Mathematics: Post your doubts here!

[kaushar]

ok,
7(i) first expand 'cos(x+45)'
cos(x+45) = cosxcos45 - sinxsin45 [<---hope you know the addition formula for cos(A +B)]
=(√2 /2)cosx - (√2 /2)sinx

so now we have to express "cos(x + 45) - (√2)sinx " in the form "Rcos(x + a)" [cos(x + 45) - (√2)sinx = Rcos(x + a)]

first i'll simplify the LHS (the one in blue):
cos(x + 45) - (√2)sinx = (√2 /2)cosx - (√2 /2)sinx - (√2)sinx
= (√2 /2)cosx - (3√2 /2)sinx

now i'm expanding the RHS(the one in green):
Rcos(x + a) = Rcosxcosa - Rsinxsina

i'll equate the RHS and the LHS :
Rcosxcosa - Rsinxsina = (√2 /2)cosx - (3√2 /2)sinx <---i can re-write this as:
(Rcosa)cosx - (Rsina)sinx = (√2 /2)cosx - (3√2 /2)sinx

now to find R:
R = √[(√2 /2)^2 + (-√3 /2)^2]
= 2.236

now to find the angle 'a' :
(Rcosa)cosx - (Rsina)sinx = (√2 /2)cosx - (3√2 /2)sinx
from the equation above^, notice the coefficients of 'cosx' and 'sinx' respectively
so ===> Rcosa = √2 /2 and Rsina= 3√2 /2
to find the angle, we need to use 'tan'
hence; tana = [3√2 /2] / [√2 /2] because tan = sin/cos
angle 'a' = 71.57

so u'll have: 2.236cos(x + 71.57) nd u've expressed it in the form they want
(btw the working is not this long)

7(ii) now we solve: cos(x + 45) - (√2)sinx = 2 [the RHS is equivalent to the answer we found in (i)]
so,
2.236cos(x + 71.57) = 2
cos (x + 71.57) = 0.8945 [to make my working easier, i let 'x + 71.57" = y]
==> cosy = 0.8945
y = 26.56
cos is positive in the first and fourth quadrant so,
y = 360 - 26.56 = 333.4
nd i also added 360 to 26.56 because the range of 'x' changed i.e 26.56 < x + 26.56 < 386.56
so another value of y= 386.56
now we've got y as 26.56, 333.4 and 386.56
so frm these we can find our 'x' where "x = y - 71.57"
===> x = 261.8 and 315.0

hope i explained properly nd didn't confuse u evn more

thnks.. i forgot to put the sin 45 and cos 45 to (√2 /2)

 

[JalalKaiser]

Can somebody seriously help with these, please? Especially question number 7.

Will someone please help me with no.6, 7ii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[Dudu]

Can somebody seriously help with these, please? Especially question number 7.

Tricky, ill try.
No answer either so i'm not sure if im right or not.

7)
So we found P=3 and and Q=2

Next its f(x)= 5 - x
Inverse is x = 5 - y
==> f^-1(x) = 5 - x

Now we need to find the domain for the inverse functions.
Range of f(x) = Domain of f^'1 (x)

5 - x for X > 3 (p =3, as we found this)

So 5 - 3 = 2
5 - 4 = 1

Range of f(x) is x < 2 and therefore Domain of Inverse is x < 2

Next one is f(x)= 11 - x^2
Inverse of this is x^2 = 11 - y
==> f^-1(x) = Square root of 11 - x

Again:
Range of f(x) = Domain of f^'1 (x)

11 - x^2 for 0 =< x =< 3 ( as p =3)
11 - 9 = 2
11 - 0 = 11

2 < x < 11 is Domain of inverse ( not really sure about this one)

 

[abdulrhaman munther]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
Guys can you please help me ... Question number 4 part (i) and (ii)
I appreciate your efforts :*

 

[$~SauD~$]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
Guys can you please help me ... Question number 4 part (i) and (ii)
I appreciate your efforts :*

Isn't paper 1 already over? :s

 

[Jaspreet Dhaliwal]

Isn't paper 1 already over? :s

4 i)
As you can see the graph of sin starts from 3 and goes to 9, that is amplitude so 9-3 = 6
For B there is a formula to use, which is 2pi/time period. (2pi divided by time period). You can see that the time period for one oscillation of the sin graph is 180 degrees which is also known as PI, so 2pi divided by PI is just 2, as pi cancel eachother out. So B is 2
For C the answer is 3, because it is the intercept, where it starts from, 3

ii) 6sin2x + 3 = 0
6sin2x = -3
sin2x= -1/2
find sinx, adn then divide by 2 to get your 2x values

 

[abdulrhaman munther]

4 i)
As you can see the graph of sin starts from 3 and goes to 9, that is amplitude so 9-3 = 6
For B there is a formula to use, which is 2pi/time period. (2pi divided by time period). You can see that the time period for one oscillation of the sin graph is 180 degrees which is also known as PI, so 2pi divided by PI is just 2, as pi cancel eachother out. So B is 2
For C the answer is 3, because it is the intercept, where it starts from, 3

ii) 6sin2x + 3 = 0
6sin2x = -3
sin2x= -1/2
find sinx, adn then divide by 2 to get your 2x values

Why is that that a= amplitude and b has a formula .. From where did you derive this formula? And what does A and B represent? I understood c but A and B

 

[Jaspreet Dhaliwal]

can someone urgently help with october november 2012 paper 11 question 7 ii) ? thanks exam is tommorow

 

[Jaspreet Dhaliwal]

can someone urgently help with october november 2012 paper 11 question 7 ii) ? thanks exam is tommorow

 

[GigsAlevel]

Hey, can you please help me with http://olevel.sourceforge.net/papers/9709/9709_s13_qp_33.pdf
No. 7 and No. 10, please. thank you

 

[Nahal88]

Please can you help me with Q 6 Paper 1 May/2013/9709/11 ?? It is urgent

 

[Syeeeer]

Can anyone explain to me when to use Fr = µR and Fr ≤ µR ?

 

[Avik181]

Can anyone help with 9709 november 11 paper 41 no.6 (ii)

 

[Munyaradzi Muchenje]

guys I need notes on kinetic energy and potential energy m1

 

[Baller23]

Hi Everyone!

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This page seems helpful for discussing A-Level problems too.
For those of you who are on Facebook do not hesitate to use this page.
Do post some of your problems there too!
Hope it helps you!
If it does share the page with your friends and let them benefit from it.

 

[Suchal Riaz]

Can anyone help with 9709 november 11 paper 41 no.6 (ii)

Please post link if paper to make it easy for us to answer ur questions

 

[Nahal88]

Can someone help me with question 6 May/June 2013 Paper 31

 

[Maz]

Could someone explain this to me? Please.

1) arg(1/3 -z)=pi/6 , what is arg(3z - 1) ?

Or similar....

 

[xino9512]

Hi I am doing my A2s and I can't do this question please help me!!
oct/nov 10 p.32 q.9
thanks in advance!

 

[Maz]

Can someone help me with question 6 May/June 2013 Paper 31

6 i) Seeing that PQ is perpendicular to the plane, PQ is therefore a normal to the plane.

A = midpoint of PQ= (OP+OQ)/2 . (7i+(-5i) +7j+j - 5+k)/2 = i +4j -2k
PQ = q-p = -5i -7i +j-7j +k+5k = -12i -6j+6k

r.n = a.n [equation of a plane]
(xi+yj+zk).(-12i -6j+6k) = (i +4j -2k).(-12i -6j+6k)
-12x-6y+6z = -48
12x+6y-6z = 48

ii) [this would be easier to solve if you make a diagram] [there are two methods of solving this, i'll be showing one]

First find "B"

Since PB is parallel to x-axis its directional vector is "i" = (1 ,0 ,0)
Therefore, The vector equation is=

r=(7i +7j -5k) + t(i)
r=((7+t i)+ 7j -5k)

Put this in the plane equation since PB is intersecting with the plane at B.
12(7+t) +6(7) -6(-5) = 48
t= -9
Put this in the earlier vector equation, the answer is the value of B = (-2,7,-5)

Use the distance equation to the the distance from A to B.