Correct Answer - C
f(x)=g(x)
`rArr {x}+{2x}=[x] " "`……………..(i)
Clearly, it is true for x=0.
we observe that LHS.i.e{x}+{2x} lies between 0 and 0 for all values of x.
`:. 0 lt [x] lt 2 rArr 1 le x lt 2`.
x=1 does not satisfy equation (i). Therefore, `1 lt x lt 2`
Now, following cases arise.
CASE I When ` 1 lt x lt 1.5`
In this case, {x}=x-1,{2x}=2x-2 and [x]=1
`:. {x}+{2x}=[x]`
`rArr x-1+2x-3=1 rArr x=(5)/(3)`
Hence, x=0,`(4)/(3),(5)/(3)` are the values of x satisfying f(x)=g(x).