Q. 153.5( 28 Votes )

# Discuss the conti

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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