# A function f(x) is defined in the following way: f(x) = -x, x ≤ 0 = x, 0 < x < 1 = 2 - x, x ≥ 1 In this case, the function f(x) is:

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A function f(x) is defined in the following way:

f(x) = -x, x ≤ 0

= x, 0 < x < 1

= 2 - x, x ≥ 1

In this case, the function f(x) is:

1. continuous at both x = 0 and x = 1
2. continuous at x = 0 and discontinuous at x = 1
3. discontinuous at x = 0 and continuous at x = 1
4. discontinuous at both x = 0 and x = 1

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Correct Answer - Option 1 : continuous at both x = 0 and x = 1

Concept:

The function is continuous if it satisfies the following conditions.

$\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{b}}{( 0-)}}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{b}}{ (0-)}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{b}} \right){\rm{}}$

Otherwise, the function is not continuous at x = b

Analysis:

To check the continuity at x = 0 and x = 1,

at x = 0,

$\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{0}}}} f(x) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{0}}}}( -x)$

= 0

at x = 0-,

f(x) = -x ; f(0-) = 0

at x = 0+

f(x) = x ; f(0+) = 0

∵ Left-hand limit = Right-hand limit = Value of the function at the point.

∴ The function is continuous at x = 0

at x = 1,

$\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{1}}}} f(x) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{1}}}}(2 -x)$

= 1

at x = 1-,

f(x) = x ; f(1-) = 1

at x = 1+

f(x) = 2 - x ; f(1+) = 1

∵ Left-hand limit = Right-hand limit = Value of the function at the point.

∴ The function is continuous at x = 1

The following properties are true in calculus:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity
• Continuity does not necessarily imply differentiability.