Mathematics: Post your doubts here!

[Jaf]

question asks for rate of change of water level means height..my ans is 1.19 but the book says 4.77

That's what I am getting too: 1.194 cm/s.
What book is this question in? The answer may be wrong.

 

[wahaajsid]

please solve Q10 of nov 11 p32... HELP NEEDED!

 

[smzimran]

can anybody give me the link to download 'list of formulae(mf9) ?? please !

http://www.xtremepapers.com/papers/...d AS Level/Mathematics (9709)/9709_y12_sy.pdf
Pg. 31 - pg.35

 

[Esme]

thanks a lot.
so all these will be provided to us during the exam ??

 

[smzimran]

thanks a lot.
so all these will be provided to us during the exam ??

Yes!

 

[anonymous123]

That's what I am getting too: 1.194 cm/s.
What book is this question in? The answer may be wrong.

http://books.google.com.pk/books?id...AA#v=onepage&q="Ap is double the 5th"&f=false
it doesnt show some pages coz its preview

 

[ffaadyy]

please solve Q10 of nov 11 p32... HELP NEEDED!

 

[ffaadyy]

please help

june 2006 p3 q 7 (iii)

arg (u/u*) = 2 arg (u)

arg (u/u*):

u/u* = 3/5 + 4i/5

arg (u/u*) = tan^-1 [(4/5)/(3/5)]
arg (u/u*) = tan^-1 (4/3)

arg u:

u = 2 + 1

arg u = tan^-1 (1/2)

arg (u/u*) = 2 arg (u)
tan^-1 (4/3) = 2 tan^-1 (1/2)

Hence, proved.

 

[Jaf]

http://books.google.com.pk/books?id=e3XhP_-1B_UC&pg=PA309&lpg=PA309&dq="Ap is double the 5th"&source=bl&ots=0YUs7HmhMk&sig=TiooH63EveQyfMux7VTZkdsRgJk&hl=en&ei=UJ_fTrWFI4bCswb5wPCJCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBkQ6AEwAA#v=onepage&q="Ap is double the 5th"&f=false
it doesnt show some pages coz its preview

Would you know if the answers in the book are normally reliable? Because I used two ways to find the answer and both ways gave me the same answer.
The first way is obviously using chain rule. The second way is by the following formula:
Rate of flow of liquid = velocity at which liquid is travelling x cross-sectional area of container through which it is flowing.
This also gives the same answer.

So unless we're missing something, it is likely that 4.77 is not the answer.

 

[Rabib_69]

Hello can anyone help me with this?

Question no 3(ii) and question no 8(ii)

 

[ffaadyy]

Hello can anyone help me with this?

Question no 3(ii) and question no 8(ii)

Q3(i):

AB = OB - OA
AB = (2)(-2)(11) - (-1)(2)(5)
AB = (3)(-4)(6)

'(3)(-4)(6)' is the direction of the plane.

3x - 4y + 6z = k

The plane passes through 'b' therefore you can find the value of 'k' using the coordinates of 'b'.

3x - 4y + 6z = k
3(2) - 4(-2) + 6(11) = k
80 = k

Put back the value 'k' in the plane equation.

3x - 4y + 6z = k
3x - 4y + 6z = 80

Therefore, the plane equation is '3x - 4y + 6z = 80'.

q3(ii):

(0) x (3)
(1) x (-4) = (1) (61)^(1/2) cos x
(0) x (6)

cos x = (-4)/[((61)^(1/2)]
x = 120.8°

120.8° - 90°
30.8°

 

[rideronthestorm]

p1 question from november 2011 varient 1.. : A curve has equation y=3x^2-6x^2+4x+2 . Show that the gradient of the curve is never negativc .................................. How do u do this , i get the differentiation part but what after that , how do u prove its never negative :S ?

 

[ffaadyy]

Hello can anyone help me with this?

Question no 3(ii) and question no 8(ii)

Q8(ii):

First of all, we'll draw an argand diagram representing the inequality 'arg(z − u) = (pi)/4'. Next, we'll draw a perpendicular bisector starting from the origin to the line of angle so that it cuts the line of angle at 90°. This is how the diagram will look like (the red lines represent this part of the question)



The perpendicular bisector is actually he least value of |z| and this is what we have to find out. Having created a right angle triangle, we can find out the value of this perpendicular bisector using the pythagoras theorem.

3^2 = x^2 + x^2
9 = 2x^2
4.5 = x^2
2.12=x

Therefore, the least value of |z| is 2.12.

 

[Rabib_69]

Q8(ii):

First of all, we'll draw an argand diagram representing the inequality 'arg(z − u) = (pi)/4'. Next, we'll draw a perpendicular bisector starting from the origin to the line of angle so that it cuts the line of angle at 90°. This is how the diagram will look like (the red lines represent this part of the question)

View attachment 6552

The perpendicular bisector is actually he least value of |z| and this is what we have to find out. Having created a right angle triangle, we can find out the value of this perpendicular bisector using the pythagoras theorem.

3^2 = x^2 + x^2
9 = 2x^2
4.5 = x^2
2.12=x

Therefore, the least value of |z| is 2.12.

Ahh!!! sweeet!! Thanks man

 

[ffaadyy]

p1 question from november 2011 varient 1.. : A curve has equation y=3x^2-6x^2+4x+2 . Show that the gradient of the curve is never negativc .................................. How do u do this , i get the differentiation part but what after that , how do u prove its never negative :S ?

y = 3x^3 - 6z^2 + 4x + 2

Differentiate with respect to 'x'.

dy/dx = 9x^2 -12x + 4
dy/dx = 9x^2 - 6x - 6x + 4
dy/dx = 3x (3x - 2) - 2 (3x - 2)
dy/dx = (3x-2)^2

No matter what value of 'x' you put in ' (3x-2)^2 ', the gradient will always be positive.

 

[rideronthestorm]

but why are we simplifying into THAT square form specifically ? ...

 

[rideronthestorm]

f(x) = 2x^2 -8x + 10 ..... Find an expression for F^-1(x) ....

 

[anonymous123]

Would you know if the answers in the book are normally reliable? Because I used two ways to find the answer and both ways gave me the same answer.
The first way is obviously using chain rule. The second way is by the following formula:
Rate of flow of liquid = velocity at which liquid is travelling x cross-sectional area of container through which it is flowing.
This also gives the same answer.

So unless we're missing something, it is likely that 4.77 is not the answer.

yea maybe the book is wrong..

 

[SalmanPakRocks]

f(x) = 2x^2 -8x + 10 ..... Find an expression for F^-1(x) ....

It's easy convert it into the general form
2(x^2 - 8x +10 )
2( (x)^2 -2(x)(4) + (4)^2 - (4)^2 + 10)
2( (x -4)^2 -16 + 10)
2(x-4)^2 -12
y = 2(x-4)^2 -12
y + 12/2 = (x-4)^2
underoot +-(y + 12)/2 +4 = x
F^-1(x) = +-(x + 12)/2 +4

 

[anonymous123]

when a question says a particle is projected and doesnt mention any force.....i have to ignore the force too right?? and what about the initial velocity?? HELP!!