Mathematics: Post your doubts here!

[Yousif Mukkhtar]

it should be accepted.
Thanks Allah for indeed he is our Creator and we need to be grateful to him.

Al hamdulilah.
Can someone explain this:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf
q(iii) I got gf(x) as 4x^2-9
Then 4x^2-9<=16
By using the quadratic formula we get x as -5/2 or 5/2
Then the values of x should -5/2<=x<=5/2

But why is it written -5/2<=x<=0 in the marking scheme?

 

[Binyamine]

Al hamdulilah.
Can someone explain this:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_13.pdf
q(iii) I got gf(x) as 4x^2-9
Then 4x^2-9<=16
By using the quadratic formula we get x as -5/2 or 5/2
Then the values of x should -5/2<=x<=5/2

But why is it written -5/2<=x<=0 in the marking scheme?

Because the domain of f(x) is less or equal to zero. Read the question well.

 

[Yousif Mukkhtar]

Because the domain of f(x) is less or equal to zero. Read the question well.

Thanks. I got it because no number above or equal to 0 is accepted.

 

[Binyamine]

Thanks. I got it because no number above or equal to 0 is accepted.

Alhamdulillah.

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w05_qp_3.pdf

Asalamoalikum.... just cant get the A in Question 2 in this paper. Anybody ? Thank you Binyamine or anyone?

 

[Binyamine]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_3.pdf

Asalamoalikum.... just cant get the A in Question 2 in this paper. Anybody ? Thank you Binyamine or anyone?

Lol, you tagged me.
First Fact : I sat for this paper as a student for A Level and i so much enjoyed life as a college student. Miss these moments and my friends...Why did i grow up so soon...???lol.

To Proceed;

y = A x^n
put log or ln on both sides

ln y = ln ( A x^n )

ln y = ln A + ln ( x^n)
ln y = n ln x + ln A

Comparing the above to y = mx + c
we see that n represents the gradient while ln A represents the intercept of the y-axis.

so join the crosses by a straigt line.
you will see that the y intercept is 0.7

so ln A = 0.7
A = e(0.7) = Evaluate this my dear

And to find n, you just take any two points on the graphs lying on the straight line and find the gradient.

 

[Binyamine]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_3.pdf

Asalamoalikum.... just cant get the A in Question 2 in this paper. Anybody ? Thank you Binyamine or anyone?

Ahhh i forgot to tell you, about your signature. We Muslims, we smile because smile is a charity. When we smile we earn good deeds. Subhanallah.And we smile to seek Allah's pleasure insha'Allah.

 

[PhyZac]

Assalamu Alikum Wa Rahatullahi Wa Barakatooho.....@Minato112

This is the full question...my problem lies on part (ii) and can i see how the graph looks like of part (iii)

For mark-scheme answer refer the picture below.. And Jazakum Allah Khairan!

 

[Silent Hunter]

Lol, you tagged me.
First Fact : I sat for this paper as a student for A Level and i so much enjoyed life as a college student. Miss these moments and my friends...Why did i grow up so soon...???lol.

To Proceed;

y = A x^n
put log or ln on both sides

ln y = ln ( A x^n )

ln y = ln A + ln ( x^n)
ln y = n ln x + ln A

Comparing the above to y = mx + c
we see that n represents the gradient while ln A represents the intercept of the y-axis.

so join the crosses by a straigt line.
you will see that the y intercept is 0.7

so ln A = 0.7
A = e(0.7) = Evaluate this my dear

And to find n, you just take any two points on the graphs lying on the straight line and find the gradient.

JazakAllah ..... thanks alot .... and any tips on doing the proving thing in the trigonometry questions?

 

[kronix6]

Help needed.
someone integrate this plz

y=2/1-sin(x)

 

[kronix6]

Need help in this diffrential equation i cant get it in terms of x. plz help

question is : solve the differential equation: dx/dt=kx(a-x) (0<x<a)

where k and a are positive constants , given x=a/2 when t=o. express x in terms of k,a and t in your answer.

 

[Binyamine]

Help needed.
someone integrate this plz

y=2/1-sin(x)

Are you sure of the question?

y = 2 [( 1 - sinx ) ^ -1]

Integrating will give us = -2 sec x ln ( 1 - sin x )

But i am not too sure of this answer that is the reason of my first question. May be someone else should confirm it...

 

[Binyamine]

Need help in this diffrential equation i cant get it in terms of x. plz help

question is : solve the differential equation: dx/dt=kx(a-x) (0<x<a)

where k and a are positive constants , given x=a/2 when t=o. express x in terms of k,a and t in your answer.

send all x to dx and the others to dt such that

dx / [ x ( a-x ) ] = k dt

we then integrate both sides.But first We see that we will have to first do partial with 1/ [ x ( a-x ) ]

1 / [ x ( a-x ) ] = A /x + B/(a-x)

1 = A ( a-x ) + B x

Let x = 0 ; Let x = a
A = 1/a ; B = 1/a

Therefore ; [ (1/a) / x + (1/a) / ( a-x) ] dx = k dt

we then integrate both sides

(1/a) ln x - (1/a) ln ( a-x ) = kt + C ; where C is a constant

(1/a) ln [ x/ (a-x) ] = kt + C

ln [ x/ (a-x) ] = akt + B ; where B is a constant

x/ (a-x) = e ^ ( akt + B )

x/ (a-x) = A e^( akt) where A is a constant

x=a/2 when t=o ; we use this information to find the value of A

(a/2) / ( a/2) = A . 1
A = 1

Hence x/ (a-x) = e^( akt)

cross multiply and make x subject of formula.

x = a e^( akt) - x e^( akt)

x + x e^( akt) = e^( akt)

x ( 1 + e^( akt) ) = e^( akt)

x = e^( akt) / ( 1 + e^( akt) )

 

[hussamh10]

help in Differentiation of

ln(sinx/cosx-1)

thanx alot

 

[Binyamine]

help in Differentiation of

ln(sinx/cosx-1)

thanx alot

y = ln(sinx/cosx-1)

recall that ln (a/b) = ln a - ln b

so we rewrite y = ln sinx - ln ( cosx - 1 )

recall d/dx [ln ( f(x)) ] = f ' (x) / f (x)

hence dy/dx = cosx/sinx - ( - sin x / ( cosx -1 ) )

= cot x + sin x / ( cosx -1 )

 

[Yousif Mukkhtar]

Got one quick question guys
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q8) ii)

Shouldn't the range be f(x)>=-4+k?
Because
(x-2)^2>=0 (square an expression is always a positive number)
Then
(x-2)^2-4+k>=-4+k
Therefore f(x)>=-4+k

But in the marking scheme it is > not equal to.

 

[Binyamine]

Got one quick question guys
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Q8) ii)

Shouldn't the range be f(x)>=-4+k?
Because
(x-2)^2>=0 (square an expression is always a positive number)
Then
(x-2)^2-4+k>=-4+k
Therefore f(x)>=-4+k

But in the marking scheme it is > not equal to.

You are right according to me.

Here is a video for the correction of the number ::

 

[Yousif Mukkhtar]

You are right according to me.

Here is a video for the correction of the number ::

Thanks man.

 

[Binyamine]

View attachment 19870
Assalamu Alikum Wa Rahatullahi Wa Barakatooho.....@Minato112

This is the full question...my problem lies on part (ii) and can i see how the graph looks like of part (iii)

For mark-scheme answer refer the picture below.. And Jazakum Allah Khairan!
View attachment 19871

(i)

a = -d ; u = 1.4 ; t = 1.2 and v = 1.1

we first find the value of a, the deceleration
v = u + at
a = ( v - u ) / t
= ( 1.1- 1.4 / 1.2
= -0.25

Hence d= 0.25
Distance ; s = (v^2 - u^2)/2 a
= ( (1.1)^2 - (1.4)^2 ) / (2 * -0.25)
= 1.5 m

(ii) Now the ball rebounds and move towards A, and while moving towards A, it stops at C i.e velocity is zero at C

v = 0 ; s = 2 ; a = -0.25

let us first find the initial speed that is the speed with which it rebounds from B towards A

u = square root of ( v^2 - 2 a s )
=square root [ (0)^2 - 2 ( -0.25) ( 2 ) ]
= 1

now we can find the time by using v = u + at

t = (v - u)/ a
= ( 0 - 1 ) / (-0.25)
= 4 seconds.

(iii) I am sure that you have understood how the graph should be. Just to confirm ;
the lines should be straight lines for they are of constant acceleration.

When t=0, velocity was 1.4 hence (0, 1.4) ;
when t=1.2 s, velocity was 1.1 hence (1.2, 1.1) ;

You join these two coordinates.

Now when t = 1.2 ; the ball rebounds at B and starts moving in the opposite direction towards A and comes at rest at C
We calculated the velocity of rebound as being 1 m/s. But since it is moving in the opposite direction; the velocity is infact -1 m/s. recall that velocity is a vector quantity.

So the coordinates (1.2, -1 )
and it comes to rest after a further 4 seconds. Velocity is Zero and the time since the beginning of the experiment is 1.2 + 4 = 5.2

hence coordinates ( 5.2, 0 )
Join these two points by a straight line.

 

[PhyZac]

(i)

a = -d ; u = 1.4 ; t = 1.2 and v = 1.1

we first find the value of a, the deceleration
v = u + at
a = ( v - u ) / t
= ( 1.1- 1.4 / 1.2
= -0.25

Hence d= 0.25
Distance ; s = (v^2 - u^2)/2 a
= ( (1.1)^2 - (1.4)^2 ) / (2 * -0.25)
= 1.5 m

(ii) Now the ball rebounds and move towards A, and while moving towards A, it stops at C i.e velocity is zero at C

v = 0 ; s = 2 ; a = -0.25

let us first find the initial speed that is the speed with which it rebounds from B towards A

u = square root of ( v^2 - 2 a s )
=square root [ (0)^2 - 2 ( -0.25) ( 2 ) ]
= 1

now we can find the time by using v = u + at

t = (v - u)/ a
= ( 0 - 1 ) / (-0.25)
= 4 seconds.

(iii) I am sure that you have understood how the graph should be. Just to confirm ;
the lines should be straight lines for they are of constant acceleration.

When t=0, velocity was 1.4 hence (0, 1.4) ;
when t=1.2 s, velocity was 1.1 hence (1.2, 1.1) ;

You join these two coordinates.

Now when t = 1.2 ; the ball rebounds at B and starts moving in the opposite direction towards A and comes at rest at C
We calculated the velocity of rebound as being 1 m/s. But since it is moving in the opposite direction; the velocity is infact -1 m/s. recall that velocity is a vector quantity.

So the coordinates (1.2, -1 )
and it comes to rest after a further 4 seconds. Velocity is Zero and the time since the beginning of the experiment is 1.2 + 4 = 5.2

hence coordinates ( 5.2, 0 )
Join these two points by a straight line.

Jazaka Allahu khairan...!!!!! Yes i now understand the question...! And yes i got the look of the graph....!! Thank you so much, Sir. !!!!!! May Allah reward you for the help u gave me and other students Ameen.