Mathematics: Post your doubts here!

[littlecloud11]

Jazaki Allah khairan!! Thank you so much for your help! , Alhamdulilah now i get it! May Allah reward you. May Allah provide you with highest grades, in this life and hereafter...!! I think your way is easier than markscheme way!

Ameen. You're welcome and thanks for all the good wishes.

 

[minie23]

https://www.google.mu/#hl=en&sclien...94,d.bmk&fp=7f3494584588b9f5&biw=1280&bih=933

Please someone help me for no. 6(i) (iii) & no. 7 (iv)

 

[littlecloud11]

https://www.google.mu/#hl=en&sclient=psy-ab&q=papers.xtremepapers.com/CIE/.../9709_s06_qp_2.pdf &oq=papers.xtremepapers.com/CIE/.../9709_s06_qp_2.pdf &gs_l=hp.3...5115.5115.13.5287.1.0.1.0.0.0.0.0..0.0.les;..0.0...1c.1.5.hp.ylcKzbdOyfY&psj=1&bav=on.2,or.r_qf.&bvm=bv.43287494,d.bmk&fp=7f3494584588b9f5&biw=1280&bih=933

Please someone help me for no. 6(i) (iii) & no. 7 (iv)

6i) x = 9e^−2x
You can consider this as the intersection between the equation y=x and y= 9e^-2x. Draw the graphs and it should look something like this.


As the graphs intersect at only one point there is only one root.

iii) For this you can derive the iterative formula using the equation given in part (i). Just rearrange it.
x = 9e^−2x
introduce ln on both sides
ln x = ln (9e^−2x)
ln x = ln 9 + ln e^-2x [ln ab = ln a +ln b]
ln x = ln 9 -2x ln e
ln x = ln 9 - 2x
2x = ln 9- ln x
x = 1/2( ln 9- ln x) [shown]

7iv) From your answer to part (iii) you should get an improper fraction. 2x+ 1 is the quotient and the remainder is -3. Quotient + reminder/dividend should give you the original exuation. So you can write this as 2x+1 + (-3)/2x+3 .
Now integrate this-
=∫ 2x +1 - 3/2x+ 3 (limit 3 and -1)
=∫ x^2 + x - (3/2) *(2/2x+3) [Take 3/2 common for the second part so that the differentiation of the denominator is present in the numerator, this allows you to use the formula a/ax+b as ln |ax+b|]
=[ x^2 + x - 3/2 ln|2x+3| ] limit 3 , -1
use the limits and you get-
=[( 9+ 3 - 3/2 ln 9) - ( 0)]
=12 - 3/2 ln 9 = 12 - 3/2 ln 3^2 = 12 - (3/2) * 2 *ln 3
= 12 - 3 ln 3

 

[minie23]

6i) x = 9e^−2x
You can consider this as the intersection between the equation y=x and y= 9e^-2x. Draw the graphs and it should look something like this.
View attachment 21971

As the graphs intersect at only one point there is only one root.

iii) For this you can derive the iterative formula using the equation given in part (i). Just rearrange it.
x = 9e^−2x
introduce ln on both sides
ln x = ln (9e^−2x)
ln x = ln 9 + ln e^-2x [ln ab = ln a +ln b]
ln x = ln 9 -2x ln e
ln x = ln 9 - 2x
2x = ln 9- ln x
x = 1/2( ln 9- ln x) [shown]

7iv) From your answer to part (iii) you should get an improper fraction. 2x+ 1 is the quotient and the remainder is -3. Quotient + reminder/dividend should give you the original exuation. So you can write this as 2x+1 + (-3)/2x+3 .
Now integrate this-
=∫ 2x +1 - 3/2x+ 3 (limit 3 and -1)
=∫x^2 + x - (3/2) *(2/2x+3) [Take 3/2 common for the second part so that the differentiation of the denominator is present in the numerator, this allows you to use the formula a/ax+b as ln |ax+b|]
=[ x^2 + x - 3/2 ln|2x+3| ] limit 3 , -1
use the limits and you get-
=[( 9+ 3 - 3/2 ln 9) - ( 0)]
=12 - 3/2 ln 9 = 12 - 3/2 ln 3^2 = 12 - (3/2) * 2 *ln 3
= 12 - 3 ln 3

Thank you so much mate
But I still can't understand how to sketch the graph for 6(i) ?
How to get the points to plot,etc.. ?

 

[littlecloud11]

Thank you so much mate
But I still can't understand how to sketch the graph for 6(i) ?
How to get the points to plot,etc.. ?

Well since the question asked you to sketch you don't have to be dead accurate with the graph but just get the basic shape right. You know how the graph of y=x looks, well in the same was y= e^-x has a basic shape. Let me elaborate. The first derivative of y= e^-x is -e^-x, so the curve always has a decreasing gradient. As x goes to minus infinity e^-x goes to positive infinity. As x goes to plus infinity, e^-x goes to 0 (so the x-axis has a horizontal asymptote). The graph y= e^-x always has the shape as below-

Whenever there is a coefficient before e^-x (like 9 in the question) the only thing that changes is the intersection at the y-axis. The coefficient of e^-x is the value of y when x=0, so only the red point in the diagram above changes. And no matter what the coefficient of the power is (like 2 in the question) the basic shape always stays the same. So just keep the shape in mind.

(If nothing helps, just remember that y= anything* e^-anything x is just a reflection of y= e^x on the y-axis )

 

[minie23]

Well since the question asked you to sketch you don't have to be dead accurate with the graph but just get the basic shape right. You know how the graph of y=x looks, well in the same was y= e^-x has a basic shape. Let me elaborate. The first derivative of y= e^-x is -e^-x, so the curve always has a decreasing gradient. As x goes to minus infinity e^-x goes to positive infinity. As x goes to plus infinity, e^-x goes to 0 (so the x-axis has a horizontal asymptote). The graph y= e^-x always has the shape as below-
View attachment 21972
Whenever there is a coefficient before e^-x (like 9 in the question) the only thing that changes is the intersection at the y-axis. The coefficient of e^-x is the value of y when x=0, so only the red point in the diagram above changes. And no matter what the coefficient of the power is (like 2 in the question) the basic shape always stays the same. So just keep the shape in mind.

(If nothing helps, just remember that y= anything* e^-anything x is just a reflection of y= e^x on the y-axis )

Wow ! Thank you soo much !
You are a genius !

 

[littlecloud11]

Wow ! Thank you soo much !
You are a genius !

No problem!

 

[Kumkum]

AOA wr wb
Can someone pls help me with question 2
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

 

[PhyZac]

AOA wr wb
Can someone pls help me with question 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Waalikum AssalamWa Rahmatullahi Wa Barakatoho
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-1.php
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-2.php
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-3.php
These will help In Sha Allah.

 

[Kumkum]

Waalikum AssalamWa Rahmatullahi Wa Barakatoho
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-1.php
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-2.php
http://www.examsolutions.net/maths-...functions/modulus/inequalities/tutorial-3.php
These will help In Sha Allah.

Jazakallah Khair
I do know how to solve. one of my answers were not correct so I wanted to see how u do it and why one solution is out
I got the value of x as -1 or 1/3 (i.e if I factorised it correctly) and in the mark scheme the only answer is x>1/3 not x<-1 or x>1/3 (this is what I got)
m.s: http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_ms_3.pdf

 

[PhyZac]

Jazakallah Khair
I do know how to solve. one of my answers were not correct so I wanted to see how u do it and why one solution is out
I got the value of x as -1 or 1/3 (i.e if I factorised it correctly) and in the mark scheme the only answer is x>1/3 not x<-1 or x>1/3 (this is what I got)
m.s: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_ms_3.pdf

You did the squaring method then ( i use graphical)...well i use the graphical method, am not good in this paper, but thought of sharing the videos..! In Shaa Allah someone will explain to u.

 

[Kumkum]

You did the squaring method then ( i use graphical)...well i use the graphical method, am not good in this paper, but thought of sharing the videos..! In Shaa Allah someone will explain to u.

oh ok....not a problem
Jazakallah Khair for the videos

 

[PhyZac]

Jazakallah Khair
I do know how to solve. one of my answers were not correct so I wanted to see how u do it and why one solution is out
I got the value of x as -1 or 1/3 (i.e if I factorised it correctly) and in the mark scheme the only answer is x>1/3 not x<-1 or x>1/3 (this is what I got)
m.s: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_ms_3.pdf

Assalamu Alikum..!!
I got it. See, ur answer is for sure wrong, the inequality says 2x > |x−1| so x can not be a negative number (because modulus are always positive)...thus x<-1 is wrong !!
and thus only x>1/3 is right!

 

[unseen95]

can anyone help me solve this question with full explanation. Thanks.

 

[unseen95]

please help me with b(ii), with explanation
please help

 

[asexamskillme111]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

question 7 part (iii)

i dont know how to go about solving the question. help pleaase

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

question 7 part (iii)

i dont know how to go about solving the question. help pleaase

7iii)
As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7

 

[Sarkerms]

Ok i am seriously stuck in this question and some help the marking scheme turned out to be. Can anyone help me solve the following question.

 

[minie23]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Please help me for no. 5

 

[Ahmedraza73]

How to find MAX OR MIN point or Stationary points in differntiation of AS level?
Please help me