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-------wrong soln------
apologies :(
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u can prove
3<log (base2) 10 <4 -----using the fact that log(base a) b= 1/(log (base b) a) and also that log 2(base 10) is +ve.
as the graph of log(base a) x is increasing for a>1 so we can say that,
log 8<log 10<log 16 ------ all base 2
i.e 3< log 10<4
hence proved!! :)
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Thomas nd Finney??
leave dat for ur 1st sem in IIT!!
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bounded fuctions are those which hve both upper nd lower bounds... like u can restrict the function between 2 finite values!
like cos/sin are bounded in there complete domain but tan/cot etc aren't!
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solve Arihant claculus nd then for revision purpose solve A Das.
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if this quad. eqn. has complex roots then they will exist in conjugate pairs
so,
let x+iy=z1 nd x-iy=z2 be the two roots
so from here u get the product of roots= c/a >1 (as given)
as the product of roots= (x^2+y^2)>1
so
lz1l,lz2l>1
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see if u hve a function f whose limit u want to find out , but it isnt coming out by the simple methods u know then u try to find functions of nature such that one function 'g' is always greater than or equal to 'f' nd another function 'g' which is always less than or equal to f.
here one thing is to be noted that f,g, h all the functions are continuous in the region aroung the point of ur concern.
now let say u are talking abt the limit of f at a point c then
as x approaches c 'g' and 'h' approach to the same limit then it is understood that f will have the same limit.
as f is continuous and less than equal to h and greater than equal to g , so by this we get the limit of f as x approaches c is same as that as for g and h.
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its not the case always!!
:)
u take ne frame of reference fixed to earth(having zero or constant velocity) as inertial although earth is rotating (and so accelerating also)
see the concept behind this is the symmetricity of the things with time, when u take ne event occuring (in earth 's frame ) then the displacement is very small if the time isn't large , so can approximate it to be the inertial frame, nd ne thing that has non zero acceleration (although may have velocity) is non inertial.
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see the concepts are the same which ever book u take up!
the difference lies in the way it is brought up, for this i feel in maths arihant (for iit jee) nd RD Sharma are both gr8 books!!!
the best book i feel for ne exam objective type is R D Sharma ,nd arihant IIT JEE objective is aslo gr8.
though RD Sharma is bit long in its exercises , it has gr8 questions (which are potential aieee questions)
no matter it's for iitjee , u can also use it to prepare for ur aieee exam.
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hey it's not asking 4 the equ. extension rather the max. extension. see when u leave this bolck when the spring is un extended then the bolck exerts sudden force on the spring which makes it go to n extension of mg/k then also moves ahead because at that instant the velocity of the block isnt zzero.
so to get the max. ext. u hve to apply energy balance to get the max. ext.
so applying the same u get ,
mgX=0.5kX^2
from here u get X=2mg/k
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well i think u wanna know what's the least rank to get into IIT.
is it??
then the answer is about 3800. but the merit list (main one) also includes admission into IT BHU nd ISM Dhanbad also, which makes the main merit list to go to about 4400.
but y r u thinking for the least rank to get in??
rather feel confident that u will surely crack JEE .
all the best!
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just solving a locus prob.
just take the coorrdinate obtained as (X,Y) nd then eliminate C, which is a variable.
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take y=c as any line || to the axis
so u get point of intersection is,
(c^2/(4a),c) and (-c^2/(4b),c)
so the mid pnt= ((c^2/8)(1/a-1/b),c)
now c is a variable so eliminate it,
so c=[8/(1/a-1/b)X]^(1/2)
and also c=Y
so u get Y^2=8/(1/a-1/b)X
which is indeed a parabola.
hence proved!
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what u require here is the union of the three sets
well i dont agree wid the answer u provided, coz u can urself crosscheck it if u r asking for atleast div. by one of three then for only div. by two case u will have 500 such numbers in 1-1000.
no req. = no that are div. by (2, 3, 5)-no. divisible by(2&3,3&5,2&5)+no. div by (2&3&5)
=500+333+200-166-66-100+33=734
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hi!!!
see fiitjee GMP is a good thing for brushing up ur concepts,coz it has many questions from which u can do ur revision work alongwith u will learn some nice methods of solving questions.
nd talking of the crash course then buddy i cant say much of it coz i dont know how does it work up with ppl trying it!
i had done the GMP nd it was nice [but of not much use in iit jee 2007 ,as u wud also affirm] ,
u can even try the RTPF they provide
it's excellent in preparing for the test coz they give around 14-15 test papers to solve at home. u can use them also. but they give it in FEB so decide upon how u gonna do ur revision nd then attack those papers!
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Hum woh patte nahi hain Joh tutt ke girr jayeen, Jaoo jake kehdo aandhiyon se Apni aukaat me rahein.
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k misunderstood the question, so , (2a-1,3-2a) lies on a line y=-x+2 seeing it graphically u can see that the line passes from (2,0) and intersects y=-  3 at x=2+ 3 so u get restriction on x as x lies between 2+ 3 and 2 so a lies between 1/2(3+ 3) and 3/2 -----this is the answer required. and regarding the area, u get area of both segments + rectangle in between = [8pi/3-2 3]+2.2 3=8pi/3+2 3 if unclear then do ask!!!
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k misunderstood the question, so , (2a-1,3-2a) lies on a line y=-x+2 seeing it graphically u can see that the line passes from (2,0) and intersects y=- 3 at x=2+ 3 so u get restriction on x as x lies between 2+ 3 and 2 so a lies between 1/2(3+ 3) and 3/2 -----this is the answer required. and regarding the area, u get area of both segments + rectangle in between = [8pi/3-2 3]+2.2 3=8pi/3+2 3 if unclear then do ask!!!
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