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Mathematics: Post your doubts here!

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MSP5491hf9f741edh5ib9f000054giee0i91cic356
help bros answer is x is less than or equal to -1

Bring -|x+1| to the right side
|1-x| = 2 + |x+1| now square both sides.
(1-x)^2 = [ (2) + (x+1) ]^2

Note: Treat [ (2) + (x+1) ]^2 as (a+b)^2 where a = 2 nd b = (x+1)

So you get, (1)^2 + (x)^2 -2(1)(x) = (2)^2 + (x+1)^2 +2(2)(x+1)
1 + x^2 - 2x = 4 + x^2 + 1 + 2x + 4x + 4
Simplify, 8x = -8
x = -1
 
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So you gotta divide the graph into three (equal) intervals, from x = 0 to π/4.

Then just figure out the area for each interval (trapezium). A = 1/2.(a+b).h

The 'h' (aka heights) will be the interval on x axis. Each height will thus be π/12 yeah? (Coz we divide π/4 into three equal bits yadda yadda)


For values of ‘breadths’ aka parallel sides, put x into equation.

Here are the corresponding values of y I got:

For:

X= 0, y = 1

X = π/12, y = 1.069

X = π/6, y = (15)^0.5/3

X = π/4, y = (3)^0.5


Interval 1: x = 0 to π/12

A = (1/2).(π/12)(1+1.069)


Interval 2: π/12 to π/6

A = (1/2).(π/12)(1.069+(15)^0.5/3)


Interval 3: x = π/6 to π/4

A = (1/2).(π/12)((15)^0.5/3+(3)^0.5)


Add ‘em up… (0.2708+0.3089+0.3957 = 0.9754 rounded off to 0.98 (2dp)

Thanks , i completely got it ..

but i think there is a formula for trapezium rule which is much faster and saves time .. if you can do it that way
 
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if this way confuses you ,
find dy/dx =0
you will get x values ,
plug this x values in the second derivative .. and they must be positive which shows its minimum

it you draw the graph , it will be parabola , and its least value ( vertex) will be ABOVE x -axis , which shows its positive and cant be negative
Hey thanx.

Thanks , i completely got it ..

but i think there is a formula for trapezium rule which is much faster and saves time .. if you can do it that way
Hmm don't know abt it. Dyu have ny idea what this formula is?
 
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upload_2014-2-22_23-2-49.png
p(x) = (x-2)(2x^3+x+2)
What's the justification for (iii)?
Ans is x>2
ER says:
upload_2014-2-22_23-3-53.png

Someone explain the 'justification' aka justify it. Thx in advance
 
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Man you've chosen some question...


(1-x)/(1+x)

dy/dx = Use the quotient rule: u/v : dy/dx = (u’.v – v’.u)/v^2

So u get: [(-1)(1-x) – (1)(1+x)]/(1+x)^2

=[-1+x – (1+x)]/(1+x)^2

=[-1+x-1-x]/(1+x)^2

=(-2)/(1+x)^2

Thus you’ve got the dy/dx for (1-x)/(1=x)

Now the bad boy question: u = [(1-x)/(1+x)]^1/2

Use the formula: (x)^n, dy/dx = x’.n.(x)^(n-1) aka just differentiate it, taking the whole thing as one.

Like this:

dy/dx = [(-2)/(1+x)^2].(1/2).[(1-x)/(1+x)]^-1/2

Simplifying it a bit, [(-1).(1+x)^1/2]/[(1+x)^2.(1-x)^(1/2)]

Now this is the dy/dx for the graph, meaning for the normal to graph’s gradient, you gotta take the negative reciprocal.

You get: grad(normal) = [(1+x)^2.(1-x)^(1/2)]/(1+x)^(1/2)

Since we’ve got (1+x) both in the numerator+denominator, so simplify that. (Subtract powers etc)

You should now get, (1+x)^(2-0.5).(1-x)^(1/2) = (1+x)^(1.5).(1-x)^(1/2)

Look at (1+x)^(1.5) now. Power of 1.5 means power of ONE and HALF. Half aka 0.5 aka ½ (power) represents square root. So… (1+x)^(1.5) = (1+x).(1+x)(1/2)

Back to the gradient:

(1+x)^(1.5).(1-x)^(1/2) is now thus: (1+x).(1+x)^(1/2).(1-x)^(1/2)

Notice that the two terms with sq. root are (a+b) and (a-b). What’s the formula involving this?

Yeah, (a+b).(a-b) = (a^2 – b^2)

Thus, (1+x)^(1/2).(1-x)^(1/2) becomes: [(1-x^2)]^(1/2), taking the square roots common.

Finally… (1+x).(1+x)^(1/2).(1-x)^(1/2) is converted to the final answer aka (1+x).[(1-x^2)]^(1/2)

It all looks wickedly messed up here... I'd advise you to grab a pen and jot down the answer as you read it... like the fractions.
Thanks a tonne bro.. I appreciate it.
Would you mind helping me with part ii of the same question? I'm getting 1.5 as the answer (The ms answer is 0.5).
Sorry for bothering man
 
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Thanks a tonne bro.. I appreciate it.
Would you mind helping me with part ii of the same question? I'm getting 1.5 as the answer (The ms answer is 0.5).
Sorry for bothering man

Hey,, no worries. Glad 2 help.

So, 9(ii):
I guess you know what we gotta do in this - we treat the gradient of normal as a line. To find the stationary (max) point on it, we find the dy/dx of this 'line'. (Technically we're findin the d^2y/dx^2)

From (i), we know dy/dx is (1+x).[(1-x^2)]^(1/2)___________ [Here, (1+x) is u and v is [(1-x^2)]^(1/2)]
Apply (u.v)' = u'.v+u.v'
d^2y/dx^2 = (1).[(1-x^2)]^(1/2) + (1+x).(1/2)[(1-x^2)]^(-1/2)(-2x)
= [(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2)
Now equate this to zero:
[(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2) = 0
I then got: (1-x^2) - x(1+x) = 0 ___________ (Coz [(1-x^2)]^(1/2).[(1-x^2)]^(1/2) = (1-x^2) )
Then: 1 - x^2 - x - x^2 = 0
-2x^2 - x + 1 = 0 (multiply with -1 in both sides)
2x^2 + x - 1 = 0
Solving this you get two sol., x = 1/2 (ANS.) or x = -1
From graph, we see dat point P ain't on x = -1
Ans. is thus x = 1/2
 
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I like the doubt it made me think for a while so here's what i did
for 8) i found the sum of n=1,n=2 and n=3 i could keep going just to be more sure but 3 are enough
i found sum of 1st n terms is -1
sum of 2nd n terms is 4
sum of 3rd n terms is 15
as we know sum of 1st term is -1 that makes it obvious first term is -1
so we have -1,y,z
y and z are 2nd and the 3rd term respectively
as we know sum of 2nd n terms is 4 that means -1+y=4
and so for the 3rd -1+y+z=15
we will get y=5 and z=11
we have -1,5,11
now lets see if its AP or GP
if its GP then un=an^(n-1) lets try that out
5=-1*2^(2-1)
5=2 that is wrong And by now you will know its not GP
lets see if its AP then un=a+(n-1)d (d is the difference between the terms here ist 5-(-1)=6)
5=-1+(2-1)6
5=5 hence proved :D
you can try for the 3rd which will be
11=-1+(3-1)6
11=11
For the next question you asked you should do the same but if you want me to do it i will do it by the way the first term is 1 and the common difference is 4
I hope i helped :) and if i did you always free to ask more questions

Thank you and sorry for not replying earlier my net went down and got fixed today :mad:
anyway i actually did it the way you did and it's wrong the proof has to be by induction
 
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Thank you and sorry for not replying earlier my net went down and got fixed today :mad:
anyway i actually did it the way you did and it's wrong the proof has to be by induction
But i got my answer right this way can i know what i did wrong and what is the right method
 
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But i got my answer right this way can i know what i did wrong and what is the right method

that solution only proves it for the first 3 terms or so, it is required to be proved for all terms, like:
Sn = [a+(n-1)d] + [a+(n-1)d] + [a+(n-1)d] ........ [a+(n-2)d] + [a+(n-1)d] = n(3n-4) and i don't know how to continue from here :unsure:
 
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that solution only proves it for the first 3 terms or so, it is required to be proved for all terms, like:
Sn = [a+(n-1)d] + [a+(n-1)d] + [a+(n-1)d] ........ [a+(n-2)d] + [a+(n-1)d] = n(3n-4) and i don't know how to continue from here :unsure:
:confused: I believe CIE examiners whould never give this question but i will ask my teacher and let you know
 
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im sorry i cannot provide the link because the 2013 past papers aren't yet uploaded. The question staes, y=mx+14 is tangent to the curve y=12/x+2 at point p. what are the cordinates of point p? :confused:
 
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