Messages posted by GOPAKUMAR

37.5 ml

4.84 ohm

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

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

(a+1)/a € (1,∞)

a+(1/a)€ [2,∞)

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

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

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

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

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

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 log_b(xy) = log_bx + log_by. and log_b(x/y) = log_bx - log_by.}

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

therefore

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

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

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

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.

THE STRUCTURE OF KH₂PO₂is

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

(n₁T₁+ n₂T₂+ n₃T₃)/( n₁+ n₂+n₃)

(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

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