|

 GOPAKUMAR's messages in the community 1 2 3 ... 12 13 14 GO Go to Page...
Message
 Discussion Forums -> This Post 17 points    (3    in 4 votes )   [?]

37.5 ml

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

4.84 ohm

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

now see

x^2+y^2+z^2=K^2

is an equation of a sphere of radius k now for real sphere k>0 regardless of x,y,z being positive number

therefore x^2+y^2+z^2> 0

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

@ guru dev do you mean (x^2+y^2+z^2)>=___________

 Discussion Forums -> This Post 17 points    (3    in 4 votes )   [?]

(a+1)/a € (1,∞)

a+(1/a)€ [2,∞)

 Discussion Forums -> This Post 20 points    (4    in 4 votes )   [?]

NOW See from dig. EF=AB=AD+GB  i.e  i.e r=4/9

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

IS UR EXAM ON 11 MAY OR WAS IT ON 1ST MAY ......IF IT IS ON 11 MAY U CAN USE THE ONLINE ADMIT CARD AVAILABLE AND GET IT ATTESTED ON THE DAY OF EXAM ELSE IF YOU HAVE ALREADY GIVEN IT ON 1ST MAY THEN SORRY iI CAN'T HELP YOU

 Discussion Forums -> This Post 17 points    (3    in 4 votes )   [?]

therefore Range of f is[3/4,1]

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

f(x)=Cos2x+sin4x =1- sin2x+ sin4x

Put  sin2x=t

Then

f(x)=1-t+t2

now since sin2x €[0,1] therefore t€[0,1]

therefore f(x) €[3/4,1]   since for  t€[0,1] the max & min value of f(x) are 1 & 3/4  at t=1/2 & t=1or0 respectively

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

pls re type ur ques I guess u used wrong symbols?

 Discussion Forums -> This Post 20 points    (4    in 4 votes )   [?]

f(x)f(4) =cos(logx)*cos(log4)

=cos(logx)*cos(log4)*2/2

= { cos(logx+log4)+cos(logx-log4) }/2

{since    2cos(a)cos(b)=cos(a+b)+cos(a-b)}

={cos(log4x)+cos(log(x/4))}/2

{since logb(xy) = logbx + logby.  and  logb(x/y) = logbx - logby.}

={f(4x)+f(x/4)}/2---------------------------------------------------------------(1)

therefore

f(x)f(4) - 1/2{f(x/4) + f(4x)}=0                                              [from(1)]

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

range={cos(2),cos(1),1}

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

see, acc. to the ques the max seperation b/w the to particles is D+A  i.e when 1 particle is at max. amplitude the other must be at mean position.

THAT WHEN ONE HAS ZERO DISPLACEMENT THE OTHER MUST HAVE MAX DISPLACEMENT SO CLEARLY THE PHASE DIFF. MUST BE

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

The tie-break criterion in the CML and category lists adopted for awarding ranks to the candidates who have scored the same aggregate marks is as follows:
For each subject, the average marks will be calculated on the basis of the marks obtained by those candidates who have scored more than or equal to MQMR in that subject.
Among the candidates having the same aggregate marks, a candidate will be ranked higher than the rest, if he/she has scored higher marks in the subject having the lowest average marks calculated as above. If there is a tie after this procedure, then the marks obtained by the tied candidates in the subject with the second-lowest average marks will be considered, and the above procedure will be repeated. Candidates tied even after this procedure will be given the same rank.

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

THE STRUCTURE OF KH2PO2 is

therefore the O.s of P is +1

while the peroxygroup the os is -1 because in this two OXYGEN atoms share a charge of -2

 Discussion Forums -> This Post 17 points    (3    in 4 votes )   [?]

 Discussion Forums -> This Post 17 points    (3    in 4 votes )   [?]

(n1T1+ n2T2+ n3T3)/( n1+ n2+n3)

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]

(x-y)(xw-y)(xw2 -y)=x3 (1-y/x)(w-y/x)(w2 –y/x)                                         _______(1)

Now if 1,w,w2 are roots of equation k3 -1=0

Then

k3 -1 =(k-1)(k-w)(k-w2)

now if k=y/x

then  (y/x-1)(y/x-w)(y/x- w2)     =  - (1-y/x)(w-y/x)(w2 –y/x)  =  (y/x)3-1

=>      (1-y/x)(w-y/x)(w2 –y/x) = 1-(y/x)3

then eqn (1) becomes

x3 (1-(y/x)3)       =x3-y3

 1 2 3 ... 12 13 14 GO Go to Page...
 Go to:  Select a forum   Board Exams - CBSE, ICSE, State Boards   Course Material   Computer Science   Physics   General Physics   Mechanics   Optics   Thermal Physics   Electricity   Magnetism   Modern Physics   Mathematics   Algebra   Trignometry   Analytical Geometry   Differential Calculus   Integral Calculus   Vectors   Chemistry   Physical Chemistry   Organic Chemistry   Inorganic Chemistry   Institutes & Courses   Coaching Institutes & Course Material   About IITs and JEE   Non IIT Institutes   Counselling Zone   Parents Corner   Parent Discussion Board   Biology (for Medical Entrance)   Botany   Zoology   Fun Zone   Lounge   Games, Puzzles and Quizzes   General Knowledge

Name

Mobile

E-mail

City

Class