Let f: R rightarrow R be defined by f(x) = be | vedclass

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

Question

(B) For $f$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^-} f(x) = f(1) = 1$. First,consider the right-hand limit:

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) For $f$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^-} f(x) = f(1) = 1$. First,consider the right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (a - \frac{\sin [x-1]}{x-1})$. For $x > 1$ and $x$ very close to $1$,$0 Thus,$\lim_{x 	o 1^+} f(x) = a - \frac{\sin(0)}{x-1} = a - 0 = a$. Since $f(1) = 1$,we have $a = 1$. Next,consider the left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (b - [\frac{\sin [x-1] - [x-1]}{([x-1])^3}])$. For $x Thus,$\lim_{x 	o 1^-} f(x) = b - [\frac{\sin(-1) - (-1)}{(-1)^3}] = b - [\frac{-\sin(1) + 1}{-1}] = b - [\sin(1) - 1]$. Since $0 The greatest integer $[\sin(1) - 1] = -1$. So,$\lim_{x 	o 1^-} f(x) = b - (-1) = b + 1$. Setting this equal to $f(1) = 1$,we get $b + 1 = 1$,which implies $b = 0$. Therefore,$a + b = 1 + 0 = 1$.

Similar Questions

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=[x^2-3]$ where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for the function $f$ in the interval $(-1,2)$ is:

If the function $f(x) = \frac{e^{x}(e^{\tan x-x}-1)+\log_{e}(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$,then the value of $f(0)$ is equal to

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases}$. Is $f$ a continuous function?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module