The function f(x) = begin{cases} x, & text{if | vedclass

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

Question

(A) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 1, & 1 < x \le 2 \end{cases}$Check continuity at $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x)

Vedclass · Accepted Answer

Continuous at all $x$,$0 \le x \le 2$ and differentiable at all $x$,except $1$ in the interval $(0, 2)$

Vedclass · Answer

(A) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 1, & 1 Check continuity at $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} (1 - h) = 1$Right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{h 	o 0} f(1 + h) = 1$Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1) = 1$,the function is continuous at $x = 1$.Check differentiability at $x = 1$:Left-hand derivative: $LHD = \lim_{h 	o 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h 	o 0} \frac{(1 - h) - 1}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$Right-hand derivative: $RHD = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h 	o 0} \frac{1 - 1}{h} = 0$Since $LHD 
eq RHD$,the function is not differentiable at $x = 1$.Thus,the function is continuous for all $x \in [0, 2]$ and differentiable for all $x \in [0, 2]$ except at $x = 1$.

The function $f(x) = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 1, & \text{if } 1 < x \le 2 \end{cases}$ is

Similar Questions

If the function $f(x) = \frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$,then $f(0) = $

The function $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$ is

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x-1, & \text{for } x \leq 1 \\ 2-x^2, & \text{for } 1 < x \leq 3 \\ x-10, & \text{for } 3 < x < 5 \\ 2x, & \text{for } x \geq 5 \end{cases}$,then the set of points of discontinuity of $f$ is

If $f(x) = \begin{cases} \frac{\sqrt{a^2-ax+x^2}-\sqrt{x^2+ax+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x \neq 0 \\ K, & x=0 \end{cases}$ is continuous at $x=0$,then $K=$

If $f(x) = \begin{cases} ax^2 + bx + 1 & \text{if } |2x - 3| \geq 2 \\ 3x + 2 & \text{if } \frac{1}{2} < x < \frac{5}{2} \end{cases}$ is continuous on its domain,then $a + b$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module