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Mathematics: Post your doubts here!

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Hi, for Q2, let 3^x= u
You now have 2|u-1|=u
Now |2u-2|=u
Square both sides, (2u-2)^2=u^2
4u^2 -2(2u)(2)+ 2^2= u^2
3u^2 -8u +4=0
Solve on calculator
u= 2 u= 2/3
But, 3^x = u
So 3^x=2 and 3^x=2/3
So xln3=ln2 and xln3= ln(2/3)
x= 0.631 and x= -0.369

Q4
x=e^-t.cost and y= e^-t.sint
Differentiate each one with respect to t, and then divide dy/dt by dx/dt to get dy/dx
Use product rule,
Dx/dt= (e^-t)(sint) - (cost)(e^-t)
Dy/dt= (e^-t)(cost) - (sint)(e^-t)
Now dy/dx = (sint - cost)/(sint + cost) cancel all (e^-t)
(tant - 1)/(tant + 1) divide by cost
But expand tan(t - 1/4pie)= (tant- tan1/4pie)/(1 + tant.tan1/4pie)
= (tant - 1)/(1 + tant)
And so they're both equal :)
If I have any mistake tell me
 
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Can someone please help me with this question?
Here you go :) if there's any more doubt tell me

Justification : Drop down a perpendicular to get AB or AC in terms of r and use simple trigonometry , then cross out r^2 and simplify , change (Costheta)^2 and 2sintheta.costheta into their identities . Then make cos2theta the subject :)
 

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Variant 42 - O/N - 2014
Q1)

I understood that when the Particle P reaches it's highest point it meets with partical Q and then they both come down together and hit at the saame time..
Screenshot (2).png
 
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9709_w07_qp4 (M1)
what would be the solution to part(ii) of this question?
pls help
upload_2015-4-26_16-20-9-png.52484
 
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Note that reversing direction means the VELOCITY changes sign from positive to negative, NOT acceleration! I've plotted a graph, where you can see the velocity becomes negative when t=12, not t=8 which is simply a local maxima. View attachment 52494
thanks alot!!!

please can u tell me that in the question 6 part 2 of following paper why the direction of line is taken as 'i' even tough its parallel to x axis,why not 7j??
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_31.pdf

thanks in advance.
 
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Here you go, if you have any more doubts tell me :)

Justification : Drop down a perpendicular to get AB or AC in terms of r and use simple trigonometry , then cross out r^2 and simplify , change (Costheta)^2 and 2sintheta.costheta into their identities . Then make cos2theta the subject :)

https://www.xtremepapers.com/community/attachments/image-jpg.52500/
For part ii) just iterate it
 
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How to do this?
We get x and y on two sides.View attachment 52533

Well you differentiate it like you normally would with an implicit function, until you get an expression which is probably x^2+y^2 = 1. I'm not sure, did this a few days ago haha. Then rearrange to get x^2 = y^2 - 1, and substitute this equation in the initial equation that you'd differentiated.
Hope that helps!
 
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Can anyone explain to me how to solve these questions? Thank You.

Intergrate the following functions with respect to x.
1. x^2 cos x
2. In (x-1)
3. (In x)^2
 
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