Mathematics: Post your doubts here!

[Tkp]

Asslamu Alikm waRahmatullahi Wa Barakatoho
...Can anyone show how to solve this question...?

The answer is
106.3 is the angle
and component bit is 7.2N

R^2=P^2+Q^2+2PQCOS theta
u will get the angle
i have prblm in the 2nd 1.like the ans would be 10+10 costheta.but y 10 iis added here.they told to find the vector in the direction of OA

 

[PhyZac]

R^2=P^2+Q^2+2PQCOS theta
u will get the angle
i have prblm in the 2nd 1.like the ans would be 10+10 costheta.but y 10 iis added here.they told to find the vector in the direction of OA

What is R? P? Q?

Second is fine....
u find R(>) tht is 10 - 10 cos (180-theta) or 12 cos (theta/2)

 

[littlecloud11]

Asslamu Alikm waRahmatullahi Wa Barakatoho littlecloud11 Dug
...Can anyone show how to solve this question...?

The answer is
106.3 is the angle
and component bit is 7.2N

There might be an easier way to solve this but this is how i did it.


Split the Left hand 10N force into its vertical and horizontal components by breaking θ into 90 degree and (θ-90).
Then resolve the vertical and horizontal component.
There is only one vertical component 10 Cos (θ-90) which is to 10Sinθ
The horizontal component would be 10- 10 sin(θ-90) or 10-10 Cosθ
Now consider these the vertices of a right angled triangle with the hypotenuse as 12N.
So, 12^2 =(10 Sinθ)^2 + (10-10Cosθ)^2
144 = 100 sin^2θ + 100 - 200 Cosθ +100 Cos^2θ
144 =100 - 200 Cosθ + 100 (sin^2θ + Cos^2θ)
144 = 100 - 200 Cosθ + 100
Solve and you get θ as 106.3

 

[Tkp]

What is R? P? Q?

Second is fine....
u find R(>) tht is 10 - 10 cos (180-theta) or 12 cos (theta/2)

.wtf.why can i upload the file
well r is the resultant one.take p as x and q as y .then solve it

 

[PhyZac]

There might be an easier way to solve this but this is how i did it.
View attachment 21926

Split the Left hand 10N force into its vertical and horizontal components by breaking θ into 90 degree and (θ-90).
Then resolve the vertical and horizontal component.
There is only one vertical component 10 Cos (θ-90) which is to 10Sinθ
The horizontal component would be 10- 10 sin(θ-90) or 10-10 Cosθ
Now consider these the vertices of a right angled triangle with the hypotenuse as 12N.
So, 12^2 =(10 Sinθ)^2 + (10-10Cosθ)^2
144 = 100 sin^2θ + 100 - 200 Cosθ +100 Cos^2θ
144 =100 - 200 Cosθ + 100 (sin^2θ + Cos^2θ)
144 = 100 - 200 Cosθ + 100
Solve and you get θ as 106.3

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!

 

[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