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(C) Given the function $f(x) = \begin{cases} 1+\frac{2x}{k}, & 0 \leq x 0$. We are given that $\lim_{x

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$4$

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(C) Given the function $f(x) = \begin{cases} 1+\frac{2x}{k}, & 0 \leq x 0$. We are given that $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$. Calculating the left-hand limit: $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (1+\frac{2x}{k}) = 1+\frac{2}{k}$. Calculating the right-hand limit: $\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (kx) = k$. Equating the two limits: $1+\frac{2}{k} = k$ Multiplying by $k$ (since $k>0$): $k+2 = k^2$ Rearranging the terms: $k^2 - k - 2 = 0$ Factoring the quadratic equation: $(k-2)(k+1) = 0$. Since $k>0$,we have $k=2$. Therefore,$k^2 = 2^2 = 4$.

If $f:[0,2) \rightarrow R$ is defined by $f(x)=\begin{cases} 1+\frac{2x}{k} & \text{for } 0 \leq x < 1 \\ kx & \text{for } 1 \leq x < 2 \end{cases}$ where $k>0$,and $f$ is such that $\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$,then find the value of $k^2$.

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