Suppose f(x) = begin{cases} [cos pi x], & x l | vedclass

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

Question

(A) Given $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \ 2\{x\} - 1, & x > 1 \end{cases}$.For $x > 1$, we have $f(x) = 2(x - [x]) - 1$. Since for $

Vedclass · Accepted Answer

right derivative is $2$

Vedclass · Answer

(A) Given $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \ 2\{x\} - 1, & x > 1 \end{cases}$.For $x > 1$, we have $f(x) = 2(x - [x]) - 1$. Since for $x$ slightly greater than $1$, $[x] = 1$, we get $f(x) = 2(x - 1) - 1 = 2x - 3$.The right-hand derivative at $x = 1$ is defined as $f'(1^+) = \lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h}$.First, $f(1) = [\cos \pi] = [-1] = -1$.Then, $f'(1^+) = \lim_{h 	o 0^+} \frac{2(1+h) - 3 - (-1)}{h} = \lim_{h 	o 0^+} \frac{2 + 2h - 3 + 1}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Thus, the right derivative at $x = 1$ is $2$.

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$, where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively, then at $x = 1$:

Similar Questions

If $f(x) = [x]$,where $[x]$ is the greatest integer not greater than $x$,then $f^{\prime}(1^{+}) = \dots$.

Let $\alpha, \beta \in R$ be such that the function $f(x) = \begin{cases} 2 \alpha (x^2 - 2) + 2 \beta x, & x < 1 \\ (\alpha + 3) x + (\alpha - \beta), & x \ge 1 \end{cases}$ is differentiable at all $x \in R$. Then $34(\alpha + \beta)$ is equal to

The function represented by the following graph is,

Let $f(x) = \begin{cases} 1, & \forall x < 0 \\ 1 + \sin x, & \forall 0 \le x \le \pi/2 \end{cases}$,then what is the value of $f'(x)$ at $x = 0$?

If the function $g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 < x \le 5 \end{cases}$ is differentiable,then the value of $k+m$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module