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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}
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..........
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