Answer :

The given function is

The function f is defined at all points of the real line.


Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1 or k < 1.


Now, Case I: k < 0


Then, f(k) = 2k


= 2k= f(k)


Thus,


Hence, f is continuous at all points x, s.t. x < 0.


Case II: k = 0


f(0) = 0


= 2 × 0 = 0


= 0



Hence, f is continuous at x = 0.


Case III: 0 < k < 1


Then, f(k) = 0


= 0 = f(k)


Thus,


Hence, f is continuous in (0, 1).


Case IV: k = 1


Then f(k) = f(1) = 0


= 0


= 4 × 1 = 4



Hence, f is not continuous at x = 1.


Case V: k < 1


Then, f(k) = 4k


= 4k = f(k)


Thus,


Hence, f is continuous at all points x, s.t. x > 1.


Therefore, x = 1 is the only point of discontinuity of f.


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