP by the way?
Original post the STEP thread has a very extensive first post which is used every year and elaborates a lot more on STEP than this one does on MAT. It would be great if you and/or other MAT candidates/people who know about MAT could make an OP which is just as informative as the STEP one (I'd help but I'm a modern languages applicant who knows pretty much nothing about MAT :lol:).
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r_gup
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#22
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#22
(Original post by Sonechka)
Original post the STEP thread has a very extensive first post which is used every year and elaborates a lot more on STEP than this one does on MAT. It would be great if you and/or other MAT candidates/people who know about MAT could make an OP which is just as informative as the STEP one (I'd help but I'm a modern languages applicant who knows pretty much nothing about MAT :lol:).
Thanks. How can I add/edit the first post?
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Quirky Object
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#23
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#23
(Original post by r_gup)
Thanks. How can I add/edit the first post?
At the bottom of the post and to the left of the orange button which says "reply", there should be an edit button.
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r_gup
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#24
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#24
(Original post by Sonechka)
At the bottom of the post and to the left of the orange button which says "reply", there should be an edit button.
Thanks. Found it.
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r_gup
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#25
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#25
I am unable to upload the past papers in the first post. TSR's websit keeps showing 'Service Unavailable'. What should I do?
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r_gup
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#26
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#26
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Found this question in the STEP thread. I feel it is relevant to the MAT.
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thotproduct
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#27
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#27
in
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r_gup
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#28
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#28
(Original post by thotproduct)
in
??
Please explain.
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thotproduct
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#29
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#29
(Original post by r_gup)
??
Please explain.
Fairly self explanatory.

I'm doing the question now, chat more when I've done.
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Lemur14
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#30
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#30
(Original post by r_gup)
I am unable to upload the past papers in the first post. TSR's websit keeps showing 'Service Unavailable'. What should I do?
Attachments often cause problems. Just link the MAT website with them all on
That OP is generally rather out of date given it's talking about the old specification MAT, old spec maths etc. I've been doing a fair bit of research into what's changed and could update the OP if you're happy with that?

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r_gup
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#31
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#31
(Original post by Lemur14)
Attachments often cause problems. Just link the MAT website with them all on
That OP is generally rather out of date given it's talking about the old specification MAT, old spec maths etc. I've been doing a fair bit of research into what's changed and could update the OP if you're happy with that?

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Yes, please.
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r_gup
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#32
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#32
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What should be the answer? Boris?
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r_gup
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#33
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#33
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What should be the answer?
Either (iii) and (iv){for a=c and b=d} or (i) and (iv) { for a=b and c=d}?
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s.xw
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#34
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#34
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Could someone please explain what this solution means? What does it mean by the second equality? It’s not clear to me how they get from sinx < x < tanx to the required result. Thanks.
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r_gup
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#35
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#35
(Original post by s.xw)
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Could someone please explain what this solution means? What does it mean by the second equality? It’s not clear to me how they get from sinx < x < tanx to the required result. Thanks.
Can you please upload the entire image or the year of the paper?
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r_gup
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#36
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#36
(Original post by r_gup)
Can you please upload the entire image or the year of the paper?
Sorry , found it . MAT 2010 Question 3.
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s.xw
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#37
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#37
(Original post by r_gup)
Can you please upload the entire image or the year of the paper?
2010 Q3 (i) requires you to derive a given result from a diagram. The rest of the question uses this result and I was able to get most of it, but I just don’t get what their solution to (i) is trying to say.
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r_gup
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#38
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#38
(Original post by s.xw)
2010 Q3 (i) requires you to derive a given result from a diagram. The rest of the question uses this result and I was able to get most of it, but I just don’t get what their solution to (i) is trying to say.
So , Area \DeltaOAC = \frac{1}{2} sin(x)
Area Sector OAC = \frac{1}{2} x
Area \DeltaOAB = \frac{1}{2} tan(x)

As Area \DeltaOAC< Area Sector OAC<Area \DeltaOAB,

\frac{1}{2} sin(x)< \frac{1}{2} x<\frac{1}{2} tan(x)

so sin(x)<x<tan(x)

Now multiplying with cos(x),
\frac{sin(2x)}{2} <xcos(x)<sin(x)
as tan(x) = \frac{sin(x)}{cos(x)} , hence cos(x) will be cancelled out.
Using sin(x)> xcos(x) ,(from the above inequality)
x>sin(x)>xcos(x)
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s.xw
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#39
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#39
(Original post by r_gup)
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What should be the answer?
Either (iii) and (iv){for a=c and b=d} or (i) and (iv) { for a=b and c=d}?
This is a guess. But they imply that
if a-d = c-b, then
ad ≠ bc.
Therefore c > a, b > d or c < a, b < d. This contradicts (i). Therefore (d) is the answer. As I say, complete guess and probably wrong! I’ve just started looking at MAT papers a few days ago.
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s.xw
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#40
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#40
(Original post by r_gup)
So , Area \DeltaOAC = \frac{1}{2} sin(x)
Area Sector OAC = \frac{1}{2} x
Area \DeltaOAB = \frac{1}{2} tan(x)

As Area \DeltaOAC< Area Sector OAC<Area \DeltaOAB,

\frac{1}{2} sin(x)< \frac{1}{2} x<\frac{1}{2} tan(x)

so sin(x)<x<tan(x)

Now multiplying with cos(x),
\frac{sin(2x)}{2} <xcos(x)<sin(x)
as tan(x) = \frac{sin(x)}{cos(x)} , hence cos(x) will be cancelled out.
Using sin(x)> xcos(x) ,(from the above inequality)
x>sin(x)>xcos(x)
Thank you so much! Forgot that I had just proved x > sinx and didn’t occur to me to reuse the fact.
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