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Step-by-Step Solution
Step 1: Understand the Function Definition
The function is given by
f(x) =
\begin{cases}
x[x], & -2 < x < 0,\\
(x - 1)[x], & 0 \le x < 2
\end{cases}
where [x] is the greatest integer less than or equal to x . We need to analyze
y = |f(x)| for discontinuities and non-differentiability within (-2,2) .
Step 2: Identify the Integer Breakpoints for [x]
The greatest integer function [x] changes its value at integers. In the given interval (-2, 2) ,
the relevant integer points are x = -1, 0, and 1 . We will specifically check the behavior
of f(x) around these points.
Step 3: Break the Function into Subintervals
Consider the intervals based on the integer values of [x] :
-2 < x < -1 : Here [x] = -2 , so f(x) = x \cdot (-2) = -2x .
-1 \le x < 0 : Here [x] = -1 , so f(x) = x \cdot (-1) = -x .
0 \le x < 1 : Here [x] = 0 , so f(x) = (x - 1)\cdot 0 = 0 .
1 \le x < 2 : Here [x] = 1 , so f(x) = (x - 1)\cdot 1 = x - 1 .
Step 4: Examine Continuity of y = |f(x)|
A function (and thus its absolute value) can only fail to be continuous where there is a jump
in the function values. We check x = -1, 0, 1 one by one.
At x = -1
Left side ( -2 < x < -1 ): f(x) = -2x . As x \to -1^- , f(x) \to -2 \cdot (-1) = 2 .
Hence |f(x)| \to 2 .
Right side ( -1 \le x < 0 ): f(x) = -x . As x \to -1^+ , f(x) \to -(-1) = 1 .
Hence |f(x)| \to 1 .
Since |f(x)| jumps from 2 to 1 , a discontinuity occurs at x = -1 .
Thus, y = |f(x)| is not continuous at x = -1 .
At x = 0
Left side ( -1 \le x < 0 ): f(x) = -x . As x \to 0^- , f(x) \to 0 ,
so |f(x)| \to 0 .
Right side ( 0 \le x < 1 ): f(x) = 0 . Hence |f(x)| = 0 for 0 \le x < 1 .
At x = 0 , |f(x)| is also 0 from the definition.
No jump occurs at x = 0 . Therefore, |f(x)| is continuous at x = 0 .
At x = 1
Left side ( 0 \le x < 1 ): f(x) = 0 . Hence |f(x)| = 0 as x \to 1^- .
Right side ( 1 \le x < 2 ): f(x) = x - 1 . As x \to 1^+ , f(x) \to 0 ,
so |f(x)| \to 0 too.
No jump occurs at x = 1 . Hence, |f(x)| is continuous at x = 1 .
Conclusion for continuity: The only point of discontinuity is x = -1 .
Therefore, the number of discontinuities m = 1 .
Step 5: Examine Differentiability of y = |f(x)|
For differentiability, we look at the points of discontinuity and possible “corners.”
A corner occurs where the left-hand derivative and the right-hand derivative do not match.
At x = -1
Since |f(x)| is discontinuous at x=-1 , it also fails to be differentiable there.
At x = 0
Left derivative for -1 \le x < 0 : f(x) = -x . The derivative of -x is -1 .
Since f(x) is negative of x , for x in (-1,0) , f(x) is positive or negative?
Actually, -x is positive for -1 < x < 0 . Thus |f(x)| = -x , whose derivative is -1.
Right derivative for 0 < x < 1 : f(x) = 0 . The derivative of the constant 0
is 0 . Hence, |f(x)| = 0 on (0,1) , and its derivative is 0 .
The left-hand derivative (-1) does not match the right-hand derivative (0) ,
so |f(x)| is not differentiable at x = 0 .
At x = 1
Left derivative for 0 \le x < 1 : f(x) = 0 , so the derivative is 0 .
Right derivative for 1 \le x < 2 : f(x) = x - 1 , whose derivative is 1 .
The left-hand derivative (0) does not match the right-hand derivative (1) ,
so |f(x)| is not differentiable at x = 1 .
Conclusion for differentiability: |f(x)| is not differentiable at x = -1, 0, and 1 .
Hence, the number of non-differentiable points n = 3 .
Step 6: Calculate m + n
• Number of discontinuities m = 1 (at x = -1 ).
• Number of points of non-differentiability n = 3 (at x = -1, 0, 1 ).
Therefore, m + n = 1 + 3 = 4.
Final Answer: 4
