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No brother i i was asking for the last part i.e (iv) . :/Put y as zero, so ; 0 = x(x-2)^2 => 0 = x^3 - 4x^2 + 4x + 0, you will find that you get two roots by solving this, one root will be 2 and second will be zero, and from graph, clearly it is seen x is not zero so x is 2, hence a = 2
It is asking for the minimum value of dy/dx... dy/dx= 3x^2 - 8x +4
hmm..okay i got it.Thanks.
Always welcome.
Change x to y and y to xhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Q9 (iii) can someone explain please
tell us the question we might help solving
let me give it a shotCan someone help me with this question?
The question is as follows:
Basically I found the area of the ABC sector and added the segments below the AB and AC lines.
This is how I went:
AB = 2r.cos(theta)
AngleBAC = 2(theta)
A of sector ABC = 1/2 x (AB)^2 x (2theta)
= 1/2 x (4r^2.cos^2theta.) x (2theta)
OA = r
A of the segments below the AC and AB lines:
2 x [ 1/2 x (OA)^2 x (theta - sintheta)
2r^2.theta - r^2.sin2theta
I added both and equated to half the area of the circle, i.e [ 1/2 x pie x r^2)
But couldn't reach the given result..
Can someone please give this question a shot? Would highly appreciate it.
p1 ?Can someone help me with this question?
The question is as follows:
Basically I found the area of the ABC sector and added the segments below the AB and AC lines.
This is how I went:
AB = 2r.cos(theta)
AngleBAC = 2(theta)
A of sector ABC = 1/2 x (AB)^2 x (2theta)
= 1/2 x (4r^2.cos^2theta.) x (2theta)
OA = r
A of the segments below the AC and AB lines:
2 x [ 1/2 x (OA)^2 x (theta - sintheta)
2r^2.theta - r^2.sin2theta
I added both and equated to half the area of the circle, i.e [ 1/2 x pie x r^2)
But couldn't reach the given result..
Can someone please give this question a shot? Would highly appreciate it.
CIRCLE RULES ANGLE BOC IS 4THETA THATS THE TWICE OF TWO THETACan someone help me with this question?
The question is as follows:
Basically I found the area of the ABC sector and added the segments below the AB and AC lines.
This is how I went:
AB = 2r.cos(theta)
AngleBAC = 2(theta)
A of sector ABC = 1/2 x (AB)^2 x (2theta)
= 1/2 x (4r^2.cos^2theta.) x (2theta)
OA = r
A of the segments below the AC and AB lines:
2 x [ 1/2 x (OA)^2 x (theta - sintheta)
2r^2.theta - r^2.sin2theta
I added both and equated to half the area of the circle, i.e [ 1/2 x pie x r^2)
But couldn't reach the given result..
Can someone please give this question a shot? Would highly appreciate it.
Yes I know that. But Im not using angle BOC im using BAC. Using BAC to find the area of so sector includes the area of the segments above OB and OC, but using angle BOC includes the area of the segments below AC and AB which is fine but that's not the flawless answer.
i think hogia mujh se
If you solved it and managed to reach the given final answer, please do upload a full stepped solution.
Change x to y and y to x
hence x = 5/1-3y
make y subject, you get y = x-5/3x
change y to f inverse.
Domain of f(x) is range of f inverse and vice-verse
so here domain of f(x) is >= 1 so range of f inverse would be >= 1
Range of f(x) is − 2.5 ≤ x < 0 so domain of f inverse will be − 2.5 ≤ x < 0
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