Messages posted by Kevin Arnold

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!!!

chill fella's...not in syllabus...gud wrk tho'...

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

I_axis = I_cm + 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...

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...

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

yes sir...

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....

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-z_i/² tend to zero...

my first mechanics thread...

SHM problem 1...

detailed solution needed...

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)

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...

grrr8888!!!

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!!!

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 = 2^mn....(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/2^mn...

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

sample space is again = 2^mn and the event space = (2^m-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) = (2^m-1)ⁿ/ 2^mn....

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!!!

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

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

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

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

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

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

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,

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

cheers!!!

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