Qualification > Math
[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !
physichemaths:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_11.pdf
question 5 part iii?
SkyPilotage:
--- Quote from: physichemaths on May 06, 2011, 07:31:56 pm ---http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_11.pdf
question 5 part iii?
--- End quote ---
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:
--- Quote from: SkyPilotage on May 06, 2011, 08:08:37 pm ---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!!
--- End quote ---
i mean part ii? typo :/
Vin:
--- Quote from: physichemaths on May 06, 2011, 08:27:37 pm ---i mean part ii? typo :/
--- End quote ---
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:
--- Quote from: Vin on May 06, 2011, 09:04:14 pm ---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.
--- End quote ---
THANK YOU! :D
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