A) for\[x=2\] only
B) for all real values of \[x\] such that\[x\ne 2\]
C) for all real values of\[x\]
D) for all integral values of \[x\] only
Correct Answer: C
Solution :
Key Idea Every polynomial function is continuous. Given, function is \[f(x)=\left\{ \begin{matrix} x-1, & x<2 \\ 2x-3, & x\ge 2 \\ \end{matrix} \right.\] Since, it is a polynomial function, so it is continuous for every value of \[x\] except at\[x=2.\] At \[x=2,\,\,LHL=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,x-1\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,2-h-1=1\] \[RHL=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,2x-3\] \[=\underset{h\to 0}{\mathop{\lim }}\,2(2+h)-3=1\] and \[f(2)=2(2)-3=1\] \[\therefore \] \[LHL=RHL=f(2)\] Thus, \[f(x)\] is continuous for all real values of\[x\].
You need to login to perform this action.
You will be redirected in
3 sec
