If `int_(0)^(x){t}dt=int_(0)^({x})t dt ("where" x gt0 neZ and and {*}` represents fractional part function), then

Options

A. `x in (0,1)`
B. `[x]=1`
C. `x in (1,6)-I`
D. none of these

Answer

A

Explanation

Let `x in (k,k+1)` where `k in Z`. Then,
`underset(0)overset(x)int{t}dt=underset(0)overset({x})int t dt`
`rArr underset(0)overset(k)int(t-[t])dt+underset(k)overset(x)int(t-[t]) dt=underset(k)overset({x})int t dt`
`rArr underset(r=1)overset(k)sum underset((r-1))overset(r)int(t[t])dt+underset(k)overset(x)int(t-[t])dt=(1)/(2){x}^(2)`
`rArr underset(r=1)overset(k)sum underset((r-1))overset(r)int{t-(r-1)}dt+underset(k)overset(x)int(t-k)dt=(1)/(2){x}^(2)`
`rarr (1)/(2)underset(r=1)overset(k)sum[-{t-(r-1)}^(2)]_(r-1)^(r)+[ (1)/(2)(t-k)^(2)]_(k)^(x)=(1)/(2){x}^(2)`
`rArr underset(r=1)overset(k)sum1+(x-k)^(2)={x}^(2)`
`rArr k+(x-k)^(2)={x}^(2)" " [ :. x in (k,k+1) :. {x}=x-k]`
`rArr k=0`
`:. x in (0,1)`