1. find domain f(x)=underroot(x-1/x-2{x}) where {.} denotes fractional part function.

2. if [x]=[x/2] +[x+1/2],here[.] is greatest integer function. and let n be a positive integer den show tht

[n+1/2] + [n+2/4] + [n+4/8] + [n+8/16] +...........=n.

Hi Manuj,

Answer: underroot(x-1/x-2{x}) is defined for any R i.e (-infiniy,+infinity) except at 0

Solution . Since Underroot(t) function is defined only for t>0, thus x-1/x-2{x} >0

this can be possible only under two cases.

Case I: x-1 and x-2{x} are simultaneously > 0

Case II: x-1 and x-2{x} are simultaneously < 0

And also x-2{x} should not be = 0 ==> x not= 2{x} ==> x not= 0 -----(1)

Evaluating Case I: x>=1 and x>2{x}

Evaluating Case II: x<1 and x<2{x}

For evaluating both the case xe need to graph rough sketches of their graphs of x and 2{x}

As clear form the graph x<2{x} for x<-1 and x>2{x} for x>=1

Note: A slight intersection tends to appear on two but the intersection doesnot happens.

So the domain is clearly (-infinity to +infinity)-{0}

Hi Manuj,

A little tricky question but i m sure u would understand

= [n+1/2] + [n+2/4] + [n+4/8] + [n+8/16] + [n+16/32] + [n+32/64] .....

= [n/2+1/2] + [n/4+1/2] + [n/8+1/2] + [n/16+1/2] + [n/32+1/2] + [n/64+1/2] ......

Now lets add [ n / 64 ] to this equation and see what happens ...

=  { [n/2+1/2] + [n/4+1/2] + [n/8+1/2] + [n/16+1/2] + [n/32+1/2] + [n/64+1/2]  ....} + [n/64]

Now we know that [x]=[x/2] +[x/2+1/2],

Thus   [n/64]+[n/64+1/2] = [n/32]

Now the equation becomes

[n/2+1/2] + [n/4+1/2] + [n/8+1/2] + [n/16+1/2] + [n/32+1/2] + [n/32]+...

Again  [n/32+1/2] + [n/32] = [n/16] and the equtaion becomes

[n/2+1/2] + [n/4+1/2] + [n/8+1/2] + [n/16+1/2]+ [n/16]+...

Again   [n/16+1/2] + [n/16] = [n/8]

[n/2+1/2] + [n/4+1/2] + [n/8+1/2] + [n/8]...

Whis process went over and over till a situation like this come into effect :

[n/2+1/2]+[n/2] = [n] = n(since n is a positive integer)

The answer will definitely be n

So u see how adding a single term in the series can reduce the entire equation. since it is an infinite series, so the last term of the series is [n/2^(infinity)+1/2 ]

Now if we add [n/2^(infinity)] (definitely be  0 as n/2^(infinity) is a very small quantitiy and would surely be ->0) to this term .  The equation will definitely get reduced to n

thanks buddy..........ur answer for 2nd question is really great..........