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Edexcel A2 C3 Mathematics 12th June 2015

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Original post by TeddyKC
Could someone help with this beast of a question?
(Part b)
ImageUploadedByStudent Room1434050875.568486.jpg


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Which paper is it from?
Original post by frozo123
I wish I knew how to use a calculator...


The struggle is real
Reply 1782
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Pitfak
Original post by Bealzibub
Do I need to learn these? Do they come up in papers?



they are quoted in the formula book;
you could be asked to prove one of them but they would probably give you a hint - let P=A+B and Q= A-B, then using the addition formulae show that sinP-sinQ =....
I do think this is unlikely to be asked though.
Original post by TeeEm
question with solution posted earlier (original post 1335)

well done to those who did it correctly ...
I did it with 5 of my students today, and they all got "eaten" ...


Ty for the worked method,

3 lines before the answer, where you took e to one side, then simplified it, and turnt it into one e^ to the power of ln's , how did you get 1/10 ln4.
Why has 3sinx from the question changed to 3cosx, how did they get to this stage ?? the model answer matches the mark scheme
Screenshot 2015-06-11 21.04.02.png375.7KB
Has anyone got an easy way to remember transforming graphs? Like what to do when it's af(x) or f(ax)
Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.
Original post by nayilgervinho
How many sig fig should I put down for solutions to contextual rcostheta/rsintheta questions.

they normally tell you how they want it, but if they dont, 3sf
Original post by Gilo98
it is five. but look below at the conventional approach using similar triangles. Apply a similar method for f(5) on the RHS branch.


Hi, could you please explain June 2013 withdrawn 8c and d, I'm confused by the format of H since the sign is the opposite
Original post by BioAgent
Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.


-ln11/20=-(ln11-ln20)=ln20-ln11=ln20/11
Original post by MrBowcat
Them asking to draw inverse trig graphs is very very very very very very very very very very very very unlikely.


That does not mean it's impossible, and a two or three marker based on that could be the difference between two grades for someone. People might as well skim them
Original post by BioAgent
Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.


Do you agree that 2ln(x)=ln(x2)2ln(x)=ln(x^2)? And do you agree that (1120)−1=2011(\frac{11}{20})^{-1}=\frac{20}{11}?
Original post by suyoof123
Ty for the worked method,

3 lines before the answer, where you took e to one side, then simplified it, and turnt it into one e^ to the power of ln's , how did you get 1/10 ln4.


1/5(....) - 1/10(....) = 1/10(....)
Original post by AbubakarB
Can you accurately draw the sin, cos and tan graphs?

From there everywhere where the graph crosses the x-axis (when y=0) draw an asymptote.

1) for the cosec graph...
every maximum point on the sin graph - draw a happy face lol
every minimum point draw a sad face.

2) same concept with the sec graph

3) the tan graph is a bit trickier to grasp but still the same concept. y=0 ... asmyptote then flip the graph

yes I can
what do you mean flip? reflection in the asymptote?
Predictions for 90 ums.
Original post by TeeEm
1/5(....) - 1/10(....) = 1/10(....)


Thank you! cant believe I missed that....:biggrin:
Original post by toddle1
Has anyone got an easy way to remember transforming graphs? Like what to do when it's af(x) or f(ax)


af(x) = multiply y coords by a
f(ax) = divide x coords by a
But how do you solve the quadratic?

I got x^2(a) +x(k) +a = 0

How would you solve a quadratic like this???
Original post by TeeEm
1/5(....) - 1/10(....) = 1/10(....)


Last minute advice/predictions?

Original post by adorablegirl1202
Hi, could you please explain June 2013 withdrawn 8c and d, I'm confused by the format of H since the sign is the opposite

check out Arsey's model answers, haven't done it myself and chilling out now :h:
Original post by TeddyKC
Could someone help with this beast of a question?
(Part b)
ImageUploadedByStudent Room1434050875.568486.jpg


Posted from TSR Mobile


set the two functions equal to each other to get an equation for their points of intersection. It should be a quadratic in x. Since they curves are tangential, you know this equation only has one solution. This means the discriminant of the equation is zero. You'll then get an equation you can solve for k.

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