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

please find the equivalent resistance for the above arrangement with ur expln..........

limit of tan(3/5^x)  / tan ( 5/3^x ) as x tends to infinity

lt_{x----infinity}    tan(3/5^x) / tan(5/3^x)

 apply L Hospital's rule i.e. diff. num and den. w.r.t. x

 =     lt        sec²(3/5^x) [-3.5^-xlog5]   /  sec²(5/3^x)[-5.3^-xlog3]

 =     lt         (3/5)^x log₃5

as x tends to infinity .........(3/5)^x tends to 0 as 3/5<1

hence the limit is 0

 guys n gals pls ans...............

please someone answer yaar........