Mathematics: Post your doubts here!

[talhajohar]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s10_qp_31.pdf


i need help in question no 4

 

[saimaiftikhar92]

http://www.xtremepapers.com/CIE/index.p ... _qp_32.pdf

QUESTION 1 ,3 4 AND 6

 

[ashemmell]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s10_qp_32.pdf

QUESTION 1 ,3 4 AND 6

Question 1:
You need to get the 2^x terms together. Multiply the denominator over to the other side, bring the 2^x terms together and then use logarithms to find x.

Question 3:
part i) a is acute, so sin and cos are both positive. Use the cos^2 + sin^2 =1 identity to find cos a. Then use the sin(A+B) identity to expand the required expansion, and use your values for cos a and sin a to evaluate.

part ii) Use tan a = (sin a)/(cos a), and then the double angle identity for tan 2a to find that value. Then use tan(3a) = tan(2a + a) and the tan(A+B) identity for the second bit.

Question 4:
part i) M is a stationary point, so dy/dx = 0 at M. Differentiate using the Quotient Rule, then solve your function =0. Remember that you only need to consider the numerator to make it 0.

part ii) M lies between pi and 2pi, so you can use any starting value in that range with the iterative formula, from the graph maybe 3pi/2 would be a decent value. Then use the iterative formula (don't forget to work in radians mode), writing each to 4dp. When they match to 3dp, your 2dp root is correct.

Question 6:
part i) Implicit differentiation to begin, don't forget that you differentiate any term involving y as normal, but then multiply by dy/dx. Then rearrange to get dy/dx on its own.

part ii) Straight lines need a gradient and a point. Use the y coordinate in the original equation to find the value of x. Then use both of these in your dy/dx function to find the gradient. Then use this point and gradient to fin the equation of the tangent.

Hope those hints help,

Cheers,

ASh.

 

[talhajohar]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s10_qp_32.pdf

QUESTION 1 ,3 4 AND 6

 

[saimaiftikhar92]

http://www.xtremepapers.com/CIE/index.p ... _qp_32.pdf

HOW TO DO NUMBER 8 ?

 

[talhajohar]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s11_qp_32.pdf

HOW TO DO NUMBER 8 ?

 

[saimaiftikhar92]

http://www.xtremepapers.com/CIE/index.p ... _qp_31.pdf QUESTION 8 ...... it is similar but i am not getting the answer

 

[talhajohar]

part 1 or 2?

 

[saimaiftikhar92]

both..

 

[talhajohar]

http://www.xtremepapers.com/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_w09_qp_31.pdf QUESTION 8 ...... it is similar but i am not getting the answer

 

[saimaiftikhar92]

thankyou so much for your help in these 5 , 6 days...... i have maths mock tomorrow ........and I really hope that it goes well......again thankyou so much! :good:

 

[SalmanPakRocks]

^^ just press the thank button to thank someone!

 

[ffaadyy]

need solution to question Q8ii and Q9ii both from Mathematics P33 May/June 2011. The questions can be found over here : http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_33.pdf. Thanks in advance.

 

[yubakkk]

how to solve this type of problem>??
please help me

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_1.pdf

seems a very simple question but still i need its explaination.

Thank You

 

[XPFMember]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_1.pdf

seems a very simple question but still i need its explaination.

Thank You

As-salam-o-alaikum!

You forgot to mention the question number!

 

[Silent Hunter]

As-salam-o-alaikum!

You forgot to mention the question number!

OH sorry its question number 3

 

[Anon]

Here is how:

Part (i)
sin θ + cos θ = 2(sin θ − cos θ)
sin θ + cos θ -2sin θ +2cos θ = 0

3cos θ = sin θ

3= sin θ
.....cos θ

sin θ = tan θ
cos θ

tanθ = 3 (expressed)

Part (ii)

Hence solve the equation sin θ + cos θ = 2(sin θ − cos θ), for 0 ≤ θ ≤ 360

so use the expression fain above:

sin θ + cos θ = 2(sin θ − cos θ)
=
tan θ= 3

θ = tan-1 (3)
= 71.6

tan is positive in first and third quadrant, so:

θ = 71.6, (180 + 71.6)

θ = 71.6° , 251.6°

 

[XPFMember]

OH sorry its question number 3

As-salam-o-alaikum!

Let me give you a small hint!

Case : 1

If sin and cos are either being added or subtracted, it’ll be possible to rearrange the equation, and hence write it in terms of tan !
Look at this!
2 sin x – 3 cos x = 0

2sin x= 3cos x​

sin x /cos x = 3/2​

​

tan x = 3/2​

​

Hence find x​

​

​

In general, bring all sin x terms to one side, and cos x to the other side, and then change it to tan x as shown above!​

​

​

For more ways and hints for solving equations, see attached! They are for the whole chapter... for the solving equations help see the last two pages!
I am adding it as a zip file, because I don't know why it showed me no option to add the word file :s

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s07_qp_1.pdf

Q11 part (i)