Mathematics: Post your doubts here!

[Alice123]

i dont understand y did u find the cross product of d1 and d2..can u pls explain this part?

cross product gives final direction
THE VECTOR PRODUCT (or CROSS PRODUCT)

The vector product is a way of combining two vectors together to get another vector quantity.
where d is a vector perpendicular to both a and b
a is the magnitude of vector a
b is the magnitude of vector b
Properties of the scalar product:
1)a × b = - b × a
2)a × b gives a vector perpendicular to both of the original vectors
u want me to show hw to do the coss product?

 

[iKhaled]

cross product gives final direction
THE VECTOR PRODUCT (or CROSS PRODUCT)

The vector product is a way of combining two vectors together to get another vector quantity.
where d is a vector perpendicular to both a and b
a is the magnitude of vector a
b is the magnitude of vector b
Properties of the scalar product:
1)a × b = - b × a
2)a × b gives a vector perpendicular to both of the original vectors
u want me to show hw to do the coss product?

nah thank you, i know how to do cross product..thanks again sooo much now i will think of the way u explained the question till i completely understand it!

 

[Alice123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_13.pdf

Q4 ... i)
Q7 ... i)
Q8 ... i)

If anyone could please explain these 3 questions.
PhyZac syed1995

Ooopsss!!! havent done p1 in a year.... divide both sides by cos2x
sin2x/cos2x+3cos2x/cos2x=0
tan2x+3=0
tan2x=-3
hope u can do it nw

 

[PANDA-]

Ooopsss!!! havent done p1 in a year.... divide both sides by cos2x
sin2x/cos2x+3cos2x/cos2x=0
tan2x+3=0
tan2x=-3
hope u can do it nw

Ooooh, lol. Thank you very much, how about the other two questions?

 

[Alice123]

Ooooh, lol. Thank you very much, how about the other two questions?

im doin them

 

[GorgeousEyes]

I'll answer your questions one by one
6i) gf(x) = 1
f(x) = 1/2 x +1/6 n [take that n as pi please ]
g(1/2 x + 1/6 n) = 1
cos ( 1/2 x + 1/6 n ) = 1
1/2 x + 1/6 n = cos-1( 1)
1/2 x + 1/6 n = 0
1/2 x = -1/6 n
x = 2(- 1/6 n )
x = -n/3

THANK YOU

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_12.pdf
Please help me sketch the graph for question no. 4(ii), I'm confused!
Thanks

 

[Alice123]

7i
dy/dx=-20(2x+1)^2
@A y=0
0=10/(2x+1)-2
x=2
A(2, 0)
put value of x in dy/dx
-2o(2*2+1)^2=-4/5
now use
y-y1=m(x-x1) where m=dy/dx
y-0=-4/5(x-2)
solve n get the ans PANDA-

 

[Alice123]

PANDA-
8i
since it is a right angled triangle, apply pythagorus rule

 

[Maz]

GorgeousEyes
8iii) Simple integration:


After that put in the value of x = -1 and find "c" i guess you could do that.
And sorry about the stupid drawing

 

[PANDA-]

Alice123
Thanks 2325590381028302802 times.
Jazaki Allahu Khairan

Just one more request, I realize I've been a burden, sorry about that
When do we add +c when integrating, and when not?

 

[Alice123]

Alice123
Thanks 2325590381028302802 times.
Jazaki Allahu Khairan

Just one more request, I realize I've been a burden, sorry about that
When do we add +c when integrating, and when not?

NO PROBLEM, jus remember me in your prayers
We do not add c when limits are given, like when u r finding area or volume. if limits r not given, we add a c

 

[Maz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf
Please help me sketch the graph for question no. 4(ii), I'm confused!
Thanks

forgive me the uneven graph but actually the curve downwards is equal to the curve upwards

 

[GorgeousEyes]

GorgeousEyes
8iii) Simple integration:
View attachment 24746

After that put in the value of x = -1 and find "c" i guess you could do that.
And sorry about the stupid drawing

Thanks for ur help

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 1, if they squared the equation, why did they only square the '6'? why not the entire bracket? (2x-3) and then why they divided by 2 ?!

 

[GorgeousEyes]

please jun010 variant 12 :no.10 ii and no.11 iv , v

 

[daredevil]

Q 6i)
and
Q9ii)

yelps for help

 

[GorgeousEyes]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 1, if they squared the equation, why did they only square the '6'? why not the entire bracket? (2x-3) and then why they divided by 2 ?!

(6)^2 /(2x-3)^2
so make the integ . --> 36 (2x-3)^-2 -- > 36(2x-3)^-1 / -2
and by using the limits 2 and 3 --> -6-(-18)= 12 pi

 

[Maz]

@GorgeousEyes
9i) if OAB is a straight line then OA is a part of OB,
Therefore, by comparing the two vectors we can that OB is twice of OA
p = 2

unit vector = each vector/ magnitude of vector
= (2i+ j + k)/(2^2 +1 +1)[square root]
= (2i+ j + k)/[square root]6

ii) Remember that the dot product of perpendicular vectors is '0'
therefore, OA.AB = 0
(pi + j + k) . ( [4-p]i, j, [p-1]k) = 0
p*(4-p) + (1*1) + (1*(p-1)) = 0
5p - p^2 = 0
solve quadricatically p= 0 or 5

iii) if you can roughly draw the diagram you will realise that AB = OC
therefore c = b-a
I hope you can work that out.

[i may have made some calculation errors in ii but the methods generally correct]

And next time could you please give the link of qp please

 

[A star]

Alice123
Thanks 2325590381028302802 times.
Jazaki Allahu Khairan

Just one more request, I realize I've been a burden, sorry about that
When do we add +c when integrating, and when not?

when integrating with limits we donot add c.
when integrating without limits we add c