Mathematics: Post your doubts here!

[saleemsamer]

Hello can some1 help me with m/j 2011 p11 q11 iv (alot of 11 )

 

[rz123]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w04_qp_1.pdf
Q. 6 iii part kindly explain me how to calculate range of these kind of functions.

 

[CaptainDanger]

M1 question...!
A car comes to a stop from a speed of 30ms in a distance of 804m. The driver breaks as to produce a deceleration of 1/2ms^2 to begin with, and then breaks harder to produce a deceleration of 3/2ms^2. Find the speed of the car the instant when the deceleration is increased, and the total time the car takes to stop. Answers are 12 m/ s and 44s...

 

[Prithvi Rajan]

October/November 2008 paper 03 question 08 please solve ASAP

 

[CaptainDanger]

M1 question...!
A car comes to a stop from a speed of 30ms in a distance of 804m. The driver breaks as to produce a deceleration of 1/2ms^2 to begin with, and then breaks harder to produce a deceleration of 3/2ms^2. Find the speed of the car the instant when the deceleration is increased, and the total time the car takes to stop. Answers are 12 m/ s and 44s...

Never mind. Managed to solve it at last!

 

[Silent Hunter]

want help in Q3 (ii) in this one http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s09_qp_1.pdf

 

[smzimran]

want help in Q3 (ii) in this one http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s09_qp_1.pdf

Write down the coefficient of x2 in terms of a by expanding the expression. (Use your answer in part i)
Then equate the coefficient of x2 = 0

 

[Sabcore]

Okay I managed to write a question in a nice style format in Photoshop. Someone please answer it, I can't solve parts (b) and (c). [If you can't see the image, simply right click on it and view image]


I tried the [Similar Triangle] idea on the part (b) to find the length of arc but the answer was not good. Oh well. Someone please help. I got no idea how to solve this problem.

Thanks.

 

[Just call me MJ]

I tried the [Similar Triangle] idea on the part (b) to find the length of arc but the answer was not good. Oh well. Someone please help. I got no idea how to solve this problem.

Thanks.[/quote]

I think in part (b), they are asking you for the length of the arc, they want the length of the elastic band in contact with ball A, so you have to find the angle that includes A in the middle of 2 point on the circumference (i.e. the part in contact with the elastic band) and use the equation (length of arc = rO)
Now go ahead and try this method, if you still don't get it just say the word and I'll solve it for you

 

[steel]

y=e^x-8e^-2x
y=0
how to change to ln form !

 

[ffaadyy]

y=e^x-8e^-2x
y=0
how to change to ln form !

y = e^x - 8e^-2x
y = 0
0 = e^x - 8e^-2x
8e^-2x = e^x
8 = (e^x)/(e^-2x)
8 = e^3x
ln 8 = ln e^3x
ln 8 = 3x
0.69 = x

 

[leadingguy]

integrate the following!
/coshx.sinh2xdx
as by my opinion it should be integrated by the IDENTITY [sin(A+B)+sin(A-B)=2sinA.cosB]
though I did so! but no achievements!!!!!!!(((((
look here;
1/2{sin(hx+2hx)+sin(hx-2hx)} opening it we gt this..
1/2{sin(3hx)+sin(-hx)} now by the law of [sin(-a= -sin(a
we get 1/2{sin(3hx)-sin(hx)} now integrate this simplified eq.
the ans. should be (2cosh3x/3 +k) here in this cosh3x, 3 is the cube (power of h, not the quofecient)

 

[leadingguy]

integrate this! /{x2(1-x)1/2}dx here this small no.'s i.e 2,3,1 represents power, they are not the multipling quofecients after complete integration put boundries; (1,0) then the ans should be (16/105) or (0.15238095)

 

[usmiunique]

integrate the following!
/coshx.sinh2xdx
as by my opinion it should be integrated by the IDENTITY [sin(A+B)+sin(A-B)=2sinA.cosB]
though I did so! but no achievements!!!!!!!(((((
look here;
1/2{sin(hx+2hx)+sin(hx-2hx)} opening it we gt this..
1/2{sin(3hx)+sin(-hx)} now by the law of [sin(-a= -sin(a
we get 1/2{sin(3hx)-sin(hx)} now integrate this simplified eq.
the ans. should be (2cosh3x/3 +k) here in this cosh3x, 3 is the cube (power of h, not the quofecient)

sinh cosh and tanh are not in the A Level syllabus as far as i know.. only the basic trig. functions like sin cos and tan and their inverse(cosec sec and cot).

 

[rz123]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s03_qp_1.pdf
need help in Q.9 iii part

 

[ffaadyy]

integrate this! /{x2(1-x)1/2}dx here this small no.'s i.e 2,3,1 represents power, they are not the multipling quofecients after complete integration put boundries; (1,0) then the ans should be (16/105) or (0.15238095)

In this question, we'll be needing to take the substitution 'x = 1 - u' but then again such questions don't come in the A Level P3's in which we have to assume a substitution ourselves. Still I 'll do it for you, here's how its done:

(x^2) * [( 1 - x )^1/2] dx

Take the substitution 'x = 1 - u'

[( 1 - u )^2] * {[ 1 - ( 1 - u ) ]^1/2} dx

Now we have to eliminate 'dx'

x = 1 - u
dx = - du

Simplify the equation '[( 1 - u )^2] * {[ 1 - ( 1 - u ) ]^1/2}' further by opening the brackets.

( 1 - 2u + u^2 ) * ( 1 - 1 + u )^(1/2) -du
( 1 - 2u + u^2 ) * ( u^1/2 ) -du
[ ( u^(1/2) - 2u^(3/2) + u^(5/2) ] -du

There's a negative sign with 'du' therefore we'll multiply the bracket with it.

[ ( - u^(1/2) + 2u^(3/2) - u^(5/2) ] du

The limits which have been provided to us are of 'x' therefore we have to find their corresponding 'u' limits.

x = 1 - u
x = 1 (upper limit)
1 = 1 - u
u = 0

x = 1 - u
x = 0 (lower limit)
0 = 1 - u
u = 1

Now the problem is that our upper limit is smaller than our lower limit. We need to do something so that 1 (lower limit) becomes the upper limit and 0 (upper limit) becomes the lower limit. This can be done by multiplying the whole equation by '-'.

- * [ ( - u^(1/2) + 2u^(3/2) - u^(5/2) ] du
[ ( u^(1/2) - 2u^(3/2) + u^(5/2) ] du

Next, we'll integrate this equation.

[ ( u^(1/2) - 2u^(3/2) + u^(5/2) ] du
[ ( 2u^(3/2)/3) - ( 4u^(5/2)/5 ) + ( 2u^(7/2)/7 ) ]

Insert the limits 1 (upper) and 0 (lower).

( 2/3 ) - ( 4/5 ) + ( 2/7) - (0)
16/105

Therefore, the final answer is '16/105' or 0.15238..

 

[ffaadyy]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s03_qp_1.pdf
need help in Q.9 iii part

The triangle ABC consists of the sides AB, AC and BC. We know that AC = 16cm and that θ = 1/3π. therefore, using these details, we'll calculate the length of sides AB and BC & then the perimeter of the triangle. To find the length of AB, we'll use the cosine rule.

AB^2 = OA^2 + OB^2 - 2 (OA) (OB) (1π/3)
AB^2 = (8)^2 + (8)^2 - 2 (8) (8) (cos 1π/3)
AB = 8cm

Now we'll find the length of the side BC.

BC^2 = AB^2 + AC^2 - 2 (AB) (AC) (cos 1π/3)
BC^2 = 8^2 + 16^2 - 2 (8) (16) (cos 1π/3)

BC = 13.856.. or 8√3

Therefore, the perimeter is:

AB + BC + AC
= (24 + 8√3) cm or 37.856.. cm

 

[rideronthestorm]

I need help on paper 1 .may june 2009 question 4 ... how do u get the values of a b and c ? ...

 

[Just call me MJ]

A= 6 ( amplitude)
B= 2 ( 2 λ)
C= 3 ( the wave starts at 3)

If you are not sure if this is correct or not then you can prove it mathematically:

Minimum value of sin(bx) = 0
therefore a x sin(bx) = 0 , this happens when sin(xb) is sin(π), thus from graph, when sin(xb) is sin(π), y = 3, thus C= 3

Maximum value of sin(bx)=1
Therefore a x sin(bx) = a , this happens when sin(xb) is sin(π/2), thus
( a + 3 ‘which is C’ = Y) , when sin(xb) is sin(π/2), ( maximum value of Y)Y = 9
Thus, a + 3 = 9, which results in A=6

B=2 that’s just standard, you have to learn it the way it is

 

[Just call me MJ]

I forgot to post the answer as a reply to you. sorry:S but this is an alert that I posted the answer to your question if you take physics this should be a little easier ​