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#126 Re: Help Me ! » the imaginary equation of a circle? is that even possible » 2014-03-15 06:12:18
thanks bob i was all messed up thinking that the question is just out of my league... but what if the circle was mod(z)=2 what would be the other two vertices z2 and z3?
#127 Re: Help Me ! » quadratic » 2014-03-15 05:24:12
hi bob bundy i really struggle at such questions whenever i see "find the possible number of values of a for blah blah equation having a as coefficient" can you suggest me any way of tackling such questions btw i did not understand your answer from the part after the graph like how did you restrict the value of a?
#128 Re: Help Me ! » the imaginary equation of a circle? is that even possible » 2014-03-15 05:07:26
well the mod of complex number gives it's distance from the origin of argand plane but the mod of z1 is sqrt(1+3)=2 so it's inside the given circle? or is my concept of modulus of a complex number wrong?
#129 Re: Formulas » Integrals » 2014-03-15 05:04:25
integral of cot(x)= -cosec^2(x)+c
??
Where did that come from?
Bob
ganesh wrote that in his list of formula's just check it once.. you will find my question and i don't think i misunderstood integration for differentiation. if you can show me how is it that we have two formulas for the integration of same function i would appreciate that alot
#130 Re: Help Me ! » the complex plane- my tragedy » 2014-03-15 04:56:23
no, i did not know can you send me the link?
#131 Help Me ! » the imaginary equation of a circle? is that even possible » 2014-03-15 04:54:01
- gourish
- Replies: 6
i read a question saying" if z1 z2 and z3 are the vertices of an equilateral triangle inscribed in a circle mod(z)=3 and z1=1+isqrt(3) what is z2+z3" but the equation of a circle doesn't seem to right (that's what i think but i am not sure) can anybody please explain me is this question even right and is it possible to have such an equation for a circle (in argand plane). i would like a simple explanation which a high school going guy can understand. thanks for the help guys appreciate it.
#132 Re: Help Me ! » the complex plane- my tragedy » 2014-03-15 04:48:00
thanks good to know that.... 
#133 Help Me ! » the complex plane- my tragedy » 2014-03-15 04:03:43
- gourish
- Replies: 5
if z=2+ia y=b+i2 x=0 are vertices of an equilateral triangle such that a, b are
real numbers such that a, b ∈(0,1) then which of the following is true?
1.)a=b=4+2sqrt(3)
2.)a=2, b=sqrt(3)
3.)a=sqrt(3),b=2
4.)a=b=-1+sqrt(2)
i found that a=b=4+2sqrt(3) or 4-2sqrt(3) but due to the condition a,b ∈(0,1) i choose 4-2sqrt(3) but it's not even in the options the answer sheet says that the right answer is 4.)
why is it so?can anybody please help me out on this one
#134 Re: Help Me ! » What am I doing wrong? #Arithmetic_Progressions » 2014-03-12 05:13:57
sorry it's actually n=2m-1 and the ratio comes out to be 6m+1/4m+5 which is the same as got by Elaina but much faster and simpler.. the idea was to get the m'th rem of the A.P. but taking n-1=2(m-1) we get that term on both numerator and denominator and cancel out the '2' which is common to both of them
#135 Re: Help Me ! » What am I doing wrong? #Arithmetic_Progressions » 2014-03-12 04:56:04
sorry to jump in but i do think that the ratio is for the sum of n terms of two series just solve it like
3n+4/2n+7=sum of n terms of Ist a.p/sum of n terms of 2nd a.p.
3n+4/2n+7=n/2*{2a+(n-1)d)/n/2*{2b+(n-1)c).....-(i)
where a and b are first terms of 1st a.p and 2nd a.p and d and c are common difference of 1st and 2nd a.p
solving that and putting n-1=2(m-1). n=2m+1 and hence using this value of 'n' the R.H.S. of eq. (i) becomes the ratio of mth terms of 1st and 2nd terms while the L.H.S. gives its value in terms of 'm' i had this problem in my higher secondary exams and this was the right answer
#136 Re: Help Me ! » IIT JEE ADVANCE and maths » 2014-03-12 04:42:14
thanks Shivam i just needed somebody like you to tell me "take it easy" thanks i will try to follow your advice
#137 Re: Help Me ! » find the answer of this series » 2014-03-12 04:38:37
oh cool! thanks bobbym.... those two forms seem quite interesting i might even challenge others to solve questions based on these...
#138 Re: Help Me ! » hello guys i have got another question » 2014-03-12 04:37:04
thanks i did too... 
#139 Re: Formulas » Integrals » 2014-03-12 02:09:58
integral of cot(x)= -cosec^2(x)+c but integral of cot(x)=log(sin(x))+c why do we have two results for the integration of the same function? @ganesh
#140 Help Me ! » hello guys i have got another question » 2014-03-12 02:04:52
- gourish
- Replies: 2
if a,b and c are in geometric progression and x and y are the arithmetic means of a and b and b and c repsectively what is the value of
(a/x)+(c/y)=?
options range from : 0,1,2,3,4,5,6,7,8,9
please explain how to solve this
#141 Re: Help Me ! » find the answer of this series » 2014-03-12 00:44:20
thanks i appreciate your help
#142 Re: Help Me ! » IIT JEE ADVANCE and maths » 2014-03-12 00:41:42
thanks Shivam i am about to write the JEE next year but i have been having trouble trying to solve the problems from M.L.Khanna is there any plan that i should follow to finish the book and it's problem in this limited time period? i would appreciate it if you can tell me how you practiced for maths and chemistry cause physics has been very easy for me
#143 Re: Help Me ! » find the answer of this series » 2014-03-12 00:36:10
beacuse i was challenged to find one innovative and simple smart way of finding the answer to the nested irrational btw thanks for the term
#144 Help Me ! » IIT JEE ADVANCE and maths » 2014-03-11 04:23:03
- gourish
- Replies: 5
hi guys... i am having tough time dealing with iit jee maths... can anybody who has attempted it and has been successful please give your precious advice to me... what do i need to practice maths for jee advance exam plus i tried M.L.KHANNA's book it's too huge and time consuming to even solve a single chapter how can i solve them all?
#145 Re: Help Me ! » find the answer of this series » 2014-03-11 03:40:00
yeah same here... but i used log on this series and made it easier... but i am looking for some innovative way of finding the answer
#146 Help Me ! » find the answer of this series » 2014-03-10 02:55:16
- gourish
- Replies: 13
hi guys.... i was given a challenge to find the answer to the series that i am going to give you all now... i was asked to find it's answer in the simplest and shortest form. i actually did find it's answer but in a 'page long' form and i wanna confirm it with you guys so please try to find it's answer and tell the shortest approach to it's answer
find the value of sqroot(9*sqroot(8*sqroot(7*sqroot(6*sqroot(5*sqroot(4*sqroot(3*sqroot(2*sqroot(1)))))))))
FYI i do know that sqroot of 1 is just one but just for the consistency of the series it's there.... i hope to get it's answer pretty soon
thank you guys
cheers....:)
#147 Re: Help Me ! » basic calculus » 2013-06-02 03:32:12
Hi;
basic calculus is all you need
Welcome! Who said that? Need for what? For physics? Yep! For computer science? Nope!
learn to apply them in any type question for exams or real life situaions
That will take a bit longer. I have been trying for 86 years!
t
well, i need it for physics and one of my friends, Mohit said me that...
#148 Re: Help Me ! » basic calculus » 2013-06-01 01:54:38
thanks guys..... i will surely look into it and where can i find "sal's videos" @Shivamcoder3013....
#149 Help Me ! » basic calculus » 2013-05-28 05:01:39
- gourish
- Replies: 49
i have heard all about extremely uneasy things about calculus... "it's a throat-chocking concept and so on" but, thank god that few people have also said things like "basic calculus is all you need and it isn't a big deal". i have heard few things about integration and differentiation and also read the topic of limits on mathisfun.com.... i would like to learn basic calculus and learn to apply them in any type question for exams or real life situaions...:)