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thinking on this quite revolting..!!!
look .. let u have P(n) describing fibonaaci series..having degree n..
now by method of difference u observe tht there is absoltely no change in series..so fibonaci series ..sum has same make-up (type of equation) as tht of series itself...
it is quite revolting tht..i think..it shud have no definite equation..
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the ques. is nothing but sm of arithmetic means is greater than sum of geometric means...
u noe. tht ..how b/w tw no.s a,b srithmetic and geometric means r inserted...
by observation...req. can be provd...
arte if useful..
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tough ..i feel..
poer functions show reversal of behaviousr only at 1..here even power is specified so -1 also comes into picture...
write ..
F(x) = f(x)/x^2n + g(x)..
now putting (1+h) ..
F(x) = f(1+h)/ inf. + g(1+h) . . .inf is to show tht denominator is tending to infinity..
at (1-h)..
F(x) = f(1-h)/ 0 + g(1+h)..
from these equations..it is clear tht..
1) ..the give func. are discontinuous ...
2) it will be continuos when f(x) = 0 ..(x-axis)..
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the given function is nothing but ...a^x .
u noe the expansion of functions by taylor's theorem..this is how it is derived...
we noe tht function a^x is continuous everywhere...(draw the graph)...
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u know the diameter end pt form of equation of a circle..
i.e.
now putting y=0..u get
so rq. equation is ..
x^2 - ax + b = 0..
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man111 ... saving nickels huh ?? :D:D...
i think u know tht method...multiplying sin(a-b) in both nr-dr.
tht's the fastest one i think...( tht's is cut & paste) :D:D
try by converting into exponential form(complex) approach..
or just ..wait for sir's reply..
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for sir .it is quite easy...but for us..it is quite a "ques.of the day"...
this typa ques. is difficult to come in jee as well...
as they have forced its max. level in complex this year..and they will not go beyond ..this max. -min. distance ques..
i think they will now force on triangle inequalities..as nth roots is not in our syllabus..
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rightly explained @vignesh
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R = abc / 4A..
R = 36a/ 4A.
R = 9a/A..
R = 9a / 1/2(h)(a)..
R = 18/ h..
R = 18/ root { 36 - (a/2)^2 }..
a = 48/5..
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..oye yaar!!..tht y is seperated..
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there is some problem i think..how can sum of +ve quantities be zero..
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100 J...
wenever there is water related problems..u can cut the tedious nature of buoyant force..by taking apparent weight into action..
so here apparent "potential energy " = W(ap)h ..now since no change of kinetic energy is observed ..
by WET..
work done by ext. force = change in pot. energy..(ap).
= 50 (1-1/3) x 3..
= 100..j
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1) ..since acceleration. req. is wrt trolley ..so make out a pseudo force...
now make fbd..
tht ma force (pseudo)..shud be horizontal opposite..now read carefully ..
since it is the case of max. deflection...so no.more angular displacement..this implies tht there is no angular velocity (already working in frame of trolley)...
so no normal(centripetal force)..here ..the only acc. is tangential..
so just resolve the two forces..pseudo and weight in the tangential direction..and get the tangential acc. of bob wrt trolley ...
cheers!!!
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operations in the last step valid only for integers...
since valid for only integers..so discontinuity..hence non-differentiability...
rate if useful..
cheers!!
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..@ abv. friends these r by no means simpler methods...
simplest method is wat the ques. is designed for..
write cotA as 1/tanA. and cotB as 1/tanB..it is over in single step...
cheers !!
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