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प्रश्न
If \[f\left( x \right) = \begin{cases}\frac{\left| x + 2 \right|}{\tan^{- 1} \left( x + 2 \right)} & , x \neq - 2 \\ 2 & , x = - 2\end{cases}\] then f (x) is
पर्याय
continuous at x = − 2
not continuous at x = − 2
differentiable at x = − 2
continuous but not derivable at x = − 2
उत्तर
(b) not continuous at x = − 2
Given:
⇒ `f(x) = {((-|x+2|)/(tan^(-1)(x+2)), x< -2),((|x+2|)/(tan^(-1)(x+2)), x> -2),(2, x = -2):}`
Continuity at x = − 2.
(LHL at x= − 2) =
\[\lim_{x \to - 2^-} f(x) = \lim_{h \to 0} f( - 2 - h) = \lim_{h \to 0} \frac{- ( - 2 - h + 2)}{\tan^{- 1} ( - 2 - h + 2)} = \lim_{h \to 0} \frac{h}{\tan^{- 1} ( - h)} = - 1 . \]
(RHL at x = −2) =
\[\lim_{x \to - 2^+} f(x = \lim_{h \to 0} f( - 2 + h = \lim_{h \to 0} \frac{( - 2 + h + 2)}{\tan^{- 1} ( - 2 + h + 2)} = \lim_{h \to 0} \frac{h}{\tan^{- 1} (h)} = 1 .\]
Also
Therefore, given function is not continuous at x = − 2
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