Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# If F ( X ) = { | X + 2 | Tan − 1 ( X + 2 ) , X ≠ − 2 2 , X = − 2 (A) Continuous at X = − 2 (B) Not Continuous at X = − 2 (C) Differentiable at X = − 2 (D) Continuous but Not Derivable at X = − 2 - Mathematics

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#### Question

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

##### Options
• continuous at x = − 2

• not continuous at x = − 2

• differentiable at x = − 2

• continuous but not derivable at x = − 2

#### Solution

(b) not continuous at x = − 2

Given:

f(x) = {(|x+2|/(tan^(-1)(x+2)), x≠ -2),(2, x= -2):}

⇒ 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

$f( - 2) = 2$
Thus,
$\lim_{x \to - 2^-} f(x) \neq \lim_{x \to - 2^+} f(x)$
$f( - 2) .$

Therefore, given function is not continuous at x = − 2

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Solution If F ( X ) = { | X + 2 | Tan − 1 ( X + 2 ) , X ≠ − 2 2 , X = − 2 (A) Continuous at X = − 2 (B) Not Continuous at X = − 2 (C) Differentiable at X = − 2 (D) Continuous but Not Derivable at X = − 2 Concept: Algebra of Continuous Functions.
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