Question

Let [.] denote the greatest integer function and${\text{f}}\left( {\text{x}} \right) = \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right]$. Then
$
  {\text{A}}{\text{. }}\mathop {{\text{lim}}}\limits_{{\text{x}} \to 0} {\text{f}}\left( {\text{x}} \right){\text{ does not exist}} \\
  {\text{B}}{\text{. f}}\left( {\text{x}} \right){\text{ is continuous at x = 0}} \\
  {\text{C}}{\text{. f}}\left( {\text{x}} \right){\text{ is not differentiable at x = 0}} \\
  {\text{D}}{\text{. f}}\left( {\text{x}} \right){\text{ = 1}} \\
$

Answer 1

Hint: In order to find out which option best describes the given function, we start off by understanding the definition of a greatest integer function and computing what it means with respect to the given function f(x). We have to check if f(x) is continuous or not, by checking its value at its limits.

Complete step by step answer:
Given Data,
${\text{f}}\left( {\text{x}} \right) = \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right]$

Given function,${\text{f}}\left( {\text{x}} \right) = \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right]$, where [.] denotes the greatest integer function.

Let us check if the function f(x) is continuous.
We do this by verifying the value of f(x) at limits which are slightly below zero, slightly above zero and equal to zero.

Now the left hand limit of f(x):
$
   = \mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ - }} {\text{f}}\left( {\text{x}} \right) \\
   = \mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ - }} \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] \\
$
Now if the value of x is slightly less than zero, i.e. it is negative. From the trigonometric table of tan function we can say that the value of tan function is slightly greater than zero, therefore the value of ${\text{ta}}{{\text{n}}^2}{\text{x}}$is slightly greater than zero. Hence the greatest integer function value of${\text{ta}}{{\text{n}}^2}{\text{x}}$, i.e. $\left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] = 0$.
$ \Rightarrow {\text{L}}{\text{.H}}{\text{.L: }}\mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ - }} \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] = 0$

Now the right hand limit of f(x):
$
   = \mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ + }} {\text{f}}\left( {\text{x}} \right) \\
   = \mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ + }} \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] \\
$

Now if the value of x is slightly greater than zero, i.e. it is positive. From the trigonometric table of tan function we can say that the value of tan function is slightly greater than zero, therefore the value of ${\text{ta}}{{\text{n}}^2}{\text{x}}$ is slightly greater than zero. Hence the greatest integer function value of${\text{ta}}{{\text{n}}^2}{\text{x}}$, i.e. $\left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] = 0$.
$ \Rightarrow {\text{R}}{\text{.H}}{\text{.L: }}\mathop {{\text{lim}}}\limits_{{\text{x}} \to {0^ + }} \left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] = 0$

When x = 0, from the trigonometric table of tan function we can say that the value of tan function is equal to zero, therefore the value of ${\text{ta}}{{\text{n}}^2}{\text{x}}$is equal to zero. Hence the greatest integer function value of${\text{ta}}{{\text{n}}^2}{\text{x}}$, i.e. $\left[ {{\text{ta}}{{\text{n}}^2}{\text{x}}} \right] = 0$.

Therefore, L.H.L of f(x) = R.H.L of f(x) = f(x = 0)
Hence we can say that the function f(x) is continuous at x = 0
Option B is the correct answer.

Note In order to solve this type of question the key is to have good knowledge in the concepts of functions.
A greatest integer function is represented by [a] which gives us the value of the real number ‘a’ which is rounded off to the nearest integer of a.
Continuity function does not have any abrupt changes in its value. It can be proved by checking the value of the function when the limit is slightly less than zero, slightly greater than zero and equal to zero. They have to give the same answer.
Every differentiable function is continuous but every continuous function need not be differentiable.