Mathematics: Post your doubts here!

[xtrasaucy]

can someone please send me a link to the MARK SCHEME of 9709/11/m/j/16

 

[krishnapatelzz]

can someone please send me a link to the MARK SCHEME of 9709/11/m/j/16

http://dynamicpapers.com/wp-content/uploads/2015/09/9709_s16_qp_11.pdf

 

[Holmes]

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Thank you DragonCub...

 

[Holmes]

can someone please send me a link to the MARK SCHEME of 9709/11/m/j/16

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_s16_ms_11.pdf
hope it helps

 

[Holmes]

would you please tell me the reason for (June 2014 /9709/12 Q 4)
why is A(aobx) = 2 A (aob)
Thank you in advance

 

[xtrasaucy]

would you please tell me the reason for (June 2014 /9709/12 Q 4)
why is A(aobx) = 2 A (aob)
Thank you in advance

Area of triangle AOB = 1/2

would you please tell me the reason for (June 2014 /9709/12 Q 4)
why is A(aobx) = 2 A (aob)
Thank you in advance

In the guestion they tell us that AOB and AXB are equal so..
Area of triangle AOB = 1/2 r^2 sin(theta)
Area of segment AXB = 1/2 r^2 (theta) - 1/2 r^2 sin(theta)

Now since their areas are equal :
1/2 r^2 sin(theta) = 1/2 r^2 (theta) - 1/2 r^2 sin(theta)

which simplifies to: 2 sin(theta)

 

[DragonCub]

would you please tell me the reason for (June 2014 /9709/12 Q 4)
why is A(aobx) = 2 A (aob)
Thank you in advance

AOBX refers to the whole shape. So the area of AOBX, or A(aobx), equals the sum of A(aob), the triagle, and A(axb), the area between the chord and the arc.
Put in an equation, this means A(aobx) = A(aob) + A(axb)
Since A(aob) = A(axb),
A(aobx) = 2 A(aob)

 

[krishnapatelzz]

can someone explain me this whole .

 

[Holmes]

AOBX refers to the whole shape. So the area of AOBX, or A(aobx), equals the sum of A(aob), the triagle, and A(axb), the area between the chord and the arc.
Put in an equation, this means A(aobx) = A(aob) + A(axb)
Since A(aob) = A(axb),
A(aobx) = 2 A(aob)

thanks again

 

[DragonCub]

View attachment 61854can someone explain me this whole .

This is a common way of solving differential equations.
Exactly which part are you confused about?

 

[sj0007]

Can someone walk me through the steps? I can't wrap my head around the part where the sign changes............

This is Q3 of - https://papers.gceguide.xyz/A Levels/Mathematics (9709)/9709_w13_qp_63.pdf this though
(The marking scheme - https://papers.gceguide.xyz/A Levels/Mathematics (9709)/9709_w13_ms_63.pdf)

 

[DragonCub]

Can someone walk me through the steps? I can't wrap my head around the part where the sign changes............
View attachment 61860
This is Q3 of - https://papers.gceguide.xyz/A Levels/Mathematics (9709)/9709_w13_qp_63.pdf this though
(The marking scheme - https://papers.gceguide.xyz/A Levels/Mathematics (9709)/9709_w13_ms_63.pdf)

 

[sj0007]

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Ooooohkayyy. So that's what I was missing out!
Thanks a lot!

 

[areeba240]

im done with 9709 pastpaper........now from where should i practise.....need advice.....should i do ocr or anyone have something better than this

 

[areeba240]

(iii)

 

[xtrasaucy]

View attachment 61864
(iii)

Whats the code for this paper?

 

[areeba240]

Whats the code for this paper?

9709

 

[DragonCub]

View attachment 61864
(iii)

Answers will be ax + by + cz = d + 70√2 and ax + by + cz = d - 70√2
The values of a, b, c and d are the same as you found in part (i).

Basically, the only change from answer to part (i) is that we add or subtract a number to d,
And this number is equal to the distance (14) times the modulus of n (|n|, which is √50).

Try to think by yourself why it can be done this way.

 

[sharyar anwer]

9
(a) The first term of an arithmetic progression is −2222 and the common difference is 17. Find the value of the first positive term. [3]
(b) The first term of a geometric progression is ï3 and the second term is 2 cos 1, where 0 < 1 < 0. Find the set of values of 1 for which the progression is convergent.
can smebody plz help

 

[sharyar anwer]

9 (a) The first term of an arithmetic progression is −2222 and the common difference is 17. Find the value of the first positive term. [3]
(b) The first term of a geometric progression is ï3 and the second term is 2 cos 1, where 0 < 1 < 0. Find the set of values of 1 for which the progression is convergent.
plz help