Mathematics: Post your doubts here!

[princeali97]

last question last part.I didnt understand.Pls explain if possible.http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[Thought blocker]

last question last part.I didnt understand.Pls explain if possible.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

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

 

[princeali97]

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

No brother i i was asking for the last part i.e (iv) . :/

 

[Thought blocker]

No brother i i was asking for the last part i.e (iv) . :/

It is asking for the minimum value of dy/dx... dy/dx= 3x^2 - 8x +4
thus yu just have to find minum value of that quadratic function
Two ways to do it .. Do completing squares or just derivate it again and equate it to zero .. u shall get x substitute that x to the dy/dx equation and that will be yur anser

 

[princeali97]

It is asking for the minimum value of dy/dx... dy/dx= 3x^2 - 8x +4
thus yu just have to find minum value of that quadratic function
Two ways to do it .. Do completing squares or just derivate it again and equate it to zero .. u shall get x substitute that x to the dy/dx equation and that will be yur anser

hmm..okay i got it.Thanks.

 

[Thought blocker]

hmm..okay i got it.Thanks.

Always welcome.

 

[qbdulsami]

Is there a solution book of mathmatics syllabus d book 2 if yes the plzzzzzzz upload the book

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_12.pdf

Q9 (iii) can someone explain please

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf

Q9 (iii) can someone explain please

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

 

[Faithix MFSH]

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 (2theta - sin2theta)
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.

 

[Rutzaba]

Is there a solution book of mathmatics syllabus d book 2 if yes the plzzzzzzz upload the book

tell us the question we might help solving

 

[Rutzaba]

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.

let me give it a shot

 

[Thought blocker]

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.

p1 ?

 

[Rutzaba]

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 THETA
oops srry fr caps lock

 

[princeali97]

Finally i got it.This is the formula for finding the angle between 2 straight lines that make angle @between them.
tan@=m1-m2/1+m1m2

 

[Thought blocker]

Finally i got it.This is the formula for finding the angle between 2 straight lines that make angle @between them.
tan@=m1-m2/1+m1m2

?

 

[Faithix MFSH]

CIRCLE RULES ANGLE BOC IS 4THETA THATS THE TWICE OF TWO THETA
oops srry fr caps lock

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.

 

[Rutzaba]

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

 

[Faithix MFSH]

i think hogia mujh se

If you solved it and managed to reach the given final answer, please do upload a full stepped solution.
Also, I didn't really understand what you said in the past comment.

 

[Rutzaba]

my answer is coming cos2Q=8Q- pi -sin2Q whole divided by 2 Q i guess its wrong