If $f:[0,2) \rightarrow[R$ is defined by $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{k} & \text { for } \quad…
If $f:[0,2) \rightarrow[R$ is defined by
$f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{k} & \text { for } \quad 0 \leq x < 1 \\ k x & \text { for } 1 \leq x < 2\end{array}\right.$ where $k>0$, and $f$ is such that $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$, then
2
1
4
$\frac{1}{4}$
Solution
Given function
$
f(x)=\left\{\begin{array}{ll}
1+\frac{2 x}{k}, & 0 \leq x < 1 \\
k x, & 1 \leq x < 2
\end{array}, \text { where } k>0\right.
$
Now, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(1+\frac{2 x}{k}\right)=1+\frac{2}{k}$ and $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(k x)=k$
Now, given that
$
\begin{array}{rlrl}
& & \lim _{x \rightarrow 1^{-}} f(x) & =\lim _{x \rightarrow 1^{+}} f(x) \\
1+\frac{2}{k} & =k \\
\Rightarrow & & \frac{k+2}{k} & =k \\
\Rightarrow & & k^2-k-2 & =0 \\
\Rightarrow \quad k^2-2 k+k-2 & =0
\end{array}
$
$\begin{array}{rlrl} & \Rightarrow k(k-2)+1(k-2) & =0 \\ \Rightarrow & (k-2)(k+1) & =0 \\ & \text { So, } & k & =2, k=-1 \\ & \text { but } & k & >0 \\ & \text { So, } & k & =2 \\ \Rightarrow & k^2 & =4\end{array}$