int e ^ (Sin^2 t )  dt       ..................i.e. integral of e raised to Sin square t with respect to t

F = m a

dF/ dt = m da/dt

2000 = m (da/dv)  (dv/dt)

2000 = m  (da/dv) a

2000 dv = ma da

2000v = ma^2 /2

4000v = ma^2

4000*25 = 1000* a^2

a = 10

applying v2 = u2 - 2as

     where v=0

hence u2 = 2 * 10 * s

          25 ^2 = 20 s

hence s= 31.25 m

if wrong please correct

 for the first question 

=            int(1 - 2Sin(x/2)Cos(x/2) ) / 2 Sin^2(x/2)

=             int {    Cosec^2(x/2) - Cot (x/2)  }

=            -int{ Cot (x/2) - Cosec^2(x/2) }

=            -  Cot(x/2)                          .....................because Integration e^x ( f(x) + f1(x) ) = f(x)

= 1

f(x) is an even function and hence

f(x) = f(x) or f(x) = f(-x)

so, when f(x) = f(x+1/x+2)

1) x = (x+1)/(x+2)

  x= (-1+- rt5)/2

2) x = - (x+1)/(x+2)

   x = (-3+-tr5)/2

since all the 4 values lie between -5 and +5 the answer is satisfied by all the options

do u get the answer as alll the above

pls help guys........

pls reply

refer the 1st fig

M >>m hence the mass 'm' is negligible w.r.to 'M' ,,,,,,, so M falls freely with accln 'g' and so the body 'm' moves up with accln 'g'

hence tension in the string = m(g+a) = m(g+g) = 2mg = T

now refer 2nd fig

tension on either side of pulley = T

hence F = 2T

= 2(2mg)

= 4mg

(4)  reger the fig below

weight of block = mg and force acting on block upwards = FSinΘ

if R is the reaction of the block on the ground then R = mg- FSinΘ

coeff. of riction = tan?

frictional force = Rtan? = (mg- FSinΘ) tan?

this force should be equal to the horizontal component of the force applied, hence

FCosΘ = (mg- FSinΘ) tan?

simplifying F = mgtan?/ (CosΘ + SInΘtan?)

hence F = mgSin?/(CosΘCos? + SinΘSin?)

            F = mgSin? / Cos(Θ-?)

(2) refer to the fig.............both the frictional forces acting on both bodies are opposite to the force F applied

   let the acceleration of the lower block be 'a' .........total mass = 2m

     hence F- ( μmg + μmg) = 2ma

hence a= F/2m        -  μg

(1) for the body to fall freely consider the fig below..........the accln of block along the directions indicated as shown........same is the case with the wedge

condition for the block to frrely is aSin? = gCos?

hence a = g Cot?

the graph of the given function may be shown as below

the required integral is the ares of the shaded region.............

area of each triangle = 1*0.5/2 = 1/4

from -100 to +100 there are 200 such triangles

hence the required area = 200 * 1/4

                                             = 50

please correct if wrong