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Question

Let $$f(x) = x |x - x^2|, x \in [-1, 1]$$. 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 C 0
Given $$f(x)=x|x-x^2|$$.................[$$x\epsilon[-1,1]$$]
Case1: $$x-x^2\geq0$$

$$\implies x(1-x)\geq0$$
Hence $$x\epsilon[0,1]$$

Case2:$$x-x^2<0$$

$$\implies x(1-x)<0$$
Hence $$x\epsilon[-1,0)$$

So,

                        $$x^2-x^3$$......for $$0\leq x\leq1$$
$$f(x)=\lbrace$$          
                        $$x^3-x^2$$......for$$-1\leq x<0$$

Since,

$$\lim_{x\rightarrow0^-}= \lim_{x\rightarrow0^+}$$.

So, function is continuous at $$x=0$$ So function is continuous in all points..


Mathematics

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