© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Definition of the Function
We are given a function
f(x) = \min \{ x - [x], \; 1 + [x] - x \}
on the interval [0, 3] , where [x] denotes the greatest integer less than or equal to x .
Note that x - [x] = \{x\} (the fractional part of x ) and
1 + [x] - x = 1 - \{x\} .
Hence on each integer interval [n, n+1) ,
f(x) = \min(\{x\}, \, 1-\{x\}) .
Step 2: Checking Continuity at Potential Points of Discontinuity
Discontinuities could possibly occur at integers (0, 1, 2, 3) because the greatest integer function [x] changes value there. We check each integer:
At x = 0 and x = 3 , these are the endpoints of the domain. The function from the right of 0 and from the left of 3 matches the value at those endpoints; no jump occurs.
At x = 1 and x = 2 , approaching from the left or right yields the same limit (both approach 0), and f(x) at the integer itself is also 0. Hence there is no discontinuity.
Therefore, f(x) is continuous at all integers in [0, 3] , and it remains continuous elsewhere since it is defined piecewise by continuous expressions.
Thus, the set of discontinuities
P = \varnothing ,
which means the number of elements in P is
|P| = 0 .
Step 3: Checking Differentiability Inside Each Sub-Interval
Within each integer interval [n, n+1] , split into two parts:
For x \in (n, n + 0.5) , we have \{x\} < 0.5 , so
f(x) = \{x\} = x - n . The derivative here is +1 .
For x \in (n + 0.5, n + 1) , we have \{x\} > 0.5 , so
f(x) = 1 - \{x\} = (n + 1) - x . The derivative here is -1 .
Hence on each interval [n, n+1] , there is a “corner” at x = n + 0.5 because the derivative changes from +1 to -1 . So, f is not differentiable at x = n + 0.5 for n = 0, 1, 2 (as long as n + 0.5 lies in (0, 3) ).
Step 4: Checking Differentiability at Integers
At each integer x = 1, 2 (inside the interval (0, 3) ), the left-hand derivative differs from the right-hand derivative:
As x \to 1^- , f(x) has slope -1 . As x \to 1^+ , f(x) has slope +1 .
Thus, f is not differentiable at x = 1 .
As x \to 2^- , f(x) has slope -1 . As x \to 2^+ , f(x) has slope +1 .
Thus, f is not differentiable at x = 2 .
At x = 0 and x = 3 , they are domain endpoints. We only consider the right-hand derivative at 0 and the left-hand derivative at 3. Each is consistent in its interval, so f is differentiable there.
Step 5: Identifying All Points of Non-Differentiability
Combining these:
Non-differentiable points inside (0, 3) are:
x = 0.5, \; 1, \; 1.5, \; 2, \; 2.5.
Hence, the set
Q = \{0.5, 1, 1.5, 2, 2.5\} ,
which means
|Q| = 5 .
Step 6: Final Answer
We have
|P| = 0
and
|Q| = 5.
The question asks for the sum of the number of elements in P and Q , which is
0 + 5 = 5.
1 - {x} = 1 - x; 0 \le x < 1
Non differentiable at x = \frac{1}{2},\,1,\,\frac{3}{2},\,2,\,\frac{5}{2}
Therefore, the required sum of the elements in sets P and Q is 5.
