Mathematics: Post your doubts here!

[kitkat <3 :P]

Phew, that took a while to write clearly and illustrating properly sorry for the delay
View attachment 41014 View attachment 41015 View attachment 41016

Y Is AOC =pi -2theta :/ like is it the formula?

 

[Thought blocker]

How did you get 3,9 ?

What was the question ?

 

[kitkat <3 :P]

Oops sorry tht was a stupid question I got it

 

[Haya Ahmed]

What was the question ?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_12.pdf
Q7 :3

 

[Haya Ahmed]

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

Q10b and Q7 and 9 (iii) (please explain the domain and range thingy cuz i'm confused with them )


http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_11.pdf

Q3

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Q7 :3

U saw ms ?

 

[Thought blocker]

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

Q10b and Q7 and 9 (iii) (please explain the domain and range thingy cuz i'm confused with them )

This 7 is easy, see my steps THAT REALLY WORKS.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_11.pdf

Q3

Expand the brackets, and multiply it, then check the coff. of X^2, add them and as given in question keep it equal to 48 hence a = 3

 

[Namehere]

I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.

 

[Namehere]

I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.

Also, how do you do Q9(ii).

 

[Talha Khatri]

I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.

Someone definitely has to help boxxy out

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Q7 :3

I gave this paper last year.. that question killed it for me

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

Q10b and Q7 and 9 (iii) (please explain the domain and range thingy cuz i'm confused with them )

Q10b

Well I used another logic here which made this question very easy for me. If you look closely.. You'll see The shaded region consists of a triangle with height OB and base OC. and the curve with limits (A,0)

From Part i you have Co-ordinates of B = (0,1) and equation of BC = y=-x/2+1
Find Co-ordinates of C.. as C lies on the x-axis.. there substitute y=0. 0=-x/2+1 // x/2=1 // x=2 Therefore C= (2,0)
Co-ordinates of A are .. as A lies on the x-axis y=0 .. sub y=0 in Curve Equation
0=√(1+4x)
0^2=1+4x
1+4x=0
x=-1/4

The length OB = 1 and the Length OC = 2 since O is origin (0,0).

Area of Triangle = 1/2*1*2 = 1unit^2

Now Area under the curve is
⌡y
⌡√(1+4x)
⌡(1+4x)^0.5
Integrating gives..
|[(1+4x)^1.5]/1.5 * 4|
Putting limits -0.25 and 0. we get Area under the curve as 1/6

Now Area of Shaded region = Area of Triangle + Area under the curve.. which equal 1+1/6 = 7/6 Unit^2 Answer.

9iii

Just Remember this.. Domain is values of x.. and range is values of y or f(x) whatever you like to call it.

And also memorize another important rule. f^-1(x), the inverse, is rewriting the equation in terms of y instead of x. like y=5/(1-3x) becomes x=(y-5)/3y

So technically as you have inversed the position of x and y.. the range (y values) of f(x) becomes the domain (x values) of f^-1(x) and the domain (x values) of f(x) becomes the range (y-values) of f^-1(x).

 

[Haya Ahmed]

I gave this paper last year.. that question killed it for me



Q10b

Well I used another logic here which made this question very easy for me. If you look closely.. You'll see The shaded region consists of a triangle with height OB and base OC. and the curve with limits (A,0)

From Part i you have Co-ordinates of B = (0,1) and equation of BC = y=-x/2+1
Find Co-ordinates of C.. as C lies on the x-axis.. there substitute y=0. 0=-x/2+1 // x/2=1 // x=2 Therefore C= (2,0)
Co-ordinates of A are .. as A lies on the x-axis y=0 .. sub y=0 in Curve Equation
0=√(1+4x)
0^2=1+4x
1+4x=0
x=-1/4

The length OB = 1 and the Length OC = 2 since O is origin (0,0).

Area of Triangle = 1/2*1*2 = 1unit^2

Now Area under the curve is
⌡y
⌡√(1+4x)
⌡(1+4x)^0.5
Integrating gives..
|[(1+4x)^1.5]/1.5 * 4|
Putting limits -0.25 and 0. we get Area under the curve as 1/6

Now Area of Shaded region = Area of Triangle + Area under the curve.. which equal 1+1/6 = 7/6 Unit^2 Answer.

9iii

Just Remember this.. Domain is values of x.. and range is values of y or f(x) whatever you like to call it.

And also memorize another important rule. f^-1(x), the inverse, is rewriting the equation in terms of y instead of x. like y=5/(1-3x) becomes x=(y-5)/3y

So technically as you have inversed the position of x and y.. the range (y values) of f(x) becomes the domain (x values) of f^-1(x) and the domain (x values) of f(x) becomes the range (y-values) of f^-1(x).

In Q9 (iii) there is 0 included in the domain not exactly
and thanks for your effort .. appreciated ^_^

 

[psychiatrist]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_13.pdf

Question 1,3,5,7i),8,9iii) how is dy/dt to be found,10 PLEASE Thought blocker

 

[Ameena Eesa]

Hi does anyone know how to use the fx-991MS calci to calculate time and indices..................? plzzzz show, i get quite confused

 

[Ameena Eesa]

can u show how to open this up and simplify?

 

[Ameena Eesa]

Awesome12

 

[Ameena Eesa]

Help me with these, how do they look tooo......

(a)what are angle properties of irregular polygons?
(c) angles in the same segment are equal look like?............
(d) angles in opposite segments are
supplementary; cyclic quadrilaterals..... what type of segments......... i cant get how they look.......................

 

[TERMINATOR]

Could anyone help me with part (iii) of this COMPLEX no. question please.

Mark scheme

 

[CЯeScɘnt]

View attachment 41037

Could anyone help me with part (iii) of this COMPLEX no. question please.

Mark scheme

View attachment 41038

i cant read the question,attachment is not viewing full size.

 

[CЯeScɘnt]

View attachment 41037

Could anyone help me with part (iii) of this COMPLEX no. question please.

Mark scheme

View attachment 41038

if u cant, just post a link of this paper