Question

Let [.]denote the greatest integer function and f(x)=[x]+[2-x],-1≤x≤4.Then

• A. f is not differentiable at x=3/2
• B. f is differentiable at x=3
• C. f is continues at x=0
• D. f is not continues at x=1
• E. f is continues at x=2

Answer 1

Answer: (E)

f is continues at x=2

Answer 2

Given that
The greatest integer function [x] represents the largest integer that is less than or equal to x.
For -1 ≤ x < 1: Since x is not an integer, both [x] and [2-x] will be 0. f(x) = [x] + [2-x] = 0 + 0 = 0
For 1 ≤ x < 2: In this interval, [x] will be 1, and [2-x] will be 0 (since 2-x is not an integer). f(x) = [x] + [2-x] = 1 + 0 = 1
For 2 ≤ x < 3: Both x and 2-x are integers in this interval. So, [x] = 2 and [2-x] = 0. f(x) = [x] + [2-x] = 2 + 0 = 2
For 3 ≤ x < 4: In this interval, [x] will be 3, and [2-x] will be 0 (since 2-x is not an integer). f(x) = [x] + [2-x] = 3 + 0 = 3
Now, let's consider the boundary points:
For $x = -1: f(-1) = [-1] + [2 - (-1)] = -1 + 3 = 2$
For $x = 1: f(1) = [1] + [2 - 1] = 1 + 1 = 2$
For x = 2: f(2) = [2] + [2 - 2] = 2 + 0 = 2$$$ For x = 4: f(4) = [4] + [2 - 4] = 4 + 0 = 4$Finally, for$x > 4, both [x]$$ and [2 - x]$will be equal to the integer part of$x$$, sof(x)$ will be the sum of these integer parts.

1. For $x > 4: f(x) = [x] + [2 - x] = x + (2 - x) = 2$
   So, the values of$f(x)$ for $-1 ≤ x ≤ 4$ can be represented as:$f(x) = 2 \text{ for } x > 4$

$f(x) = 4 \text{ for } x = 4$
$f(x) = 3 \text{ for } 3 ≤ x < 4$
$f(x) = 2$ for $2 ≤ x < 3$
$f(x) = 2$ for $x = 2$
$f(x) = 1$ for $1 ≤ x < 2$
$f(x) = 2$ for $x = 1$
$f(x) = 2$ for $x = -1$
So, the correct option is (D) : f is continues at x=2.