Let [t] denote the greatest integer less than | vedclass

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

Question

(A) Given $f(x) = x - [x]$ and $g(x) = 1 - (x - [x])$. Let ${x} = x - [x]$. Then $f(x) = {x}$ and $g(x) = 1 - {x}$.$h(x) = \min \{ {x}, 1 - {x} \}$.Fo

Vedclass · Accepted Answer

continuous in $[-2, 2]$ but not differentiable at more than four points in $(-2, 2)$

Vedclass · Answer

(A) Given $f(x) = x - [x]$ and $g(x) = 1 - (x - [x])$. Let ${x} = x - [x]$. Then $f(x) = {x}$ and $g(x) = 1 - {x}$.$h(x) = \min \{ {x}, 1 - {x} \}$.For $x \in [n, n+1)$,${x} = x - n$. So $h(x) = \min \{ x-n, 1-(x-n) \} = \min \{ x-n, 1-x+n \}$.The graphs of $f(x)$ and $g(x)$ intersect when ${x} = 1 - {x}$,which implies $2{x} = 1$,or ${x} = 0.5$.In each interval $[n, n+1)$,the function $h(x)$ increases from $0$ to $0.5$ and then decreases from $0.5$ to $0$.Since $h(x)$ is continuous everywhere and the graph shows sharp corners at $x = n$ (where ${x}=0$) and at $x = n + 0.5$ (where ${x}=0.5$),we count the points of non-differentiability in $(-2, 2)$.The points are $x = -1.5, -1, -0.5, 0, 0.5, 1, 1.5$. There are $7$ such points.Since $7 > 4$,option $A$ is correct.

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f(x)=x-[x]$,$g(x)=1-x+[x]$,and $h(x)=\min \{f(x), g(x)\}$ for $x \in [-2, 2]$. Then $h$ is :

Similar Questions

If $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

If the function $f(x) = \begin{cases} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $p = $

Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is

The value of $p$ for which the function $f(x) = \begin{cases} \frac{(4^x - 1)^3}{\sin(\frac{x}{p}) \log(1 + \frac{x^2}{3})}, & x \ne 0 \\ 12(\log 4)^3, & x = 0 \end{cases}$ is continuous at $x = 0$ is:

If $f(x) = \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$ for $x \neq 0$ and $f(x) = k$ for $x = 0$,is continuous at $x = 0$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module