Mathematics: Post your doubts here!

[farhan141]

Just the circled ones. Thanks

 

[farhan141]

 

[Rifat Fawzia]

I dont have it. Can I ask why do u need it?

It's bcz em facing difficulties solving it so just wanted to see it!! if sm1 can help i wud really appreciate it...

 

[The Sarcastic Retard]

It's bcz em facing difficulties solving it so just wanted to see it!! if sm1 can help i wud really appreciate it...

Be chillofied.. Everyone stuck at somewhere or the other. Just bring on the doubts u have.

 

[The Sarcastic Retard]

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3) Y = Ae^-kx^2
ln Y = ln A - kx^2
ln Y = -kx^2 + ln A
This is now rearrange into Y = mX + C form.
So k = gradient
ln A = y intercept.

 

[farhan141]

qwertypoiu

Waiting for ur magic

 

[Fiery987]

Those Questions sure are tough! I am also stuck at Complex no. (first one) and Q4 (ii).

 

[Anum96]

Can u guys upload the questions again? Only the ones you need. Ill see if i can help.

 

[Fiery987]

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For Vector question, I used this approach but got weird values of b and c. Must be a silly mistake. I am just writing the steps as the actual answer is way too long!

• First, write down the vector for normal of plane p and q
• Use Dot product { np.nq = |np| x |nq| x cos 60 }
• Obtain equation containing variables b and c
• Next, { d = a.nq } and 'a' can be Pnt A or B from Part (i)
• Obtain a second equation { I got d = 3+b-c }
• Solve both eq. simultaneously to find b or c
• If you get a Quadratic equation, take the positive answer as b or c
• Find d and subs. the values into plane q eq.
• Problem Solved (I think)

I hope this method is right. Otherwise, please point out my mistake.
BTW, was this variant 3?

 

[Fiery987]

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For Polynomial Question:

Use Factor theorem {subs. x = -0.5 into p(x) = 0} and obtain a-2b = -8
Use Remainder theorem {subs. x = 0.5 into p(x) = 1} and obtain a+2b = -12
Solve both of them simultaneously and you get a = 10 and b = -11

 

[Bilal.khan]

Hey guys.
Can someone please help me with this question


I got the answer for the first part


But the problem is i don't get why did they not calculate mg sinθ as resistive force in the second part although they used it in the first part of the question. The engine would also do work against this component of weight so why dont we use it in our calculation.

please help....

 

[Fiery987]

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OK! Here's the solution to Complex no. question (#7).

Part (i) is easy while in Part (ii), you should remember that the given eq. can be proven if you work the other way round (from answer to question). But both ways are OK.

P.S. Our Teacher discussed this question with us in school and we proved Part (ii) in like, 40 min!

 

[farhan141]

OK! Here's the solution to Complex no. question (#7).

Part (i) is easy while in Part (ii), you should remember that the given eq. can be proven if you work the other way round (from answer to question). But both ways are OK.

P.S. Our Teacher discussed this question with us in school and we proved Part (ii) in like, 40 min!

In part (i) why is |z| = sqrt(a^2 + b^2)^2? Shouldnt it be sqrt(a+bi)^2?.
Firstly why are there 2 times square when removing modulus? Secondly u missed the iota as z is a complex number

 

[farhan141]

OK! Here's the solution to Complex no. question (#7).

Part (i) is easy while in Part (ii), you should remember that the given eq. can be proven if you work the other way round (from answer to question). But both ways are OK.

P.S. Our Teacher discussed this question with us in school and we proved Part (ii) in like, 40 min!

In second part i dont get it from 3rd step onwards. After squaring on both sides why is (z-2i)^2 equal to (z-2i)(z-2i)*. I dont get the conjuguate part (*)

 

[i_try9621]

It's bcz em facing difficulties solving it so just wanted to see it!! if sm1 can help i wud really appreciate it...

so did you get the papers? because I do have two of them

 

[i_try9621]

doubt in 7) iv) .
It's from http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_41.pdf

 

[Fiery987]

In part (i) why is |z| = sqrt(a^2 + b^2)^2? Shouldnt it be sqrt(a+bi)^2?.
Firstly why are there 2 times square when removing modulus? Secondly u missed the iota as z is a complex number

Whoa, man! To find mod of a complex no., you use this formula:

If z = a+bi
then |z| = sqrt(a^2+b^2)

because if you plot z on an Argand diagram, you can find the length of the line (from origin to z) using Pythagoras theorem as it makes a triangle. And that length is mod of z. You getting it?

And what's iota? The 'i'? Well, i just differentiates between the real and unreal part of the complex no. so that is why it isn't written during calculations (e.g. when finding arg z, you don't write i in the formula)

Hope you understand it because this my first time explaining something on xtremepapers!

 

[farhan141]

Whoa, man! To find mod of a complex no., you use this formula:

If z = a+bi
then |z| = sqrt(a^2+b^2)

because if you plot z on an Argand diagram, you can find the length of the line (from origin to z) using Pythagoras theorem as it makes a triangle. And that length is mod of z. You getting it?

And what's iota? The 'i'? Well, i just differentiates between the real and unreal part of the complex no. so that is why it isn't written during calculations (e.g. when finding arg z, you don't write i in the formula)

Hope you understand it because this my first time explaining something on xtremepapers!

thanks buddy. What abt part ii question i asked

 

[Fiery987]

In second part i dont get it from 3rd step onwards. After squaring on both sides why is (z-2i)^2 equal to (z-2i)(z-2i)*. I dont get the conjuguate part (*)

DON'T read the first part where I crossed it. The answer starts from |z-2i| = 4

Well, I used the relationship given in part (i). As they made us prove it, there must be some use of it, right? But I did it this way (as I proved in reverse) :

because |z|^2 = zz* {Part (i) question}
then |z-2i|^2 = (z-2i)(z-2i)* {z is a complex no. so z-2i must also be a complex no. because add/subt of a complx no. = another complx no.}

 

[Fiery987]

Um, I just posted the Part (ii) explanation. Refresh your browser.