Mathematics: Post your doubts here!

[Browny]

is that method important?

 

[kitkat <3 :P]

I don't remember doing it :/

 

[CЯeScɘnt]

How to do it using midpoint?

((2+0)/2, (14+10)/2)=(1,12) => midpoint of AC
let coordinates of D be x,y
midpoint of AC =the MP of BD
(1,2) = ( (-2+x)/2,(8+y)/2)
now compare both x-coordinates n y-coordinates to find x and y

 

[Faaiz Haque]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 2 ii , please preferred on paper so I can understand better each step.
Brief explanation if necessary pleaseee,

 

[Faaiz Haque]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 3 ii, on paper please.

 

[bigboy]

can anyone please help with no10 part2 http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_qp_32.pdf thanks in advance

 

[Talha Khatri]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 2 ii , please preferred on paper so I can understand better each step.
Brief explanation if necessary pleaseee,

 

[Faithix MFSH]

Can someone please help me finish off this question? I sort of got stuck..

The question is as below:

I solved and reached:

Now the last step I'm supposed to:

which should be quite simple, don't know what I'm missing.

Can someone please help me out?

 

[Talha Khatri]

Can someone please help me finish off this question? I sort of got stuck..

The question is as below:

I solved and reached:

Now the last step I'm supposed to:

which should be quite simple, don't know what I'm missing.

Can someone please help me out?

Since tan 0.25pi = 1

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 3 ii, on paper please.

When you expand ((1+x^2)^(2)) you get is x^4 + 2x^2 + 1

Now from part a you got the expansion like ; 32 + 80x^2 + 80x^4

Now multiply each brackets.

(x^4 + 2x^2 + 1)(32 + 80x^2 + 80x^4) = 32x^4 + 80x^6 + 80x^8 + 64x^2 + 160x^4 + 160x^6 + 32 + 80x^2 + 80x^4

Now we need coff. of x^4, hence, add 'em, 32 + 160 +80 = 272

 

[Talha Khatri]

Do you know how to do the question using vector move step?

the vector move step is pretty easy, you have to look at A and B look at how many x steps it takes and how many y steps it takes and in what direction for you to move from A to B or vice versa for example, look at A to B, it moves from 2 to -2 in x direction, so use this:
Final position - initial position
(-2) - (2) = -4 so 4 steps back in x direction
and for y direction:
8 - 14 = -6 so six steps down.
Now you can find the co ordinates for D using equation of line and stuff and then use this translation to find the co ordinates for C.

 

[Faaiz Haque]

View attachment 40921

Thanks so much man.

 

[Faaiz Haque]

When you expand ((1+x^2)^(2)) you get is x^4 + 2x^2 + 1

Now from part a you got the expansion like ; 32 + 80x^2 + 80x^4

Now multiply each brackets.

(x^4 + 2x^2 + 1)(32 + 80x^2 + 80x^4) = 32x^4 + 80x^6 + 80x^8 + 64x^2 + 160x^4 + 160x^6 + 32 + 80x^2 + 80x^4

Now we need coff. of x^4, hence, add 'em, 32 + 160 +80 = 272

Thank you

 

[Faaiz Haque]

Need help in identities
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 2 I

 

[Faithix MFSH]

Since tan 0.25pi = 1
View attachment 40924

I dont understand the last step, the one right before the final answer..
How does the denominator finalize to become 1? I know that the numerator is the answer we're trying to reach, okay. But the denominator should be 1, now how does 1 + (tant) (tan 1/4pi) equate to 1? (tan 1/4pi) is equal to 1 not zero, so how did tant get cancelled off? Shouldn't it be, tant - (tan 1/4pi) / (tan 1/4pi) + tant ?
I don't think it's possible to substitute 1 with (tan 1/4pi) only in the numerator..

Please do explain.

 

[CЯeScɘnt]

Need help in identities
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 2 I

2tan^2 x.cos x=3
2(sin^2 x/cos^2 x).cos x=3
cos cancelled the power of cos^2 x
2(sin^2 x/cos x)=3

put sin^2 x=1-cos^2 x
then simplify it n u will get the answer.

 

[Thought blocker]

Need help in identities
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf

Question 2 I

2(1-cos^2 theta)
------------------- = (3/cos theta)
cos^2 theta

2-2cos^2 theta = ( 3 cos^2 theta / cos theta )

-(2-2cos^2 theta) = 3cos theta

2cos^2 theta + 3cos theta - 2 = 0

 

[princeali97]

Dude, first off, you really need to cool off and let go of that khuari a bit. Give the lady some respect. Secondly, what you are asking is pretty simple. Look, you need to find the angle between a curve and a tangent, right? That would require you to know two things: The gradient of the tangent to the curve and the other is the point on the curve to which you need angle from. Next you differentiate the equation of the curve and find the gradient of that point by substitution and now all you need to know is the fact that a gradient is just simple the ratio of the height of an point to the width of it, that means,

Rise/Run. Which is just like the tan formula, Tan@ = perp/base so if you take the tan inverse of the gradient of that point and then take the tan inverse of the gradient of the tangent, you will have two angles, subtract the larger angle from the smaller angle and you have the angle between a tangent and a point on the graph. Lastly, I would humbly ask you to remove that comment from the thread, it looks extremely degrading on such an intellectual forum.

I dont find anything "degrading on this intellectual forum".But anyways i dont bother all this.Nice explanation .But my question is that the gradient of tangent and curve will be the same at that point ryt?.So how are u supposed to find the angle my friend?

 

[Rutzaba]

I dont understand the last step, the one right before the final answer..
How does the denominator finalize to become 1? I know that the numerator is the answer we're trying to reach, okay. But the denominator should be 1, now how does 1 + (tant) (tan 1/4pi) equate to 1? (tan 1/4pi) is equal to 1 not zero, so how did tant get cancelled off? Shouldn't it be, tant - (tan 1/4pi) / (tan 1/4pi) + tant ?
I don't think it's possible to substitute 1 with (tan 1/4pi) only in the numerator..

Please do explain.

i think he multiplied it with a conjugate

 

[Browny]

the vector move step is pretty easy, you have to look at A and B look at how many x steps it takes and how many y steps it takes and in what direction for you to move from A to B or vice versa for example, look at A to B, it moves from 2 to -2 in x direction, so use this:
Final position - initial position
(-2) - (2) = -4 so 4 steps back in x direction
and for y direction:
8 - 14 = -6 so six steps down.
Now you can find the co ordinates for D using equation of line and stuff and then use this translation to find the co ordinates for C.

Are you sure that is the method?