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The function \mathrm{f(x)=x-\left|x-x^2\right|,-1 \leq x \leq 1} is continuous on the interval

 

 

Option: 1

[-1,1]
 


Option: 2

 (-1,1)
 


Option: 3

[-1,1] \backslash\{0\}
 


Option: 4

(-1,1) \backslash\{0\}


Answers (1)

best_answer

Given, \mathrm{f(x)=x-\left|x-x^2\right|,-1 \leq x \leq 1 }
\mathrm{\Rightarrow f(x)= \begin{cases}2 x-x^2 & ,-1 \leq x \leq 0 \\ x^2 & , 0 \leq x \leq 1\end{cases} }

Now,  \mathrm{ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(2 x-x^2\right)=0 and \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} x^2=0 }
Also, \mathrm{f(0)=0 }
\mathrm{\therefore f(x)\, \, is \, \, continuous\, \, in\, \, the \, \, interval\, \, [-1,1]. }

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manish

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