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Mathematics: Post your doubts here!

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could you please show the steps ^
(tan9 + tan7) + (tan5+tan3) + 4(tan7+tan5)
tan n + tan (n+2) = 1/(n+1)
so for tan9 + tan7 it will be 1/(7+1)
for tan3 + tan5 = 1/(3+1)
for tan5 + tan7 = 1/(5+1)
1/(7+1) + 1/(3+1) + 4*1/(5+1)
 
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dx/dt = 2t
dt = dx/2t
4t^3ln(x) dx/2t
2t^2 ln(x) dx
2(x-1)lnx dx
(2x-2)lnx dx
thanks a lot,i made the mistake of setting t as the subject and then differentiating.Could you please solve the second part as well.
 
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Can somebody link me to good notes on integration by substitution. I really don't understand how it works!!!!:cry:
 

Jaf

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Q5, for the second and third part we use C, for eg. 5C2 but not the first. I don't get when we are supposed to use it. also for part iv, why is have they taken n as 5?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_6.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_ms_6.pdf
For the first part they ARE supposed to use 5C0 , but since 5C0 = 1, they are not showing it (like how (1/3)^0 is also = 1, which they're supposed to show but don't because they're just lazy lol).

They're using it because they're applying binomial theorem (distribution) in all the parts.

n = 5 because of the same reason n=5 in the above three parts (because you're choosing a total of 5 discs).
 
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thanks a lot,i made the mistake of setting t as the subject and then differentiating.Could you please solve the second part as well.
it will be done by parts.
u = lnx dv = 2x-2
du/dx = 1/x v = 2x^2/2 - 2x
du = dx/x v = x^2 - 2x
uv - integral of vdu
lnx(x^2 - 2x) - integral[ (dx/x) (x^2 - 2x)]
lnx(x^2 - 2x) - integral [ (x - 2)dx]
lnx(x^2 - 2x) - (x^2/2 - 2x)
lnx(x^2 - 2x) - (x^2)/2 + 2x
apply the limits. ans is 15ln5 - 4
 
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can anyone pls upload the formula sheet, which will be given to us during the exam (i need the formula sheet for P3 and M1)
 
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I found the mean
but could you please explain for the others in detail

You have to assume that the probability up to wind speed of 63 km/h is 0.75, because the data is approx/ normally distributed. Use the Critical value in Normal Distribution Function Table to find value of z , which is 0.674. Then use the formula z = (x-µ)/σ to find σ.
 
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You have to assume that the probability up to wind speed of 63 km/h is 0.75, because the data is approx/ normally distributed. Use the Critical value in Normal Distribution Function Table to find value of z , which is 0.674. Then use the formula z = (x-µ)/σ to find σ.
aha got it , thanx a lot
 
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it will be done by parts.
u = lnx dv = 2x-2
du/dx = 1/x v = 2x^2/2 - 2x
du = dx/x v = x^2 - 2x
uv - integral of vdu
lnx(x^2 - 2x) - integral[ (dx/x) (x^2 - 2x)]
lnx(x^2 - 2x) - integral [ (x - 2)dx]
lnx(x^2 - 2x) - (x^2/2 - 2x)
lnx(x^2 - 2x) - (x^2)/2 + 2x
apply the limits. ans is 15ln5 - 4
thank you again.will bother you again soon.:)
 
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1/8 - (cos4x)/8
1/8 - [2(cos^2)2x - 1]/8
1/8 - [2(cos2x)^2 - 1]/8
1/8 - {[2(2(cos^2)x - 1)^2] - 1}/8
1/8 - {[2(4(cos^4)x - 4(cos^2)x +1] -1}/8
1/8 - {8(cos^4)x - 8(cos^2)x +2 -1}/8
1/8 - (cos^4)x + (cos^2)x -(1/8)
-(cos^4)x + (cos^2)x
cos^2x (-cos^2x +1)
cos^2x(sin^2x)
 
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