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Mathematics: Post your doubts here!

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Same method here again bruh ... It will be useful if you can remember this formula (it's not given in the formula sheet) :

(1+x)^-1 = 1 - (x) + (x)^2 - (x)^3

Saves a lot of time and effort than the original one, and it's what I've used in solving these problems.

12729200_1679359148994164_4095977479260756425_n.jpg

Messy, but understandable right?
 
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Same method here again bruh ... It will be useful if you can remember this formula (it's not given in the formula sheet) :

(1+x)^-1 = 1 - (x) + (x)^2 - (x)^3

Saves a lot of time and effort than the original one, and it's what I've used in solving these problems.

12729200_1679359148994164_4095977479260756425_n.jpg

Messy, but understandable right?
You saved me, I also did silly mistake.. :p
I hope I will score full marks, only this was my problem, rest other is as easy as piece of a cake. Moreover its as easy to say as binomial expansion, its just one silly mistakes can cost u many marks.. xD :p
I will tell u my marks afternoon. ;)
 
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You saved me, I also did silly mistake.. :p
I hope I will score full marks, only this was my problem, rest other is as easy as piece of a cake. Moreover its as easy to say as binomial expansion, its just one silly mistakes can cost u many marks.. xD :p
I will tell u my marks afternoon. ;)
Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do :)
 
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Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do :)
:D :D Paper was really damn easy, I think I made a mistake in binomial expansion xD That's it. As expected... :p completed paper in just 45 mins.. :D
This was the question, can u please show me the working of it? :p
 

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Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do :)
Vectors are really easy to solve, just jot down all the concepts in one piece of a paper, and revise it everytime u come through that paper. And then see where u are making the mistakes in vector questions and work on it. Yes I also use to do mistakes till my 3 to 5 papers. But now, I have made my weak points to be strong points. ;) Similarly do this with any conceptual questions. ^_^
 
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:D :D Paper was really damn easy, I think I made a mistake in binomial expansion xD That's it. As expected... :p completed paper in just 45 mins.. :D
This was the question, can u please show me the working of it? :p
So yeah ... not as easy as just saying 'binomial expansion' :p
The 2nd part?

Vectors are really easy to solve, just jot down all the concepts in one piece of a paper, and revise it everytime u come through that paper. And then see where u are making the mistakes in vector questions and work on it. Yes I also use to do mistakes till my 3 to 5 papers. But now, I have made my weak points to be strong points. ;) Similarly do this with any conceptual questions. ^_^
uhuh ... thanks man! will try to do but imma be lazy :/
 
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Just the last two , thanks a lot ! :)View attachment 59312
let <ADE = a
then,
<ADE = <DCA = a

also,
<DCA = <DBA = a (<s subtended by arc at circumference are equal)

<ADB = 90 (angle subtended by diameter at circumference is 90)

<ADE + <EDB = <ADB
a + <EDB = 90
<EDB = 90 - a

now consider triangle DEB,
the sum of interior angles in a triangle is 180, so

<DEB + <DBA + <EDB = 180
<DEB + a + 90 - a = 180
<DEB = 90 (proved)

(ii) <ACB = 90 (angle subtended by diameter at circumference is 90)

in a cyclic quadrilateral, the sum of two angles facing each other is 180.

So,
<ACB = <FCB

<FCB + <FEB = 90+90 = 180

in a quadrilateral, the sum of all interior angles is 360, so

<FEB + <FCB + <CFE + <EBC = 360
180 + <CFE + <EBC = 360 ( <FCB + <FEB = 180 )
<CFE + <EBC = 180

Hence shown that FEBC is a cyclic quadrilateral.
 
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let <ADE = a
then,
<ADE = <DCA = a

also,
<DCA = <DBA = a (<s subtended by arc at circumference are equal)

<ADB = 90 (angle subtended by diameter at circumference is 90)

<ADE + <EDB = <ADB
a + <EDB = 90
<EDB = 90 - a

now consider triangle DEB,
the sum of interior angles in a triangle is 180, so

<DEB + <DBA + <EDB = 180
<DEB + a + 90 - a = 180
<DEB = 90 (proved)

(ii) <ACB = 90 (angle subtended by diameter at circumference is 90)

in a cyclic quadrilateral, the sum of two angles facing each other is 180.

So,
<ACB = <FCB

<FCB + <FEB = 90+90 = 180

in a quadrilateral, the sum of all interior angles is 360, so

<FEB + <FCB + <CFE + <EBC = 360
180 + <CFE + <EBC = 360 ( <FCB + <FEB = 180 )
<CFE + <EBC = 180

Hence shown that FEBC is a cyclic quadrilateral.
thanks a lot! and the last problem if it isn't too much trouble....
 
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