For function fx-tan-1 1x - x0 which of the following statement is TRUE

Clearly, f(x) is not defined at x=0 and lim_x→ 0^+ nbsp;f(x)=π2 lim_x→ 0^ nbsp;f(x)= π2 ⇒lim_x→ 0 nbsp;f(x) doesn't exists ∴ f(x) has finite type non rem ...

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Standard XII

Mathematics

Integral as Antiderivative

For function ...

Question

For function $f (x) =$ ${tan}^{- 1} (\frac{1}{x}), x \neq 0$ which of the following statement is TRUE

f (x)

has missing point discontinuity

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f (x)

has finite type discontinuity

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f (x)

has infinite type discontinuity

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f (x)

has isolated point discontinuity

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Solution

The correct option is B $f (x)$ has finite type discontinuity
Clearly, $f (x)$ is not defined at $x = 0$ and
$lim x \to 0^{+} f (x) = \frac{π}{2}$
$lim x \to 0^{-} f (x) = - \frac{π}{2}$
$\Rightarrow lim x \to 0 f (x)$ doesn't exists
$∴ f (x)$ has finite type non-removable discontinuity.

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For function f(x)=tan−1(1x),x≠0 which of the following statement is TRUE

The correct option is B f(x) has finite type discontinuityClearly, f(x) is not defined at x=0 and limx→0+f(x)=π2 limx→0−f(x)=−π2 ⇒limx→0f(x) doesn't exists ∴f(x) has finite type non-removable discontinuity.

For function $f (x) =$ ${tan}^{- 1} (\frac{1}{x}), x \neq 0$ which of the following statement is TRUE

The correct option is B $f (x)$ has finite type discontinuity
Clearly, $f (x)$ is not defined at $x = 0$ and
$lim x \to 0^{+} f (x) = \frac{π}{2}$
$lim x \to 0^{-} f (x) = - \frac{π}{2}$
$\Rightarrow lim x \to 0 f (x)$ doesn't exists
$∴ f (x)$ has finite type non-removable discontinuity.