A) \[4\]
B) \[3\]
C) \[2\]
D) \[1\]
Correct Answer: D
Solution :
Given, \[f(x)=\left\{ \begin{matrix} x+a, & if & x\le 1 \\ 3-{{x}^{2}}, & if & x>1 \\ \end{matrix} \right.\] LHL= \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\,\,f(x)\,=\underset{h\to 0}{\mathop{\lim }}\,\,\,1-h+a=1+a\] RHL= \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\,\,f(x)\,=\underset{h\to 0}{\mathop{\lim }}\,\,\,3-{{(1+h)}^{2}}=2\] Since, \[f(x)\] is continuous at \[x=1.\] \[\therefore \] \[LHL=RHL\] \[\Rightarrow \] \[1+a=2\,\,\,\Rightarrow \,\,a=1\]
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