If a real-valued function f(x) = begin{cases} | vedclass

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

Question

(B) Since $f(x)$ is continuous at $x=3$,we have $\lim_{x \rightarrow 3} f(x) = f(3) = l$.The limit is $\lim_{x \rightarrow 3} \frac{2x^2+(k+2)x+9}{3x^

Vedclass · Accepted Answer

$\frac{124}{11}$

Vedclass · Answer

(B) Since $f(x)$ is continuous at $x=3$,we have $\lim_{x \rightarrow 3} f(x) = f(3) = l$.The limit is $\lim_{x \rightarrow 3} \frac{2x^2+(k+2)x+9}{3x^2-7x-6}$.Since the denominator $3x^2-7x-6$ becomes $3(9)-7(3)-6 = 27-21-6 = 0$ at $x=3$,for the limit to exist,the numerator must also be $0$ at $x=3$.Thus,$2(3)^2 + (k+2)(3) + 9 = 0$.$18 + 3k + 6 + 9 = 0 \Rightarrow 3k + 33 = 0 \Rightarrow k = -11$.Now,substituting $k=-11$ into the function,we get $\lim_{x \rightarrow 3} \frac{2x^2-9x+9}{3x^2-7x-6}$.Using $L$'$H$ôpital's rule: $\lim_{x \rightarrow 3} \frac{4x-9}{6x-7} = \frac{4(3)-9}{6(3)-7} = \frac{12-9}{18-7} = \frac{3}{11}$.So,$l = \frac{3}{11}$.Finally,$l-k = \frac{3}{11} - (-11) = \frac{3}{11} + 11 = \frac{3+121}{11} = \frac{124}{11}$.

If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

Similar Questions

The function $f(x) = 2x - |x - x^2|$ is

The function $f: R - \{0\} \to R$,given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at $x = 0$ by defining $f(0)$ as:

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$

If $f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module