If the function defined by f(x) = begin{cases | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the function $f(x) = \begin{cases} \frac{2^x - 2^{-x}}{x}, & x 
eq 0 \ k, & x = 0 \end{cases}$.Since $f(x)$ is continuous at $x = 0$, we m

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the function $f(x) = \begin{cases} \frac{2^x - 2^{-x}}{x}, & x 
eq 0 \ k, & x = 0 \end{cases}$.Since $f(x)$ is continuous at $x = 0$, we must have $\lim_{x ightarrow 0} f(x) = f(0)$.Therefore, $k = \lim_{x ightarrow 0} \frac{2^x - 2^{-x}}{x}$.Applying $L$'Hospital's rule, we differentiate the numerator and denominator with respect to $x$:$k = \lim_{x ightarrow 0} \frac{\frac{d}{dx}(2^x - 2^{-x})}{\frac{d}{dx}(x)} = \lim_{x ightarrow 0} \frac{2^x \ln 2 - 2^{-x} \ln 2(-1)}{1} = \lim_{x ightarrow 0} (2^x \ln 2 + 2^{-x} \ln 2)$.Substituting $x = 0$:$k = 2^0 \ln 2 + 2^0 \ln 2 = \ln 2 + \ln 2 = 2 \ln 2 = \ln(2^2) = \ln 4$.We need to find $e^k$.$e^k = e^{\ln 4} = 4$.

If the function defined by $f(x) = \begin{cases} \frac{2^x - 2^{-x}}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$, then $e^k$ is equal to:

Similar Questions

If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi \end{cases}$ is twice differentiable,then the ordered pair $(k_{1}, k_{2})$ is equal to

If the function $f$ defined by $f(x) = \begin{cases} \cos x, & \text{if } x \leq 0 \\ 3x + \alpha, & \text{if } 0 < x < 2 \\ \beta x + 3, & \text{if } 2 \leq x \leq 4 \\ 11, & \text{if } x > 4 \end{cases}$ where $\alpha$ and $\beta$ are real constants,is continuous on $R$,then $\alpha^2 + \beta^2 =$

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

If $f(x) = \begin{cases} k, & \text{for } x = 1 \\ \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}, & \text{for } x \neq 1 \end{cases}$ is continuous on $[0, \infty)$, then $k =$

Consider the function $f(x) = (x - 2) \log_e x$. Then the equation $x \log_e x = 2 - x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module