Question

The points of discontinuity of the function

f(x) = `1/(x - 1)` if 0 ≤ x ≤ 2

= `(x + 5)/(x + 3) if 2 < x ≤ 4` in its domain are.

पर्याय

• x = 1, x = 2

• x = 0, x = 2

• x = 2 only

• x = 4 only

Answer 1

x = 1, x = 2

Explanation:

f(x) = `{(1/(x - 1) "," if 0 ≤ x ≤ 2),((x + 5)/(x + 3) "," if 2 < x ≤ 4):}`

Clearly, f(x) is not defined at x = 1

Hence, f(x) is discontinuous at x = 1

At x = 2

`lim_(x -> 2^-) "f"(x) = lim_(x -> 2^-) 1/(x - 1) = 1/(2 - 1)` = 1

`lim_(x -> 2^+) "f"(x) = lim_(x -> 2^+) (x + 5)/(x + 3) = (2 + 5)/(2 + 3) = 7/5`

∴ `lim_(x -> 2^-) "f"(x) ne  lim_(x -> 2^+)  "f"(x)`

∴ f(x) is discontinuous at x = 2