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Let $$f(x)=x-\left | x-x^{2} \right |, x\epsilon \left [ -1, 1 \right ]$$. Then the number of points at which $$f(x)$$ is discontinuous is


A
1
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B
2
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C
0
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D
none of these
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Solution

The correct option is A $$0$$
case 1:
$$x-x^{2}\geq 0 \Rightarrow x\epsilon [0,1]$$
$$ f(x)=x^{2}$$
case 2:
$$x-x^{2}\leq 0 \Rightarrow x\epsilon [-1,0)$$
$$f(x)= 2x-x^{2}$$
At x=0 both the functions tend to zero. So the function f(x) is continuous at x=0.
Both the functions are continuous in respective intervals.
Hence the number of discontinuous are zero.

Mathematics

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