Question

If `f(x) = {{:(-x^2",", "when"  x ≤ 0),(5x - 4",", "when"  0 < x ≤ 1),(4x^2 - 3x",", "when"  1 < x < 2),(3x + 4",", "when"  x ≥ 2):}`, then

विकल्प

• f(x) is continuous at x = 0

• f(x) is continuous at x = 2

• f(x) is discontinuous at x = 1

• None of these

Answer 1

f(x) is continuous at x = 2

Explanation:

`lim_(x ->0) f(x)` = 0; `f(0)` = 0, `lim_(x -> 0^+) f(x)` = – 4

`f(x)` discontinuous at x = 0

And `lim_(x -> 0) f(x)` = 1 and `lim_(x -< 1^+) f(x)` = 1, `f(1)` =  1

Hence `f(x)` is continuous at x = 1.

Also `lim_(x -> 2^-) f(x)` = 4(2)² – 3.2 = 10

`f(2)` = 10 and `lim_(x -> 2^+) f(x)` = 3(2) + 4 = 10

Hence `f(x)` is continuous at x = 2.