Question

In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]

Answer 1

Given:

 \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]

If  \[f\left( x \right)\]  is continuous at x = 0, then 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]

\[\Rightarrow \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( k\left( \left( - h \right)^2 - 3h \right) \right) = \lim_{h \to 0} \left( \cos 2h \right)\]
\[ \Rightarrow 0 = 1 \left[ \text{It is not possible} \right]\]

Hence, there does not exist any value of k, which can make the given function continuous.