Question

Let $\left[.\right]$ denote the greatest integer function $f\left(x\right)=\left[{\tan }^{2}x\right]$, then:

• A. $li{m}_{x\to 0}f\left(x\right)$ does not exist
• B. $f\left(x\right)$ is continuous at $x=0$
• C. $f\left(x\right)$ is not differentiable at $x=0$
• D. $f^{\text{'}}\left(x\right)=1$

Answer 1

The correct option is B

$f\left(x\right)$ is continuous at $x=0$

Explanation for correct option:

Option B: $f\left(x\right)$ is continuous at $x=0$

Consider the given equation as: $f\left(x\right)=\left[{\tan }^{2}x\right]$

Condition for the $f\left(x\right)$ is continuous at $x=0$

$f\left(0\right)=f\left(0^{+}\right)=f\left(0^{-}\right)$

Check the condition for the given Equation.

$\Rightarrow f\left(0\right)=\left[{\tan }^{2}0\right]=0$ [since, $x=0$]

$$\begin{gathered} \Rightarrow f\left(0^{+}\right)=li{m}_{h\to 0}\left[{\tan }^2\left(0+h\right)\right] \\ \Rightarrow f\left(0^{+}\right)=li{m}_{h\to 0}\left[{\tan }^{2}h\right] \\ \Rightarrow f\left(0^{+}\right)=\left[{\tan }^{2}0\right] \\ \Rightarrow f\left(0^{+}\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow f\left(0^{-}\right)=li{m}_{h\to 0}\left[{\tan }^2\left(0-h\right)\right] \\ \Rightarrow f\left(0^{-}\right)=li{m}_{h\to 0}\left[{\left(-\tan h\right)}^2\right] \\ \Rightarrow f\left(0^{-}\right)=\left[{\tan }^{2}0\right] \\ \Rightarrow f\left(0^{-}\right)=0 \end{gathered}$$

$\therefore f\left(0\right)=f\left(0^{+}\right)=f\left(0^{-}\right)=0$

Thus, option (B) is the correct answer

Explanation for incorrect option(s):

Option A: $li{m}_{x\to 0}f\left(x\right)$ does not exist

$$\begin{gathered} \Rightarrow li{m}_{x\to 0}f\left(x\right)=li{m}_{x\to 0}\left[{\tan }^2\left(x\right)\right] \\ \Rightarrow li{m}_{x\to 0}f\left(x\right)=\left[{\tan }^{2}0\right] \\ \Rightarrow li{m}_{x\to 0}f\left(x\right)=0 \end{gathered}$$

Hence, $li{m}_{x\to 0}f\left(x\right)$ exist

Thus, option (A) is an incorrect answer

Option C: $f\left(x\right)$ is not differentiable at $x=0$

$$\begin{gathered} f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h} \\ f\text{'}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^{2}h\right]-\left[{\tan }^{2}0\right]}{h} \\ f\text{'}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^{2}h\right]}{h}=0 \end{gathered}$$

$$\begin{gathered} f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a-h\right)-f\left(a\right)}{-h} \\ f\text{'}\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^2\left(-h\right)\right]-\left[{\tan }^{2}0\right]}{-h} \\ f\text{'}\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^{2}h\right]}{-h}=0 \end{gathered}$$

$\therefore f\left(0^{+}\right)=f\left(0^{-}\right)$

Thus, Option (C) is the correct answer

Option D: $f^{\text{'}}\left(x\right)=1$

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^{2}h\right]-\left[{\tan }^{2}0\right]}{h}\left[x=0\right] \\ f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\left[{\tan }^{2}h\right]}{h} \\ f\text{'}\left(0\right)=\frac{\left[{\tan }^{2}0\right]}{h} \\ f\text{'}\left(0\right)=0 \end{gathered}$$

$f^{\text{'}}\left(x\right)\ne 1$

Thus, Option (D) is an incorrect answer.

Therefore, the correct answer is Option (B) .