Question
What is the value of $$k$$ for which the following function $$f\left( x \right)$$ is continuous for all $$x\,?$$
\[f\left( x \right) = \left\{ \begin{array}{l}
\frac{{{x^3} - 3x + 2}}{{{{\left( {x - 1} \right)}^2}}},{\rm{ for\,\, }}x \ne 1\\
\,\,\,\,\,\,\,\,k,\,\,\,\,\,\,\,\,{\rm{ for\,\, }}x = 1
\end{array} \right.\]
A.
$$3$$
B.
$$2$$
C.
$$1$$
D.
$$ - 1$$
Answer :
$$3$$
Solution :
Let \[f\left( x \right) = \left\{ \begin{array}{l}
\frac{{{x^3} - 3x + 2}}{{{{\left( {x - 1} \right)}^2}}},{\rm{ }}\forall {\rm{ }}x \ne 1\\
\,\,\,\,\,\,\,\,k,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\forall {\rm{ }}x = 1
\end{array} \right.\] and $$f\left( x \right)$$ is continuous.
$$\eqalign{
& \therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = k \cr
& \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 3x + 2}}{{{{\left( {x - 1} \right)}^2}}} = k \cr
& \Rightarrow k = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 3}}{{2\left( {x - 1} \right)}}\,\,\,\,\,\,\,\,\,\left[ {{\text{By L'Hospital rule}}} \right] \cr
& \Rightarrow k = \mathop {\lim }\limits_{x \to 1} \frac{{6x}}{2}\,\,\,\,\,\,\,\,\left[ {{\text{By L'Hospital rule}}} \right] \cr
& \Rightarrow k = 3 \cr} $$