If $$f(x)=$$|$$x+2$$|$$tan$$ −$$1$$⁡(x+2), x≠−$$22$$, $$x=-2then$$ $$f(x)$$ is

Correct option is 2not continuous at x=−2

Questions

$If f (x) = \{\begin{cases} \frac{| x + 2 |}{\tan_{}^{- 1} (x + 2)}, x \neq - 2 \\ 2, x = - 2 \end{cases} then f (x) is$

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Continuous at x=−2

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continuous but not differentiable at x=−2

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detailed solution

Correct option is B

limx→−2−f(x)=limx→−2−x+2tan−1x+2=limx→−2−−x+2tan−1x+2=−1 ∵limx→0−tan−1xx=1limx→−2+f(x)=limx→−2+x+2tan−1x+2=limx→−2+x+2tan−1x+2=1∴limx→−2f(x) does not exist.

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If f(x)=|x+2|tan −1⁡(x+2), x≠−22, x=-2then f(x) is

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free study material

$If f (x) = \{\begin{cases} \frac{| x + 2 |}{\tan_{}^{- 1} (x + 2)}, x \neq - 2 \\ 2, x = - 2 \end{cases} then f (x) is$