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 Kevin Arnold's messages in the community 1 2 3 4 GO Go to Page...
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 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

isit D)

u can do dis in 2 ways...1-P(1or 2 numbers appearing) which i guess has been done by sir..

so i shall tell the 2nd method..

we can select those 4 numbers in 6C4 ways...now 4 of the 6 tries are gonna contain these 4 numbers..the rest of the 2 tries also contain 2 of

the same 4 numbers...these can be 2 similar numbers or 2 different numbers...so we have 2 cases 1) 2 similar numbers, 2) different numbers

for 1) we have  6C4 * 4C1 *6!/3!...

for 2) we have  6C4 * 4C2 *6!/(2!)2

so event space= 1) +2)

sample space is 66

this way i guess the answer comes as D) if there has not been any agonising silly mistake on my part..

cheers!!!

 This Post 0 points    (0    in 0 votes )   [?]
chill fella's...not in syllabus...gud wrk tho'...
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

Sirs.. im sorry for spoiling this problem...im sure a lot of u mite be working on it...but iam leaving this site for some time...so ill post the solution...

to be honest this problem wasnt solved by me and was given to me by my teacher...

it has very little to do with maninmasation...infact very little to do with maths...

just imagine that u have 4 bodies...each of mass 4, 13, 12, and 23 units...located at (1,2) ,(4,5) , (2,7), and (0,6)..

we are interested in finding the point about which the moment of inertia of the system is minimum...

now by parallel axis theoram, we have that

Iaxis = Icm + m (d)2    d the distance between the reqd. axis and the axis passing thru' the centre of mass...

so clearly minimum moment of inertia is achieved at the centre of mass of the system

(becoz, d =0)

here the centre of mass is (80/52 , 295/52)

so the value of Z for which F(Z) is minimum, is 80/52 + 295/52 i

hope there's no calculation error...even if there is, the method is the main thing...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

2 balls are connected by a inextensible string...(initial seperation L),  of length '2L'.....light and massless....the masses of the 2 balls are 'M' one ball is given a velocity in the upward direction of magnitude 'V'....describe all that happens after this...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

these functions( 2logx and log x2 ) differ in their domains hence they are not equivalent functions...

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

yes sir...

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]

hey guyz...till now i was trying hard to figure out a way of approaching these problems...i tried to work this one out and am getting my answer as 2pi root(8m/3K)

this is how i did it...let the extension of the upper most spring the midle spring and the bottom most spring be x1,x2 and x3 resp.

and the net distance travelled by the block from the eqbm. position be x...

kinematically we have,

2(x1/2  + x2) + x3 = x....(if upper most spring extends by x1 then the top most pt of the middle spring comes down by x1 /2....and if the middle spring extends by x2, then the lower most part of the mid spring comes down by ( x1 /2 + x2)....if this happens the lower most pulley moves down by the same ...now if the extension in the lower most spring =x3 then the block reletive to its initial eqbm. position wud go down by

2(x1/2  + x2) + x3 )

and for force balance we have,

2 (Kx1) = (2K) x2....1)

2(3K)x3= (2K)x2.......2)

solving the above 2 eqn.s and the above eqn, we have x3 = x /8

and we have tension acting on the block as 3/8K x= M a

so time period is 2pi root(8M/3K)...

i dont know whether its rite though....

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

hari sir and ankit...

here / z / means mod(z)...i apologise for usage of bad text...

and sir i didnt get ur point...

if x and y tend to infinity, then how does the summation/ Z-zi /2 tend to zero...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

SHM problem 1...

detailed solution needed...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

nice problem of maxima minima visualisation.....find the value of Z for which we have minimum value of F(Z)...

where F(Z)= 4/ Z-1-2i /2 + 13/Z-4-5i /2 + 12/ Z-2-7i /2 + 23/ Z- 6i /2

and find the value of F(Z)...(second part not a compulsion)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

maron is a wonderful buk...after doing it u do feel the joy of conquering the subject...even arihant differential and integral calculus are good buks...they start from the basics and take u high...

 This Post 0 points    (0    in 0 votes )   [?]
grrr8888!!!
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

x+y+z=1 and x,y,z>o

=> (1-x),(1-y),(1-z) >0

3-(x+y+z)=2

(1-x)+(1-y)+(1-z)=2

AM >=GM

=> (1-x)+(1-y)+(1-z)/3 > {(1-x)(1-y)(1-z)}^1/3

or 8/27>(1-x)(1-y)(1-z)

or max value of (1-x)(1-y)(1-z) = 8/27

cheers!!!

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

OK as no one except a few of them are interested to solve this one, i shall post the solution...

for E1  we need that there shud be no mutual intersections between the sets...

here the sample space = 2mn....(each set can make 2n selections of objects and there are a total of m sets

the event space is (m+1)n....(each object has m choices to go to any one of the m sets, and one choice of not going to any of the sets)

so P(E1) = (m+1)n/2mn...

for E2 we need that there shud be no element such that it shud lie in all the sets....

sample space is again = 2mn and the event space = (2m-1)n...(each element can make any choices amongst the m sets except one choice when it chooses to go to every set)

so P(E2) = (2m-1)n/ 2mn....

for 2 events to be mutually independent P(A^B)=P(A)*P(B)..

hare P(A^B)= nothing but (m+1)n/2^mn....

so putting the values we have n= anything, but m=1...

cheers!!!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

every box that contains a ball contains only 1 ball...so we have to choose m amongst the 2m boxes that will contain balls...this can be done in

2mCm ways...i am not able to read out what u have written in the second question...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

pink ele has given the right and apt explanation tina...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

http://www.goiit.com/posts/list/algebra-find-the-range-of-sinx-cosx-tanx-cosecx-secx-cotx-912096.htm

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

http://www.goiit.com/posts/list/algebra-find-the-range-of-sinx-cosx-tanx-cosecx-secx-cotx-912096.htm

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

http://www.goiit.com/posts/list/algebra-find-the-range-of-sinx-cosx-tanx-cosecx-secx-cotx-912096.htm

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]

you dont need to post the same problem a hundred times...

the max value of the expression is infinity, which occurs at pi/2...

now for the minimum value,

|sinx|,|cosx|,|tanx/,|cosecx|,|secx|,|cotx| are positive numbers,

so AM >= GM

=> {|sinx|+|cosx|+|tanx|+|cosecx|+|secx|+|cotx|} /6 > {|sinx|*|cosx|*|tanx|*|cosecx|*|secx|*|cotx|}1/6

=> |sinx|+|cosx|+|tanx|+|cosecx|+|secx|+|cotx| >= 6

so f(x)=|sinx|+|cosx|+|tanx|+|cosecx|+|secx|+|cotx| lies between [6, infinity)

cheers!!!

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