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Mathematics: Post your doubts here!

A

***amd***

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holoholo said:
can you explain why it isn't 5.5 ?
Click to expand...
what is the lower limit? ">0" ? i.e. Zero is NOT included, but 0.000000001 is. so we just round off the min possible number in that group to Zero, which eventually becomes our lower limit.
and mid pt = (lower limit + upper limit) / 2 = (0 + 10) / 2 = 5
 
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***amd***

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Magnanimous!!! said:
Assalamalikum.
can anyone please help me in this question
may june 2104 paper33 question 5
Jazak Allah!!
Click to expand...
rearrange the eq u get
(2x + 1)^-0.5 dx = 1/2 *(1/ cos^2 O) dO (writing theta as O)
as cos O = 1/ sec O, we get
(2x + 1)^-0.5 dx = (1/2) * (sec^2 O) dO

integrate both sides
[(2x + 1)^(-0.5 + 1)] / [(-0.5 + 1)*(2)] = 0.5 * tan O
(2x + 1)^0.5 = 0.5 * tan O

just simplify and u'll get ur answer
 
qwertypoiu

qwertypoiu

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Maayee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_31.pdf
Pls can someone help me with question 4, thank you:)
Click to expand...

y = e^-t * sint
dy/dt = (e^-t) * (cost) + (-e^-t) * (sint) = e^-t *(cost - sint)

x = e^-t * cost
dx/dt = (e^-t) * (-sint) + (-e^-t) * (cost) = -e^-t *(sint + cost)

dy/dx = dy/dt ÷ dx/dt = [e^-t *(cost - sint)] / [-e^-t *(sint + cost)] = (sint - cost) / (sint+cost)

We want tan in there somewhere, so divide everything by cos:

= (tant - 1) / (tant + 1)

remember that tan(pi/4) = 1, above expression becomes:

= (tant - tan(pi/4)) / (1 + tant*tan(pi/4))

Looking familiar? If not, time to revise the tangent sum rule thingy!

= tan(t - pi/4)
 
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Maayee

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qwertypoiu said:
y = e^-t * sint
dy/dt = (e^-t) * (cost) + (-e^-t) * (sint) = e^-t *(cost - sint)

x = e^-t * cost
dx/dt = (e^-t) * (-sint) + (-e^-t) * (cost) = -e^-t *(sint + cost)

dy/dx = dy/dt ÷ dx/dt = [e^-t *(cost - sint)] / [-e^-t *(sint + cost)] = (sint - cost) / (sint+cost)

We want tan in there somewhere, so divide everything by cos:

= (tant - 1) / (tant + 1)

remember that tan(pi/4) = 1, above expression becomes:

= (tant - tan(pi/4)) / (1 + tant*tan(pi/4))

Looking familiar? If not, time to revise the tangent sum rule thingy!

= tan(t - pi/4)
Click to expand...
Thank you so much!!!:):):):)
 
The Sarcastic Retard

The Sarcastic Retard

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Anum96 said:
Now use any equation to equate. Lets take the first one.
x+2>2x+1
x - 2x > 1-2
-x>-1
x<1 :)

Or.
You can even take -x-2 and equate. You will get the same answer. But remember to change the sign of RHS as well.
Click to expand...
For the first time I saw this. :3
Anyways ty. Though I am not getting why we did like this. :p

Can u help me wid other doubts as well?

Thanks.
 
Anum96

Anum96

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The Sarcastic Retard
Heres the real explanation :p I was too sleepy yesterday. Sorry.
So you have an absolute value function here. Remember that any absolute value function gaph is a simple v.
Here inside the function you have plus 2 which means your vertex which was originally 0,0 Will mave 2 units to the left and so you will get -2,0 as your vertex.
Second, You have a co-efficient of 1 meaning your slope is 1 and no y intercept so it will be a straight V-shaped graph.
Second draw the line of the second part of the inequality. Both of these functions will intersect at x equals 1.
Now, Since they intersect and you can solve the inequality 'otherwise' as mentioned in your question, Just equate them.
x+2>2x+1
x - 2x > 1-2
-x>-1
x<1 :)
Im sorry for confusing it up but here it is.
 
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