If f(x)=x−[x],x(≠0)∈R , where [x] is the greatest integer less than or equal to x , then the number of solutions of f(x)+f(1x)=1  is

Given that f(x)=x−[x], ⇒f(1x)=1x−[1x] We have: f(x)+f(1x)=1 ⇒x−[x]+1x−[1x]=1 ⇒x+1x−1=[x]+[1x] ⇒x2+1−xx=  Integer k  (say)⇒x2−(k+1)x+1=0 Since x  is real,∴  (k+1)2−4≥0     (∵  b2−4ac≥0) ⇒k2+2k−3≥0   ⇒(k+3)(k−1)≥0 ⇒k≤−3  or k≥1 .Hence the number of solutions is infinite.