Mathematics: Post your doubts here!

[Sk. Shahriar Hossain]

I have solved it... uploading in few min... but first do u knw the answers? are they
i. 3x-4y+6z=80
ii.30.81

Uploadin the full procedure will take some tym just w8!

yes!!!....they are correct!!!.........sorry for not posting the year or the answer...........thanks a lot!!...

 

[Anika Raisa]

The values of mod Z is a line through the origin passing through the centre of the circle.So the greatest and least values of mod z is the length that corresponds to the two points that the line touches the circle with.and since the diameter is four,you know that the greatest mod is four more than the vaule for the min mod Z
Hope this helps you.

, post: 533181, member: 5635"]The values of mod Z is a line through the origin passing through the centre of the circle.So the greatest and least values of mod z is the length that corresponds to the two points that the line touches the circle with.and since the diameter is four,you know that the greatest mod is four more than the vaule for the min mod Z
Hope this helps you.[/quote]
Iadmireblue
Thank u!
That cleared the confusion 2 some extent!thank u!

 

[Anika Raisa]

yes!!!....they are correct!!!.........sorry for not posting the year or the answer...........thanks a lot!!...

Mention not!its ok!

 

[Sk. Shahriar Hossain]

Here is the answer:


Direction of plane=n= direction of line
n=(a b c)
Direction of line:
AB⃗ =OB⃗-OA⃗
=r⃗ .n⃗ =a⃗ .n⃗
=(2 -2 11)-(-1 2 5)
=(3 -4 6)
= (a b c)

Therefore equation of plane= 3x-4y+6z=d

Let r= (x y z) and n=(3 -4 6)

Therefore, (x y z). (3 -4 6)=(2 -2 11).(3 -3 4)
ð 3x-4y+6z= 6+8+66=80

Now Solution to part 2:

At y axis xand z is equal to zero
Therefore, Subsituting values In the plane equation:
We get… 4y=80
ð y=20

Now using dot product, we have…
(0 20 0).(3 -4 6)= |(0 20 0)|.|(3 -4 6)|.cos O
=> -80=(Ö202).(Ö32+42+62) .cos 0
=> cos O = -80/(20Ö61)
Or a=cos-1(80/(20Ö61))=59.193 o (Look that – sign has been ignored)
As there is a minus sign so we have to consider 2nd quadrant (Remember Trigonometry! ; ))
so we get, O =120.807 o
Now O is the angle between the normal of plane and y
To get require angle we need to subtract 90o from it n so we have
the angle=30.81

Hope it helped!

THANKS A LOT!!!!....

 

[Sk. Shahriar Hossain]

what year is this? pls when posting question post the paper so i can check the mark scheme after solving it to make sure i am not giving u the wrong answer!

Sorry.. i'll keep this in mind next time....

 

[Anika Raisa]

THANKS A LOT!!!!....

Mention not!

 

[Anika Raisa]

CIE Puremath3:Help needed!
Iadmireblue : or anyone sitting for A2
Could u possibly help me with dis too...(Both question n its answer frm ms r given below

(I could do till iii but got badly stuck with part iv)
Q.The complex number −2 + i is denoted by u.
(i) Given that u is a root of the equation x
3 − 11x − k = 0, where k is real, find the value of k. [3]
(ii) Write down the other complex root of this equation. [1]
(iii) Find the modulus and argument of u. [2]
(iv) Sketch an Argand diagram showing the point representing u. Shade the region whose points
represent the complex numbers z satisfying both the inequalities
|z| < |z − 2| and 0 < arg(z − u) < 1/4π.

Thank u! May u be blessed for helping!
BESIDES here is the mark scheme below... to me it sounded completely gibberish so i couldnot work out the last part of the sum!
(iv) Show point representing u in relatively correct position in an Argand diagram B1
Show vertical line through z = 1 B1
Show the correct half-lines from u of gradient zero and 1 B1
Shade the relevant region B1 [4]
[SR: For parts (i) and (ii) allow the following alternative method:
State that the other complex root is –2 – i B1
State quadratic factor x2 + 4x + 5 B1
Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using
3-term quadratic, factorise cubic and obtain k M1
Obtain k = 20 A1]

Would be grateful if anyone can help!

 

[Anika Raisa]

help guys in solving this .. (2x-1)ln5=ln2 + xln3 .. i have difficulities in this sort of questions

Hey tried solving it bt cnt! Sorry
Btwn 1 doubt: Are u sure the question is dis way or is the last part 3lnx?
the x b4 ln 3 seems unusual to me!(as in i didnt come across such a sum!) n that is wat creates the big problem while solving it!
Please let knw asap!!

 

[Just visiting]

I need your help guys in Question 6 Part ii
BTW this is may 2012

 

[Anika Raisa]

help guys in solving this .. (2x-1)ln5=ln2 + xln3 .. i have difficulities in this sort of questions

ignore ma previous post!
I cud do it... dnt knw hw it came to me now but here it is:
The solution is given by:


(2x-1)ln5=ln2+xln3
=>2xln5-ln5=ln2+xln3
=> x(2ln5-ln3)=ln2+ln5
=> x=ln10 divided by ln25/3
=>x=1.086

formula used:

lna+lnblna−lnbalnb=lnab=lnab=lnba​

Golden rule:One trick while doing algebra always expand or simplify as much as u cn! n ys u gt it!

Uff dat was quiet challenging!(Btwn i dnt knw if the answer is ryt!)
Hope it helped!

​

 

[Sanis]

ignore ma previous post!
I cud do it... dnt knw hw it came to me now but here it is:
The solution is given by:


(2x-1)ln5=ln2+xln3
=> x(2ln5-ln3)=ln2+ln5
=> x=ln10ln253

formula used:

lna+lnblna−lnbalnb=lnab=lnab=lnba​

Golden rule:One trick while doing algebra always expand or simplify as much as u cn! n ys u gt it!

Uff dat was quiet challenging!(Btwn i dnt knw if the answer is ryt!)
Hope it helped!

thanks alot u did help me

 

[Sanis]

ignore ma previous post!
I cud do it... dnt knw hw it came to me now but here it is:
The solution is given by:


(2x-1)ln5=ln2+xln3
=> x(2ln5-ln3)=ln2+ln5
=> x=ln10ln253

formula used:

lna+lnblna−lnbalnb=lnab=lnab=lnba​

Golden rule:One trick while doing algebra always expand or simplify as much as u cn! n ys u gt it!

Uff dat was quiet challenging!(Btwn i dnt knw if the answer is ryt!)
Hope it helped!

thanks alot u did help me

 

[Sanis]

ignore ma previous post!
I cud do it... dnt knw hw it came to me now but here it is:
The solution is given by:


(2x-1)ln5=ln2+xln3
=> x(2ln5-ln3)=ln2+ln5
=> x=ln10ln253

formula used:

lna+lnblna−lnbalnb=lnab=lnab=lnba​

Golden rule:One trick while doing algebra always expand or simplify as much as u cn! n ys u gt it!

Uff dat was quiet challenging!(Btwn i dnt knw if the answer is ryt!)
Hope it helped!

thanks alot u did help me

 

[Anika Raisa]

thanks alot u did help me

ma pleasure!

 

[Anika Raisa]

I need your help guys in Question 6 Part ii
BTW this is may 2012

I will give u the general points:

• Differentiate the equation of the curve this is the gradient of the curve
• Gradient of curve n line should equal if line is tangent to curve!
  (Equation of line: y=mx+c; m=gradient of line)
• Or equations n then write b^2-4ac=0 (BEST METHOD!)
• n knw u got k!
• Use this value of k and write down the equation of both line n curve replacing k with its value.
• now equate both the equations n get the value of x
• then place this value of x in either equation of line or curve n you will have the value of y!
• this value of x n y r the co ordinate of the point of intersection of tangent n curve
• As here the co ordinate is asked remember to write in the format (x,y).
• Follow the steps n u wil find the ans!

Hope it helped!

 

[iKhaled]

hey can someone pls show me a sketch of how to find the max value of arg z in this question 7(iii) i really wanna see a sketch not just word explanation..thanks a lot !!

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

 

[Esme]

hey can someone pls show me a sketch of how to find the max value of arg z in this question 7(iii) i really wanna see a sketch not just word explanation..thanks a lot !!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

I had a doubt in a similar question in another paper. I don't know how we find the max value or arg z. I hope someone can explain

 

[Esme]

the whole thing sounded gibberish to me....then i realized its paper 3

So you're giving p4 but not p3 ??

 

[$$AK$$]

hey,
can any one please tell me how to solve sequence questions

 

[Anika Raisa]

hey can someone pls show me a sketch of how to find the max value of arg z in this question 7(iii) i really wanna see a sketch not just word explanation..thanks a lot !!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

I had a doubt in a similar question in another paper. I don't know how we find the max value or arg z. I hope someone can explain

If u r not clear by the diagram only:

• I have used tangent theorem
• first i found out the angle uOX
• then multiplied by two!
• Well i did it dis way cz u c both OM n OX r tangent to d circle n so Ou bisects the angle MOX.

One question on ma part: If min arg ws to be found which angle would it hv been?wud it hv bn angle uOX?

n guyzz cn u help me wid the sum here i nid help here desperately!:
https://www.xtremepapers.com/community/threads/pure-mathematics-3-help.25895/

Thank u!N hope i was helpful!