Correct Answer - Option 3 : -1
Concept:
1. A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.
2. f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)
Calculation:
Given, \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} { x\ +\ 4}&{{\rm}}&{{\rm{x}} \le -1}\\ {{\rm{x^2}} {\rm{}}}&{{\rm{}}}&{-1 < {\rm{x}} < 1}\\ 2\ -\ x&{{\rm}}&{{\rm{x}} \ge 1} \end{array}} \right.\)
Using the relation
\(\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)
Hence, left-hand limit (LHL) at x = -1
\(\mathop {\lim }\limits_{x \to {{\rm{-1}}^ - }} f( x ) = \mathop {\lim }\limits_{x \to {{\rm{-1}}^ - }} (x\ +\ 4)\)
\(\Rightarrow \ \mathop {\lim }\limits_{x \to {{\rm{-1}}^ - }} f( x ) = -1 +\ 4\ =\ 3\) .......(1)
and right-hand limit (RHL) at x = -1
\(\mathop {\lim }\limits_{x \to {{\rm{-1}}^ + }} f( x ) = \mathop {\lim }\limits_{x \to {{\rm{-1}}^ + }} (x^2)\)
\(\Rightarrow \ \mathop {\lim }\limits_{x \to {{\rm{-1}}^ + }} f( x ) = \mathop {\lim }\limits_{x \to {{\rm{-1}}^ + }} (1^2)\ =\ 1\) .......(2)
We can see that, LHL ≠ RHL, therefore function is discontinuous at x = -1
Hence, option 3 is correct.
