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Question
The point of discountinuity of the function `f(x) = {{:(2x + 3",", x ≤ 2),(2x - 3",", x > 2):}` is are
Options
2
C < Z
C > Z
Not defined
MCQ
Solution
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.
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