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Mathematics Paper 3: Post your doubts!

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Here's a platform for you to clarify your doubts in paper 3!
I shall give a start with this question:
q.JPG
 
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ok for 1) the equation is r= OA + t( AB) where AB is OB - OA
2) you do (1+2t, 2+2t, 2-2t) dot (1,2,2) = 0 which gives t to be -1/6. then OP is (2/3, 5/3, 7/3)
3) OP is the normal since it is perpendicular to AB then you do r.n=a.n to give u 2x+5y+7z=26.. post if u need more explaination :):)
 
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gu6lw

tomorrow ill give the solution to this.. i just want u to give it a go.. this is the hardest question ive ever seen in p3 ;)
 
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gu6lw

tomorrow ill give the solution to this.. i just want u to give it a go.. this is the hardest question ive ever seen in p3 ;)
Hmmm...but where is the question dr..?!? i think it is a broken file...!! please post it...i want to see the hardest question in p3...
Thanks :)
 
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Guys, I need notes for Vectors p3. It's really complicated for me. Please share it here asap! thanks in advance
 
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For number 9:

area = integral of y from 1 to 2.

Use intergration by parts. u=lnx ; du/dx= 1/x; dv/dx= x^3 and v = 1/4 x^4
if we use the formula, the integral is [(lnx*1/4x^4)-(x^4 ?16)]

we know that at the x axis, y=0, so the curve meets the x axis at lnx=0 therefore x=1.

Substituting 2 and 1 for x we get [4(ln2)-1]-[0-1/16]

simplifying this we get 4ln2-1+1/16

further simplifying we get 4ln2-15/16

Hope u find it helpful
 
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For number 9:

area = integral of y from 1 to 2.

Use intergration by parts. u=lnx ; du/dx= 1/x; dv/dx= x^3 and v = 1/4 x^4
if we use the formula, the integral is [(lnx*1/4x^4)-(x^4 ?16)]

we know that at the x axis, y=0, so the curve meets the x axis at lnx=0 therefore x=1.

Substituting 2 and 1 for x we get [4(ln2)-1]-[0-1/16]

simplifying this we get 4ln2-1+1/16

further simplifying we get 4ln2-15/16

Hope u find it helpful
Okay, but why did u use limits 1 and 2??? o_O
 
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4 iii) let x0 =0.7
x1= 0.7394
x2= 0.7239
x3= 0.7299
x4= 0.7275... it stays at 0.73 so thats the answer

5) take thita to be A,
from the identity cos2A=1 -2sin^2 a we get 4 sin^2A to be 2-2cos2A
when we integrate it we get 2A -sin2A. substite with the limits we get the required answer

6iii) arg of a is pi/6 ( by doing tan inverse (1/sqrt 3)
b is 1,sqrt3i so arg b is pi/3 ( same method as for a)
reuired andgle is arg b-arg a which is pi/3 -pi/6 which is pi/6

7)i) the equation is r= OA + t( AB) where AB is OB - OA
ii) you do (1+2t, 2+2t, 2-2t) dot (1,2,2) = 0 which gives t to be -1/6. then OP is (2/3, 5/3, 7/3)
iii) OP is the normal since it is perpendicular to AB then you do r.n=a.n to give u 2x+5y+7z=26

8) expression is (2x+1)(1+2x^2)^-1 -(1+x)^-1

(2x+1)(1-2x^2 +4x^4) -(1-x+x^2-x^3)

which simplifies to that answer given
9) i think it is answered in earlier post..

a thanks would be appreciated :p
 
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