Question ID: #398
Let $[x]$ denote the greatest integer function, and let $m$ and $n$ respectively be the numbers of the points, where the function $f(x) = [x] + |x – 2|, \quad -2 < x < 3$ is not continuous and not differentiable. Then $m+n$ is equal to:
- (1) 6
- (2) 9
- (3) 8
- (4) 7
Solution:
We define the function $f(x) = [x] + |x-2|$ for $x \in (-2, 3)$.
We break the function into intervals based on the integer values of $x$:
$$ f(x) = \begin{cases} -2 + (2-x) = -x, & -2 < x < -1 \\ -1 + (2-x) = 1-x, & -1 \le x < 0 \\ 0 + (2-x) = 2-x, & 0 \le x < 1 \\ 1 + (2-x) = 3-x, & 1 \le x < 2 \\ 2 + (x-2) = x, & 2 \le x < 3 \end{cases} $$
Checking for continuity at the integer points (breaking points):
At $x = -1$: LHL $= -(-1) = 1$; RHL $= 1-(-1) = 2$. (Discontinuous)
At $x = 0$: LHL $= 1-0 = 1$; RHL $= 2-0 = 2$. (Discontinuous)
At $x = 1$: LHL $= 2-1 = 1$; RHL $= 3-1 = 2$. (Discontinuous)
At $x = 2$: LHL $= 3-2 = 1$; RHL $= 2$. (Discontinuous)
So, $f(x)$ is not continuous at 4 points.
Thus, $m = 4$.
Since the function is discontinuous at these 4 points, it is also not differentiable at these 4 points.
Thus, $n = 4$.
$$ m + n = 4 + 4 = 8 $$
Ans. (3)
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