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

A
continuous at x = - 2
B
not continuous at x = - 2
C
differentiable at x = - 2
D
continuous but not derivable at x = - 2

Solution:

We have,

$lim_{x \to - 2^{-}} f (x) = lim_{x \to - 2^{-}} \frac{| x + 2 |}{\tan^{- 1} (x + 2)} = lim_{x \to - 2^{-}} \frac{- (x + 2)}{\tan^{- 1} (x + 2)} = - 1$

and,

$\begin{array}{l} lim_{x \to - 2^{+}} f (x) = lim_{x \to - 2^{+}} \frac{| x + 2 |}{\tan^{- 1} (x + 2)} = lim_{x \to - 2^{+}} \frac{x + 2}{\tan^{- 1} (x + 2)} = 1 \\ ∴ lim_{x \to - 2^{-}} f (x) \neq lim_{x \to - 2^{+}} f (x) . \end{array}$ .

So, $f (x)$ is neither continuous nor differentiable at x=-2.

If $f (x) = \{\begin{array}{cc} \frac{| x + 2 |}{\tan^{- 1} (x + 2)}, & x \neq - 2 \\ 2, & x = - 2 \end{array}$ , then $f (x)$ , is

Solution:

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