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  • Let <img alt="\mathrm{f(x)=x-\left|x-x^2\right|, x \in[-1,1]}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3Dx-%5Cleft%7Cx-x%5E2%5Cright%7C%2C%20x%20%5Cin%5B-1%2C1%5D%

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

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

\mathrm{f(x)=x-|x| \cdot|1-x|}. We know that \mathrm{x,|x|,|1-x|}  are continuous everywhere. As the product and algebraic sum of continuous functions are continuous, \mathrm{f(x)} is continuous everywhere.

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