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Mathematics: Post your doubts here!

studen12345

studen12345

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MustafaMotani said:
from part i) u would have got
y = 70 e^(e^(-3t) -1)

p = mass at time t / initial mass x 100
p = 70 e^(e^(-3t) -1) / 70 x 100
p = e^(e^(-3t)-1)) x 100


now if yu see if t becomes very large the expression e^(-3t) would approach zero .
look at it this way
e^(-3t) = 1/e^(3t)
see if denominator becomes very large the fraction becomes too small i.e. it approaches zero.
thus if e^(-3t) aproaches zero then e^(-3t)-1 approaches -1
thus e^(e^(-3t)-1) approaches e^(-1)

so p = e^(-1) x 100

take yur time reading this post..
Click to expand...
Plz Can U Help Me With q6 (i) Part And Q 9 First How To Solve Them Realy Confused Of This Paper..
 
Soldier313

Soldier313

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Aoa wr wb

Can someone please tell me, how do we know whether we have to use
5 rather than 4.5 etc for these sort of qns?

Screen shot 2013-05-17 at 9.14.23 PM.png

Thank you.
 
D0cEngi

D0cEngi

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I'm not able to solve a single question of vectors and argand diagram..I'm screwed.i'm so messed up. I'm crying. :cry: Don't know what to do. Plz help me. :cry:
 
MustafaMotani

MustafaMotani

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studen12345 said:
Plz Can U Help Me With q6 (i) Part And Q 9 First How To Solve Them Realy Confused Of This Paper..
Click to expand...

bro i did Q6(i) but its too long and too complicated for me to type in the post ...

i shall give u little starting point

tan(2x+x) = (tan2x + tanx)/ (1 - tan(2x)tan(x)) = k tanx


i then substituted tan2x = (2tanx)/ (1-tan^2(x))

and after doing some algebraic i got the answer
 
MustafaMotani

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studen12345 said:
Plz Can U Help Me With q6 (i) Part And Q 9 First How To Solve Them Realy Confused Of This Paper..
Click to expand...
in Question
9i) just equate both the equations

equation of l : r1 = (3-λ)i + (-2+2λ)j + (1+λ)k
equation of m : r2 = (4 +µa)i + (4 +µb)j + (2-µ)k

since they intersect r1 = r2
(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k
compare the coeficients
and after some substitutions u shall get the desired result .

ii) since they are perpendicular ... dot the direction vector of both the lines and equate it to zero ..
u shall get another eqution with a and b . Solve them simultaneously to get values of a and b.

iii) easiest.. when you got values of a and b ... u would have got some equations in terms of a and b for λ and µ in part (i)
calculate either λ or µ and then substitute that value in either equation of l or m respectively ...

its difficult to understand by reading .. just do it to get it :)
 
Esme

Esme

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D0cEngi said:
I'm not able to solve a single question of vectors and argand diagram..I'm screwed.i'm so messed up. I'm crying. :cry: Don't know what to do. Plz help me. :cry:
Click to expand...

These might help, but do as many past paper questions on these topics as you can.
And most importantly pray to Allah, insha'Allah you will get it. :)
 

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studen12345

studen12345

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MustafaMotani said:
bro i did Q6(i) but its too long and too complicated for me to type in the post ...

i shall give u little starting point

tan(2x+x) = (tan2x + tanx)/ (1 - tan(2x)tan(x)) = k tanx


i then substituted tan2x = (2tanx)/ (1-tan^2(x))

and after doing some algebraic i got the answer
Click to expand...
Oh I Also Did The Same But Might Have Done Smthng Wrong In Algebra Thanku Very Much :D
 
studen12345

studen12345

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MustafaMotani said:
in Question
9i) just equate both the equations

equation of l : r1 = (3-λ)i + (-2+2λ)j + (1+λ)k
equation of m : r2 = (4 +µa)i + (4 +µb)j + (2-µ)k

since they intersect r1 = r2
(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k
compare the coeficients
and after some substitutions u shall get the desired result .

ii) since they are perpendicular ... dot the direction vector of both the lines and equate it to zero ..
u shall get another eqution with a and b . Solve them simultaneously to get values of a and b.

iii) easiest.. when you got values of a and b ... u would have got some equations in terms of a and b for λ and µ in part (i)
calculate either λ or µ and then substitute that value in either equation of l or m respectively ...

its difficult to understand by reading .. just do it to get it :)
Click to expand...
 
Esme

Esme

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MustafaMotani said:
from part i) u would have got
y = 70 e^(e^(-3t) -1)

p = mass at time t / initial mass x 100
p = 70 e^(e^(-3t) -1) / 70 x 100
p = e^(e^(-3t)-1)) x 100


now if yu see if t becomes very large the expression e^(-3t) would approach zero .
look at it this way
e^(-3t) = 1/e^(3t)
see if denominator becomes very large the fraction becomes too small i.e. it approaches zero.
thus if e^(-3t) aproaches zero then e^(-3t)-1 approaches -1
thus e^(e^(-3t)-1) approaches e^(-1)

so p = e^(-1) x 100

take yur time reading this post..
Click to expand...

Yep understood perfectly ! Thank you soo much ! :D
 
studen12345

studen12345

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MustafaMotani said:
in Question
9i) just equate both the equations

equation of l : r1 = (3-λ)i + (-2+2λ)j + (1+λ)k
equation of m : r2 = (4 +µa)i + (4 +µb)j + (2-µ)k

since they intersect r1 = r2
(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k
compare the coeficients
and after some substitutions u shall get the desired result .

ii) since they are perpendicular ... dot the direction vector of both the lines and equate it to zero ..
u shall get another eqution with a and b . Solve them simultaneously to get values of a and b.

iii) easiest.. when you got values of a and b ... u would have got some equations in terms of a and b for λ and µ in part (i)
calculate either λ or µ and then substitute that value in either equation of l or m respectively ...

its difficult to understand by reading .. just do it to get it :)
Click to expand...
CAn Not dothefirst Part Have Done Therest I Know Its Silly But Can U Plz Show Da Wrking For Part 1 Iam Trying For So Long THANX
 
MustafaMotani

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studen12345

studen12345

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MustafaMotani

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studen12345 said:
CAn Not dothefirst Part Have Done Therest I Know Its Silly But Can U Plz Show Da Wrking For Part 1 Iam Trying For So Long THANX
Click to expand...


(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k

comparing coefficients of k
(1+λ) = (2-µ)

λ = 1 - µ

comparing coefficients of i

(3-λ) = (4 +µa)
3 -1 + µ = 4 + µa
µ = 2/(1-a)

comparing coefficients of j

(-2+2λ) = (4 +µb)
-2 + 2 - 2µ = 4 + µb
0 = 4 + µ(2+b)
0 = 4 + (4+2b)/(1-a)

2a-b = 4
 
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