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| r | > 1. That's what I used to solve for y. What did you do?

y > 1/2

You'll probably get a mark for differentiating it properly. I got y > 1/2 except y =/= 1. Probably wrong.

Annoyed with myself for realising that 20 mins after the exam. How much was iii worth? 2 or 3 marks?

Yeah same. But I did a = 1, l = 12, n = 12, then did Sn = n/2(a+l). Then multiplied the answer by two.

I got something else. Idk. I hope it was.

Wait, what? Omg. So depressed right now. Another 3 marks gone.

Because they all have 2 hearts. HH - 2 normal hearts HB - 1 normal and 1 sparkly heart BH - 1 sparkly and 1 normal heart BB - 2 sparkly hearts

No. HH HB BH and BB It was good. Just didn't finish the last question - part iii. But now I realised all you had to do was find the derivative and sub in pi/12. ><

Intramolecular: covalent bonds. Ethanol has both a polar end(hydroxyl group) and a non-polar end (ethyl group), allowing it to dissolve in both polar and non-polar substances. The hydroxyl group can form hydrogen bonds with polar substances (e.g. water), and the ethyl group can form dispersion...

Normally you would do 360 - 45, right? For the second revolution: 360 + 360 is 720. So basically, 720 - 45 = 675 But since its tan, -675. then just divide by two.

But when you divide 405 by 2 you get 202'30'. But the limit is between -180 and 180. It's right, but just not what the question is asking for. You have to answer the question.

Um... What strategies/tactics were implemented in order to break the stalemate?

Whenever there's a number infront of the x, e.g. tan(2x), there's more than one revolution. In this case, there's 2 revolutions. At the end, you divide the answer by 2. So you would've gotten: 2x = 45', 225, -135' and -315' But since you're solving for x, you divide everything by 2. Hope that...

Here you go. Hope you understand what I did. The picture is cutting off from the bottom. But what you do is sub m (20 and -44) into the equation, y = mx + 2 and you will get the anwer. y = 20x + 2 and y = -44x + 2.

If you're confident with let's say Q 1 - 8, then just do 9 and 10. What I do with the older papers is go through them and identify all the questions I think I'll find challenging/hard and do those, and leave the easy ones out. Saves time.

1995 - 2010 are alright. But 2011 and 2012 are more challenging. I hope this year's isn't too bad.

Yeah, question 10s can be tricky at times. You should then just maybe do Q 9s and 10s? The only way you can study for it is to practice it. Just try your best. If you've tried it a million times, look at the answer and try to understand it. And then what you would've learnt here, you can apply...

What exactly is the question asking? Oh, okay I understand the question now. "Any tangent lines that pass through 0,2". I'll try it now.

This is where you started to go wrong. This not the slope, but the equation to find the slope at a specific point. So the question is asking for the equation of the tangent at (0,2). What is the gradient of the tangent at (0,2)? Sub 0 into the derivative, y = 4x^3 -12. You get -12. Where m is...