Mathematics: Post your doubts here!

[Binyamine]

Great thanks alot!
there is one question bothering me , its related to chapter FUNCTION
I don't know how to find domain & range and also stuck in the last part
I have uploaded the question and awating for your reply.(This question is from my school test paper)View attachment 19827

Have a look on the following videos.

Question 8 June 2012 Paper 1 9709/11/M/J/12
Question 8 :

Question 10 JUNE 2012 Paper 1 9709/12/M/J/12

Question 10 :

Question 11 JUNE 2012 Paper 1 9709/11/M/J/12

Question 11 :

Question 11 BES, 2013, Paper 1, specimen paper, cie, 9820


Question 11 :

 

[Yousif Mukkhtar]

Question 2.

When you differentiated you got dy/dx = 9 x^2 - 12 x + 4

What you should next do is to complete the square.

= 9 ( x^2 - 4/3 x ) + 4
= 9 [ ( x - 2/3 ) ^ 2 - 4/9 ] + 4
= 9 ( x - 2/3 ) ^ 2 - 4 + 4
= 9 ( x - 2/3 ) ^ 2

which will be positive for any value of x since when we square any number it become positive. And a positive number multiplied by 9 is still positive.

Question 7 (i)
You were on the right track until you made a division mistake.
x+3y+4x+y+3x+2y=48
8x+6y=48
6y=48-8x
y=(48-8x)/6

Divide every term by 2

y = ( 24 - 4 x ) / 3
or y = 8 - (4/3) x

7 (ii) Area = 3xy + 3 xy
= 6 xy
= 6 x ( (48-8x)/6 )
= x ( 48-8x)
= 48 x - 8 x^2

7(iii) THis you can do, you find dA/dx, and the value of x at stationary point. Find an expression for d A^2/ dx^2 and substitute the value of x for which we have a stationary point in it. You will see that d A^2/ dx^2 is negative, hence a maximum.

So y=8-(4/3)x is accepted even though it's not written in the marking scheme?

Anyways thanks man!

 

[Binyamine]

So y=8-(4/3)x is accepted even though it's not written in the marking scheme?

Anyways thanks man!

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

 

[kronix6]

Th

6 sin^2Q + 6 sinQ cosQ = sinQ cosQ - cos^2Q
6 sin^2Q + cos^2Q = -5 sinQ cos Q

(Divide by cos^2Q everywhere) : 6 tan^2Q + 1 = -5 tan Q

6 tan^2Q + 5 tanQ + 1 = 0
(3 tanQ + 1) (2 tanQ + 1) = 0
tan Q = (-1/3) or tan Q = (-1/2)

Hope It Helps

anks man . i appreciate it

 

[Minato112]

Th

anks man . i appreciate it

No Problem!

 

[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 )