[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

[elemis]

You would equate the discriminant as follows :

b²-4ac > 0

[I Jimmy I]

Lovely!
Thank You

[I Jimmy I]

Ok can some1 help me solve question 4?

To find the values of a,b and c

I know c is the y-intercept which is 3.
I know b is the number of cycles which is 2.
But I dunno how we got a=6 altho the maximum value is 9 and minimum value is -3
Cheers

[elemis]

Ok can some1 help me solve question 4?

To find the values of a,b and c

I know c is the y-intercept
So how do I find "a" and "b"
Cheers

360/b = the period of the curve. Since, in this case, it repeats itself every 180 degrees, b = 2.

C represents how many units the normal sine curve has been translated upwards from its normal position. Can you figure this out for yourself ?

[elemis]

But I dunno how we got a=6 altho the maximum value is 9 and minimum value is -3

'a' represents the vertical stretch factor. that has been applied to the normal sine curve.

Translate the curve in the question back to its original position and consider the max and min points..... Can you see the stretch factor ?

[I Jimmy I]

Ahaaaa!

Why thank u very much sir

[physichemaths]

june 2011 paper 11 question 5 part ii?

urgent. anyone please help me?

[bulono]

question 4i...help....
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_11.pdf

[Vin]

june 2011 paper 11 question 5 part ii?

urgent. anyone please help me?

June 2011. :/ Right. :/

question 4i...help....
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_11.pdf

Question paper would help better my friend.
Anywho,

LHS

Multiply numerator and denominator by 1 + cos x

Solve it carefully.

You should reach the stage where you get :

sin x tan x + sin x tan x cos x
-----------------------------
             sin^2 x

separate to give two fractions and cancel out the sine

you'll get:

tan x       tan x cos x
-----  +  ------------
sin x           sin x

1/ cos x + (equate tan x = sin x / cos x) and cancel out everything to get 1

Et Voila!

One tip. Write tan x = sin x / cos x at the top of the page ALWAYS while solving identities. Helps
Good Luck for Monday.

[iluvme]

CIE November 2010 Variant 2
Q7 iii)

I got the answer using a very very long method, does anyone have a shorter method since its 1 mark only?

[physichemaths]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_11.pdf

question 5 part iii?

[SkyPilotage]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_11.pdf

question 5 part iii?

After you found a+bcos^2x use that to substitute f(x) and add 1 and equate to zero...
-Rearrange making cos subject.. then square root (remeber its + or -) and find values of x.....             P.S: if cosx>1 or cos<-1 then its rejected!!

[physichemaths]

After you found a+bcos^2x use that to substitute f(x) and add 1 and equate to zero...
-Rearrange making cos subject.. then square root (remeber its + or -) and find values of x.....             P.S: if cosx>1 or cos<-1 then its rejected!!

i mean part ii? typo :/

[Vin]

i mean part ii? typo :/

Minimum value of x in the domain is 0, try that first
Substitute in 2- 5cos^2 x you get 2 - 5 = -3
try with the max value, pi, you get the same, right? since its cos^2 x, it cant take -ve values,  min = 0 and max = 1 (for cos^2 x NOT fx)
So minimum value of fx is -3

max is always 2, the constant. at x = pi/2 cos is 0, so 2.

[physichemaths]

Minimum value of x in the domain is 0, try that first
Substitute in 2- 5cos^2 x you get 3 - 5 = -3
try with the max value, pi, you get the same, right? since its cos^2 x, it cant take -ve values,  min = 0 and max = 1 (for cos^2 x NOT fx)
So minimum value of fx is -3

max is always 2, the constant. at x = pi/2 cos is 0, so 2.

THANK YOU!