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

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$H00t!N& $t@r:
oh ok no problem

$H00t!N& $t@r:
wow am I the only one asking questions?  :-[

Anyway, I have another question!

Q) The equation of a curve is y = 3 cos 2x. The equation of a line is x + 2y = pie. On the same diagram, sketch the curve and the line for 0<= x<= pie. 

The question is from O/N 2009 paper

Thanks  ;D

samih.mabsout:
ddnt okosodo solve this on the board?? :P

samih.mabsout:
y=4x^4+4x+9
i) Find the coordinates of the stationary point on the curve and determine its nature.
ii) Find the area of the region enclosed by the curve, the x-axis and the lines x=0 and x=1

* paper 1/oct/nov/2009 question 4*

Alpha:

--- Quote from: $H00t!N& $t@r on November 17, 2010, 10:18:33 am ---how do you find the range of a fuction?  ??? i read the book but i dont understand it  :-\

can someone help me with m/j 05 q7 (i) and (iii)  please explain the answer in (iii) Thanks

--- End quote ---

(i)
f : x -> 3 - 2 sinx

Amplitude is -2. y= - 2 sinx will have maximum and minimum points at 2 and -2. And since,  f : x -> 3 - 2 sinx => curve is translated 3 units upward, minimum and maximum points will become:

2+3=5
and -2+3=1.

The range in which x is found therefore is: 1 </= f(x) </= 5 (Because of the equal to sign here: 0o </= x </= 360o.)

(iii)

You must have drawn the curve. For any curve to have an inverse, it has to be a one-one function, i.e. every horizontal line drawn across it should cut at only one point. From the graph, you'll notice that as you move from 90o onwards, any horizontal line through that curve will cut twice. Therefore, the curve is one-one in the range: 0o </= x </= 90o.

=> A= 90o.

N. B: Curve is also one-one in the ranges: 90o </= x </= 270o and 270o </= x </= 360o

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