Limit Continuity And Differentiability Ques 88

Answer:

Correct Answer: 88.(a)

Solution:

88. Given, $f(x)=\left[\tan ^{2} x\right]$

Now, $-45^{\circ}<x<45^{\circ}$

$$ \begin{array}{lc} \Rightarrow & \tan \left(-45^{\circ}\right)<\tan x<\tan \left(43^{\circ}\right) \\ \Rightarrow & -\tan 45^{\circ}<\tan x<\tan 45^{\circ} \\ \Rightarrow & -1<\tan x<1 \\ \Rightarrow & 0<\tan ^{2} x<1 \\ for 0<x<\frac{\pi}{2} \Rightarrow & {\left[\tan^{2} x\right]=0} \end{array} $$

i.e. $f(x)$ is zero for all values of $x$ from $x=-45^{\circ}$ to $45^{\circ}$. Thus, $f(x)$ exists when $x \rightarrow 0$ and also it is continuous at $x=0$. Also, $f(x)$ is differentiable at $x=0$ and has a value of zero.

Therefore, (b) is the answer.