Question

The point of discountinuity of the function `f(x) = {{:(2x + 3",", x ≤ 2),(2x - 3",", x > 2):}` is are

पर्याय

• 2

• C < Z

• C > Z

• Not defined

Answer 1

2

Explanation:

`f(x) = {{:(2x + 3",", x ≤ 2),(2x - 3",", x > 2):}`

At `x` = 2, L.H.L = `lim_(x -> 2) (2x + 3)` = 7

`f(2)` = 2 × 2 + 3 = 7

R.H.L = `lim_(x -> 2) + (2x - 3) = 2 xx 2 - 3` = 1

∴ `f` is discontinuous at `x` = 2

At `x = C < 2,  lim_(x -> 0) (2x + 3) = 2C + 3 = f(c)`

∴ `f` is continuous at `x` = C < 2

At `x = C > 2,  lim_(x -> 0) (2x - 3) = 2C - 3 = f(c)`

∴ `f` is continuous at `x` = C > 2

⇒ Point of discontinuty is `x` = 2.