If the function f(x) = begin{cases} frac{(e^{ | vedclass

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(A) Since $f(x)$ is differentiable at $x=0$, it must be continuous at $x=0$. $\lim_{x \rightarrow 0} f(x) = f(0) = P$. $\lim_{x \rightarrow 0} \frac{(

Vedclass · Accepted Answer

$P=0$, $f^{\prime}(0)=\frac{k^2}{4}$

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(A) Since $f(x)$ is differentiable at $x=0$, it must be continuous at $x=0$. $\lim_{x ightarrow 0} f(x) = f(0) = P$. $\lim_{x ightarrow 0} \frac{(e^{kx}-1) \sin kx}{4 	an x} = \lim_{x ightarrow 0} \frac{(e^{kx}-1)}{x} \cdot \frac{\sin kx}{x} \cdot \frac{x^2}{4 	an x} = (k) \cdot (k) \cdot (0) = 0$. Thus, $P = 0$. Now, $f^{\prime}(0) = \lim_{x ightarrow 0} \frac{f(x) - f(0)}{x} = \lim_{x ightarrow 0} \frac{(e^{kx}-1) \sin kx}{4x 	an x}$. $f^{\prime}(0) = \lim_{x ightarrow 0} \left( \frac{e^{kx}-1}{x} ight) \left( \frac{\sin kx}{x} ight) \left( \frac{x}{	an x} ight) \cdot \frac{1}{4} = (k) \cdot (k) \cdot (1) \cdot \frac{1}{4} = \frac{k^2}{4}$.

If the function $f(x) = \begin{cases} \frac{(e^{kx}-1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x=0 \end{cases}$ is differentiable at $x=0$, then

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