If f(x) = begin{cases} k, & text{for } x = 1 | vedclass

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

Question

(D) Since $f(x)$ is continuous at $x = 1$, we have $k = \lim_{x \rightarrow 1} f(x)$.$k = \lim_{x \rightarrow 1} \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Since $f(x)$ is continuous at $x = 1$, we have $k = \lim_{x \rightarrow 1} f(x)$.$k = \lim_{x \rightarrow 1} \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$.Using $L$'$H$ôpital's rule, we differentiate the numerator and denominator with respect to $x$:Numerator: $\frac{d}{dx}[(9x-1)(\sqrt{x}-1)] = 9(\sqrt{x}-1) + (9x-1)(\frac{1}{2\sqrt{x}})$.Denominator: $\frac{d}{dx}[3x^2+2x-5] = 6x+2$.Evaluating the limit as $x \rightarrow 1$:$k = \frac{9(1-1) + (9-1)(\frac{1}{2})}{6(1)+2} = \frac{0 + 8(\frac{1}{2})}{8} = \frac{4}{8} = \frac{1}{2}$.

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 $k =$

Similar Questions

$A$ point in the domain of a function where the discontinuity cannot be removed by redefining the function at that point is called:

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

Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

If $f(x) = \begin{cases} \frac{8^x-4^x-2^x+1}{x^2}, & \text{if } x > 0 \\ e^x \sin x + x + \lambda \log 4, & \text{if } x \leqslant 0 \end{cases}$ is continuous at $x = 0$,then the value of $1000 e^\lambda = $

$A$ function $f(x)$ is defined as $f(x) = \begin{cases} x^m \sin \frac{1}{x} & x \neq 0, m \in N \\ 0 & x = 0 \end{cases}$. The least value of $m$ for which $f'(x)$ is continuous at $x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module