Question

Which of the following function is not continuous for all real x?
a)e^×
b)tanx
c)tan^-1(x)
d)e^-x​

Answer 1

Answer:

I think option (c) is a answer

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Answer 2

Answer

The function tan(x) is not continuous ∀ x ∈ R.

Concept

Continuity of a function: Let f be a real valued function. Let a be a point in its domain. Then, f is continuous at x = a if the right hand and left hand limits at x = a are finite and are equal to the value of the function at x = a. Mathematically,

               $\lim_{x\rightarrow a^-} f(x)= f(a) = \lim_{x\rightarrow a^+} f(x)$,

where both limits are finite.

Solution

Calculate the right hand limit at x = a for the function eˣ.

RHL = $\lim_{x\rightarrow a^+}f(x) = \lim_{h \rightarrow 0}f(a + h) = \lim_{h \rightarrow 0}e^{a + h} = e^a\lim_{h \rightarrow 0}e^h = e^a e^0$

⇒  RHL = $e^a$.

Calculate the left hand limit at x = a.

LHL = $\lim_{x\rightarrow a^-}f(x) = \lim_{h \rightarrow 0}f(a - h) = \lim_{h \rightarrow 0}e^{a - h} = e^a\lim_{h \rightarrow 0}e^{-h} = e^a e^0$

⇒  LHL = $e^a$.

Both LHL and RHL are finite and equal to f(a) = $e^a$.

So, eˣ is continuous for all x in R.

At all odd multiples of $\pm \pi/2$, tan(x) is ± $\infty$. So, the value of the function is not finite at infinitely many values of x in R.

tan(x) is, thus, not continuous for all x in R.

Consider the function $tan:\;[-\pi/2,\;\+\pi/2]\; \rightarrow\; R$. The function tan is continuous in the interval $(-\pi/2,\;+\pi/2)$ with the range $(-\infty, +\infty)$. So, the function $\tan^{-1}:\, (-\infty,+\infty)\;\rightarrow\; (-\pi/2,+\pi/2)$ being the inverse of the same is continuous.

Moreover,

$\lim_{x \rightarrow \infty}\tan^{-1} x = \pi/2 = \tan^{-1} (\infty)$,

and $\lim_{x\rightarrow -\infty}\tan^{-1}x = -\pi/2 = \tan^{-1} (-\infty)$.

So, $\tan^{-1} x$ is continuous at x = $\pm \infty$.

Thus, $\tan^{-1} x$ is continuous for all x in R.

e⁻ˣ = 1/ eˣ. Since eˣ ≠ 0, 1/eˣ is also continuous.

So, e⁻ˣ is a continuous function for all x in R.

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