Let [x] be the greatest integer leq x. Then t | vedclass

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(D) The function is defined as $f(x) = |[x]| + \sqrt{x - [x]}$.We examine the points of discontinuity in the interval $(-2, 1)$. The function $[x]$ is

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$2$

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(D) The function is defined as $f(x) = |[x]| + \sqrt{x - [x]}$.We examine the points of discontinuity in the interval $(-2, 1)$. The function $[x]$ is discontinuous at all integers. In the given interval,the integers are $-1$ and $0$.Case $1$: At $x = -1$:$f(-1) = |[-1]| + \sqrt{-1 - [-1]} = |-1| + \sqrt{0} = 1$.$f(-1^+) = \lim_{h 	o 0^+} (|[ -1 + h ]| + \sqrt{-1 + h - [-1 + h]}) = |-1| + \sqrt{0} = 1$.$f(-1^-) = \lim_{h 	o 0^+} (|[ -1 - h ]| + \sqrt{-1 - h - [-1 - h]}) = |-2| + \sqrt{-1 - h - (-2)} = 2 + \sqrt{1 - h} = 2 + 1 = 3$.Since $f(-1^+) 
eq f(-1^-)$,the function is discontinuous at $x = -1$.Case $2$: At $x = 0$:$f(0) = |[0]| + \sqrt{0 - [0]} = 0 + 0 = 0$.$f(0^+) = \lim_{h 	o 0^+} (|[ 0 + h ]| + \sqrt{0 + h - [0 + h]}) = |0| + \sqrt{0} = 0$.$f(0^-) = \lim_{h 	o 0^+} (|[ 0 - h ]| + \sqrt{0 - h - [0 - h]}) = |-1| + \sqrt{-h - (-1)} = 1 + \sqrt{1 - h} = 1 + 1 = 2$.Since $f(0^+) 
eq f(0^-)$,the function is discontinuous at $x = 0$.Thus,the function is discontinuous at $x = -1$ and $x = 0$. The total number of points is $2$.

Let $[x]$ be the greatest integer $\leq x$. Then the number of points in the interval $(-2, 1)$,where the function $f(x) = |[x]| + \sqrt{x - [x]}$ is discontinuous,is $........$.

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