Answer:
option (a) and (c).
Step-by-step explanation:
(a) The function f(x) is given as:
which could also be represented as:
clearly the limit of the function at x=2 exist (since f(x)=x+2, and limit x→2 f(x)=2+2=4)
and also f(x) is a polynomial function and hence is continuous at x=2.
(b) The function f(x) is given as:
when x≠2 ( done in part (a))
when x=2
<em>for checking the continuity of a function at a point x=a we must know that:</em>
<em>The left hand limit(L.H.L) of f(x)(at x=a)=Right hand limit(R.H.L.) of f(x)(at x=a)=f(a)</em>
so from here we get that L.H.L=R.H.L=4 (since f(x)=x+2,and limit x→2 f(x)=2+2=4)
but f(2)=3
Hence, the function is not continuous.
(c) The function f(x) is given as:
when x≠2
when x=2
Now the L.H.L=R.H.L=4 (since f(x)=x+2, and limit x→2 f(x)=2+2=4)
Also f(2)=4
Hence as L.H.L=R.H.L=f(2)
Hence, the function is continuous.
(d) As function in part (b) is not continuous.
Hence option (d) is incorrect ( All are continuous at x=2)
Hence, the answer to this question are option (a) and (c).