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Mathematics: Post your doubts here!

Pie-man

Pie-man

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Anika Raisa said:
CIE Puremath3: Some help here please!​
​
Please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you​
​
​
Click to expand...
i like how you emphasized that it was paper 3 xD


the people that take paper 3 in this forum (of which i know about) will be online soon so solve more equations until then. good luck :)
 
Sanis

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

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Anika Raisa

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Sk. Shahriar Hossain said:
View attachment 25695

Plz help me solve this math.....i will be really grateful...:)
Click to expand...

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

(Note that the Os with two dots above them are square roots n O with a strikethrough is theta!Weird it turns out dis waay wen i copy n paste it here frm ms word!)
Hope it helped!
 
Anika Raisa

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I

Iadmireblue

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Anika Raisa said:
@iKhaled
Cud u help wid this...
9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you

Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Thank you
Click to expand...
Anika Raisa said:
@iKhaled
Cud u help wid this...
9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you

Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Thank you
Click to expand...
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.
 
Sk. Shahriar Hossain

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Anika Raisa said:
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!
Click to expand...
yes!!!....they are correct!!!....:D.....sorry for not posting the year or the answer....;).......thanks a lot!!...:)
 
Anika Raisa

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[quote="Iadmireblue said:
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.
Click to expand...

, 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! :)
 
Sk. Shahriar Hossain

Sk. Shahriar Hossain

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Anika Raisa said:
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!
Click to expand...
THANKS A LOT!!!!....:D
 
Anika Raisa

Anika Raisa

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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

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Sanis said:
help guys in solving this .. (2x-1)ln5=ln2 + xln3 .. i have difficulities in this sort of questions
Click to expand...
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!!
 
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