Paper 12 maths

[confused123]

nope im damn sure its 12 pi

orryte congrats then :|

 

[Aarjit]

but but .. it said, "the plate was made by removing a triangular piece"... or somethin like that! ='[ .. I wrote 146 .. then re read the question and changed both answers I thin *sobsniffcry*

I'm sorry but the plate was not made by removing the triangular piece. In fact, it was not a triangular 'piece' at all - it was a hint. The dotted lines merely showed the radii of the circular plate which met at the ends of chord AB, making it easier for the candidates to use trigonometry to determine its length.

 

[OrlandoBloom<3]

The paper was actually quiet tough compared to the previous papers , but if you had all your concepts clear for all the topics then you would have managed to do fine , relying only on the past papers wouldnt have helped so much for this paper . Anyways I had problems with the vector question and the arithmetic question where they had given some formula for sum of terms :/ .Otherwise I could do all questions but ofcourse silly mistakes are always bound to happen for sure no matter how hard you try to avoid

 

[Aarjit]

what was the solution for the trigo question?? I spent lot of time on it but not sure whether its right.

Okay. Let me see how far I can recall..

tanx + 1/tanx = (sinx/cosx)+(cosx/sinx) = (sin^2 x + cos^2 x)/sinx.cosx = 1/sinx.cosx Q.E.D

now,
solve for 2/sinx.cosx = (an expression with tanx); 0 < x < 180
>> 2*(tanx + 1/tanx) = (RHS) [from the first part]
factorize to get (tanx - 1)(tanx + 2) = 0
>> tanx = 1 ; tanx = -2
>> x =45 ; x = 180 + (-63.43...)
= 116.56 ~ 116.6​

 

[shanky631]

h

Okay. Let me see how far I can recall.. ​

​

tanx + 1/tanx = (sinx/cosx)+(cosx/sinx) = (sin^2 x + cos^2 x)/sinx.cosx = 1/sinx.cosx Q.E.D​

​

now,​

solve for 2/sinx.cosx = (an expression with tanx); 0 < x < 180​

>> 2*(tanx + 1/tanx) = (RHS) [from the first part]​

factorize to get (tanx - 1)(tanx + 2) = 0​

>> tanx = 1 ; tanx = -2​

>> x =45 ; x = 180 + (-63.43...)​

= 116.56 ~ 116.6​

hey thanks dude. i made a mistake in the signs so i got +63.43

 

[angelicsuccubus]

I'm sorry but the plate was not made by removing the triangular piece. In fact, it was not a triangular 'piece' at all - it was a hint. The dotted lines merely showed the radii of the circular plate which met at the ends of chord AB, making it easier for the candidates to use trigonometry to determine its length.

I could so kill the cie people right now -_- I must've lost 5 marks there!

 

[Aarjit]

I could so kill the cie people right now -_- I must've lost 5 marks there!

Don't worry. They'll credit your bid to calculate the perimeter of the greater arc (AB) and the chord length too. The examiner will penalize you for the area and you'll lose the A1 mark. ~ 2 marks. I'm sure you've done well.

 

[FarahT]

didn't the last part of the last question have something to do with fg(x), as in we had to place either g or ginverse into f which was then equated with kx? i definitely remember that there was f(g)x....??

 

[123add1]

who is giving m1 ???

 

[FarahT]

Were everyone's graphs for f(x) and f inverse of x reflected in y=x.
f(x) was 2x + 5 and its inverse was not a reflection in y=x???

 

[Munther Hassan]

I did very well in my exam yesterday, i was really happy when i came out of the exam hall

 

[Aarjit]

Were everyone's graphs for f(x) and f inverse of x reflected in y=x.
f(x) was 2x + 5 and its inverse was not a reflection in y=x???

f(x) and f-1(x) are always reflections of each other in the line y = x, since you're interchanging y and x the very instant you commence deducing the inverse (often by making 'x' the subject).

P.S. f(x) = 2x +5 and f-1(x) = (x-5)/2 are reflections of each other in the line y = x.

 

[Prisonbreak94]

Crap, guys in the first question where we were given the curve and were asked to find the volume obtained, i wrote down integral pi y^2 but i forgot to square the function while taking the integral, I went on to take the integral of the normal function and got an answer like that (3pi ln3), how many marks out of 4 do you think I can get? 2?

 

[hard2face]

i made a mistake in finding the shaded area in integration, i got 1 limit correct but the other was wrong but i showed all the steps would i get any marks for it (if any then how much out of 4)

 

[Khalifa]

Guys I don't remember the GP question at all. I remember the AP part and remember getting a=9 and d=2. If someone remembers the question, can they please post it here, the GP one?

 

[muhammadbinaziz]

for that question..
part (i) yu had to use cosine rule=14.9
(ii) 45.9
(iii) 145.9

part (iii) you will be deducted a mark for not writing answer in 3 s.f.

 

[muhammadbinaziz]

My answer was 5 in the binomial question but im not so sure. I think I messed up the formula. Can someone else please confirm?

yeah ur rite a=5 for binomial...chill..

 

[muhammadbinaziz]

haha in the trig question, I wrote about 5 different degrees in the range of -180 to 180

two answers only
equated by tan(x)

 

[Sanis]

3) a=5 ,, trigonometry angles were 45 & 116.6 ,,, 9) area was 28/3,, 10) k<0 or k>64/9 ,,, 2) dy/dt was 0.12*7/8 ,, i was unable 2 solve ap & gp

 

[Aarjit]

Guys I don't remember the GP question at all. I remember the AP part and remember getting a=9 and d=2. If someone remembers the question, can they please post it here, the GP one?

the second term if a GP is 9 less than the first term. The sum of the second and third terms is 30. Given that all terms are positive*, find the value of the first term.

a.r = a - 9 ; a.r + a.r^2 = 30
>> r = (a - 9)/a

>> 30 = (a - 9) + a.[(a-9)/a]^2
>> 30 = a - 9 + (a-9)^2/a
>> 39.a = (a^2) + a^2 - 18.a + 81
>> 2.a^2 - 57.a + 81 = 0
>> factorize to get (a - 27)(2.a - 3) = 0

a = 27 or a = 3/2
since r= -5 when a= 3/2, the terms become negative*(see question);

Ans: a = 27