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Check the third step. There should be a 2 there
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Plz make sure u didn't forget to include the limits of integration in Q2
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Well I went way wrong with that. Sorry dude ... the occasional slip !
The number of ways in which each element in the domain can be linked to either of the two elements in the co-domain = 2n
Now, there are two possibilities that we must overrule to make it onto, that no element in the co-domain is devoid of a pre-image. The above linkings will have 2 such cases, when either one of the elements in the codomain is unlinked.
So total number of onto functions = 2n -2
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Excellent solution Vinu
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For the ball thrown up
-h = ut1 - (1/2)gt12
h = -ut1 + (1/2)gt12 --------(1)
For the ball thrown down
-h =- ut2 - (1/2)gt22
h = ut2 + (1/2)gt22 --------------(2)
For the freely falling ball
-h = - (1/2)gt2 -----------(3)
Multiply (1) by t2 & (2) by t1 and add
h(t1 + t2) = (1/2)gt1t2 ((t1 + t2)
Cancelling (t1 + t2) and substituting the value of h from (3)
(1/)gt2 = (1/2)gt1t2
t =  t1t2
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Start from rest.
Keep applying the force and move in a closed path.
If the force is conservative, it won't do any work & by work energy theorem, there will not be any change in KE. So when you reach the starting point your particle will again be at rest.
If this doesn't happen, the force in non conservative.
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Sneha has decided that she won't rate me!!!
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The answer given in the book seems to be incorrect. My approach is a bit different from urs.
The left portion of the string is a variable mass system in which mass is entering.
By Merchersky's equation,
m(dv/dt) = Fext + urel (dm/dt) = Fext + urel (dm/dx)(dx/dt)
As the left portion of the chain is in equilibrium,
0 = {T-(m/L)[(L/2)+(x/2)]}j +(- 2gxj)(m/L)( 2gx)
T = (mg/L)[(L/2) + (5x/2)]
= (W/2)[1+(5x/L)]
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This problem is best solved by a v-t graph
 
The slope of the first line is 'a' and that of the second is "-b"
Let the maximun velocity be vm
Then, vm/T = a  vm/a = T ---------(1)
& vm/t-T = b vm/b = t-T ---------(2)
Adding (1) and(2)
vm = abt/(a+b)
As the v-t graph is entirely above the t axis,
So distance travelled = Displacement = Area under v-t curve = (1/2)vmt
=abt2 /2(a+b)
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Let the bottom of the inclined plane be the origin of a polar coordinate system
Then
d2 r/dt2 = -gsin
d/dt = w
dr/dt = 0 t (-gsin )dt = g0 t sin(0 + wt) dt = (g/w)[1-cos(0 + wt)]
where 0 is the initial angle of incline.
Acceleration of the block
a = (d2 r/dt2 - r(d/dt)2 )er + (2(dr/dt)(d/dt) + r (d2/dt2))e
= -(gsin + rw2)er + (2(g/w)[1-cos(0 + wt)]w + r (dw/dt))e
= -(gsin + rw2)er + (2g[1-cos(0 + wt)] + r (dw/dt))e
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5h/9 is the distance travelled in the last one second and not upto T-1th sec
So distance travelled till T-1 th sec = h - 5h/9
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If you want to understand HOW this occurs, Huygens construction (given in every standard textbook) will prove useful.
Light is an electromagnetic wave comprising of electric and magnetic components. We usually study light using the electric component. In higher physics, reflection & refraction can be explained in terms of this electric vector. But don't try to break ur head trying to understand this. Read up Huygen's construction. That should be fairly good explaination
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The addition on the left hand side makes sense only if x is an +ve integer, for we can add something to itself only in steps of +ve integers
So the sum = x2
But the graph of this function will be a collection of points at integral values of x
So the sum isn't differentiable.
Thats the fallacy ...we are differentiating something that is non differentiable
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Are you sure of the relevance of this question??? Where did u pick it up?
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Well bro ... if you have done all the books you say you have done, you don't need to worry. Best strategy is to review the work you have done, identify you weak areas and work on them.
Get any previous year question bank. Work it out thoroughly. That should help a lot.
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1)
{1/[3+(8tan(x/2) / (1+tan2(x/2))]} dx
{sec2(x/2)/[3+3tan2(x/2)+(8tan(x/2) ]} dx
Substitute tan(x/2) = t
2{1/[3+3t2+8t]} dt
(2/3) {1/[t2+ 8t/3 + 1]} dt
(2/3) {1/[(t + 2/ 3)2- (1/ 3)2]} dt
(2/3)(1/2 3)ln[( 3t + 3)/( 3t + 1)]
(1/3 3)ln[( 3tan(x/2) + 3)/( 3tan(x/2) + 1)]
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