If the function f defined as f(x)=1x-k-1e2x-1, x≠0 is continuous at x = 0, then the ordered pair (k, f(0)) is

f(x)=1x-k-1e2x-1, f is continuous at x=0 ⇒limx→0f(x) existslimx→0f(x)=limx→01x-k-1e2x-1 =limx→0e2x-1-k-1xxe2x-1 =limx→0e2x-1-k-1x2x2 since limx→0e2x-12x=1 =limx→02e2x-k-14x exists ⇒2-k-1=0⇒k=3f(0)=limx→0f(x)=2e2x-122x=1k, f(0)=3,1

If the function f defined as f(x)=1x-k-1e2x-1, x≠0 is continuous at x = 0, then the ordered pair (k, f(0)) is

f(x)=1x-k-1e2x-1, f is continuous at x=0 ⇒limx→0f(x) existslimx→0f(x)=limx→01x-k-1e2x-1 =limx→0e2x-1-k-1xxe2x-1 =limx→0e2x-1-k-1x2x2 since limx→0e2x-12x=1 =limx→02e2x-k-14x exists ⇒2-k-1=0⇒k=3f(0)=limx→0f(x)=2e2x-122x=1k, f(0)=3,1

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If the function f defined as f(x)= $\frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}, x \neq 0$ is continuous at x = 0, then the ordered pair (k, f(0)) is

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$(3, 2)$

$(3, 1)$

$(\frac{1}{3}, 2)$

$(2, 1)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} f (x) & = & \frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}, f is continuous at x = 0 \\ \Rightarrow & \lim_{x \to 0} f (x) exists \\ \lim_{x \to 0} f (x) & = & \lim_{x \to 0} \frac{1}{x} - \frac{(k - 1)}{e^{2 x} - 1} \\ = & \lim_{x \to 0} \frac{(e^{2 x} - 1) - (k - 1) x}{x (e^{2 x} - 1)} \\ = & \lim_{x \to 0} \frac{(e^{2 x} - 1) - (k - 1) x}{2 x^{2}} since \lim_{x \to 0} \frac{e^{2 x} - 1}{2 x} = 1 \\ = & \lim_{x \to 0} \frac{2 e^{2 x} - (k - 1)}{4 x} exists \\ \Rightarrow & 2 - (k - 1) = 0 \Rightarrow k = 3 \\ f (0) & = & \lim_{x \to 0} f (x) = \frac{2 (e^{2 x} - 1)}{2 (2 x)} = 1 \\ (k, f (0)) & = & (3, 1) \end{array}$