If (1+x)∈[1,2] then f(1+x)=1+(1+x)=2+x Similarly If ((1+x)∈(2,3) then f(1+x)=1+(1+x)=2+x Similarly, if (3−x)∈(0,1) then f(3−x)=1+(3−x)=4−x $g(x)=\{\table 2+x,{ ; },0 ≤ x<1;2-x,{ ;} ,1 At x=1 LHL−
lim
x→1−
g(x)=
lim
x→1−
(2+x)=3 RHL=
lim
x→1+
g(x)=
lim
x→1+
(2−x)=1 Here, g(x) is not continuous at x=1 At x=2 LHL=
lim
x→22
g(x)=
lim
x→2−
(2−x)=0 RHL=
lim
x→2+
g(x)=
lim
x→2+
(4−x)=2 Here, g(x) is not continuous at x=2 Therefore, the function g(x) is continuous for all x∈[0,1)∪(1,2)