0606 ADDITIONAL MATHS P2 SUCKED BIG TIME!

[Serial_Ripper]

More or less 24 hours have passed since the examination - it's more adequate to call it eternal damnation - So, yeah, am I the only one afflicted by the unscrupulousness of this examination? And if so, does this mean that it was merely a piece of cake that the candidates seemed to slurp with utter relish? Or that my stupidity knows no bounds? For "hard" is an underestimation ... one I'm constantly obliged to resort to for the sake of obviating this entire ramble. It seems as though I've MURDERED the examiner's entire family (even if my username suggests that I already had and that's merely his revenge! All my hopes of guaranteeing an A* have been so gruesomely mutilated that scoring an A (and not shredding the paper into bits) seemed (and still does, I assure you) like an impossible feat. I'm well-aware of the fact that I wasn't the only one who has suffered...and hence all I ask for is a small account of the torment every Add Maths student has suffered from during those two hours...

 

[Dark Destination]

It was so easy.
Which question did you find hard?

 

[Awesome12]

It was so easy.
Which question did you find hard?

You gave IGCSE Add-Maths??????

Did you get anything wrong?

 

[Serial_Ripper]

It was so easy.
Which question did you find hard?

EASY! ARE YOU HIGH? (sorry...but easy is just too much)
THAT SURD QUESTION, FOR EXAMPLE! (where possible integer values of a and b are to be found) I COULDN'T DO IT!

 

[Dark Destination]

EASY! ARE YOU HIGH? (sorry...but easy is just too much)
THAT SURD QUESTION, FOR EXAMPLE! (where possible integer values of a and b are to be found) I COULDN'T DO IT!

I know, that was a 'bit' trickier than other questions.

The question was:

(a + 3√5)₂ + a - b√5 = 51

You had to expand the left side of the equation.

a^2 + 45 + 6a√5 + a - b√5 = 51

It becomes (a^2 + a + 45) + √5(6a - b) = 51

Then it was simple. Comparing both sides. Equate the (a^2 + a + 45) to 51.. and the co-efficient of √5 to zero, since it is not on the right hand side.
The answers were:

a = 2, 12
b = -3, 18

Get it? xD

 

[Serial_Ripper]

Great...I got one mark.........out of six
Thanks...

 

[Dark Destination]

Great...I got one mark.........out of six
Thanks...

If the paper was hard, the percentile would probably be lower this time. So you can still get an A*.
And Believe me. CIE is all miracles. xD Unexpected things happen! *_*
It all depends on how the candidates performed this time.

That Circular measure question was pretty annoying though. ._.
It was getting hard to keep track of all the different lengths and sides.

 

[Harsheys]

I know what you mean! It was a horrible paper,
All my friends even said that they hated it big time!
That surds question sucked, I don't think I got many marks for it either.

But there was also that trig questions that said "using part (a), solve sec(x)cosex(x) = 3cot(x)"
I didn't know how to use part (a), so I just cross multiplied sin(x)cos(x) and the 3 from the 3cot(x)...

 

[Serial_Ripper]

If the paper was hard, the percentile would probably be lower this time. So you can still get an A*.
And Believe me. CIE is all miracles. xD Unexpected things happen! *_*
It all depends on how the candidates performed this time.

That Circular measure question was pretty annoying though. ._.
It was getting hard to keep track of all the different lengths and sides.

I REALLY hope so!
And you're right abut the Circular measure question!

 

[Serial_Ripper]

I know what you mean! It was a horrible paper,
All my friends even said that they hated it big time!
That surds question sucked, I don't think I got many marks for it either.

But there was also that trig questions that said "using part (a), solve sec(x)cosex(x) = 3cot(x)"
I didn't know how to use part (a), so I just cross multiplied sin(x)cos(x) and the 3 from the 3cot(x)...

I did the same thing...and tried to verify the answer with Function 7 on my calculator...I hope it's correct.

 

[Dark Destination]

I know what you mean! It was a horrible paper,
All my friends even said that they hated it big time!
That surds question sucked, I don't think I got many marks for it either.

But there was also that trig questions that said "using part (a), solve sec(x)cosex(x) = 3cot(x)"
I didn't know how to use part (a), so I just cross multiplied sin(x)cos(x) and the 3 from the 3cot(x)...

Do you remember the identity we had to prove in part (a)?
What was it?

 

[Harsheys]

Do you remember the identity we had to prove in part (a)?
What was it?

Yea I did, it was sec(x)coesc(x) - cot(x) = tan(x)

 

[Dark Destination]

Yea I did, it was sec(x)coesc(x) - cot(x) = tan(x)

Yes, so i converted it into:

sec(x)cosec(x) = tan(x) + cot(x)

Substituted this into the next part, and solved to get four answers. Do you think it's correct?

 

[Harsheys]

Yes, so i converted it into:

sec(x)cosec(x) = tan(x) + cot(x)

Substituted this into the next part, and solved to get four answers. Do you think it's correct?

One of my friends did that, I think it's correct.
I got 4 answers as well, but I didn't substitute it that way...do you think I'll still get marks for my answers?

 

[Dark Destination]

One of my friends did that, I think it's correct.
I got 4 answers as well, but I didn't substitute it that way...do you think I'll still get marks for my answers?

Well, i think you would.