Let $f(x) = \begin{cases} \dfrac{\sin 3x}{5x} + k & x 0 \end{cases}$. If $f(x)$ is continuous at $x=0$, then $k$ equals

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Q.Let $f(x) = \begin{cases} \dfrac{\sin 3x}{5x} + k & x < 0 \ \dfrac{1 + \cos 2x}{\sqrt{3} x} & x > 0 \end{cases}$ . If $f(x)$ is continuous at $x=0$ , then $k$ equals

\dfrac{1}{5}

\dfrac{2}{5}

\dfrac{3}{5}

\dfrac{4}{5}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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