Question

Let $g(x) = 1 + x - [x]$ (where $[]$ represents GIF) and f(x) = \left\\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} { - 1}&:&{x < 0} \end{array}} \\\ {\begin{array}{*{20}{c}} 0&:&{x = 0} \end{array}} \\\ {\begin{array}{*{20}{c}} 1&:&{x > 0} \end{array}} \end{array}} \right.
Then $\int\limits_0^4 {f\left( {g(x)} \right)}$ is
A. 1
B. 0
C. 2
D. 4

Answer 1

This problem deals with the greatest integer function. The greatest integer function is also known as the floor function. It is written as $f(x) = [x]$. The value of $[x]$ is the largest integer that is less than or equal to $x$. For all real numbers, $x$, the greatest integer function returns the largest integer, less than or equal to $x$. In essence, it rounds down a real number to the nearest integer. For example:
$\Rightarrow \left[ 1 \right] = 1$
$\Rightarrow \left[ {1.5} \right] = 1$
$\Rightarrow \left[ {3.7} \right] = 3$
$\Rightarrow \left[ {4.3} \right] = 4$

Complete step-by-step answer:
Given that the function $g(x)$ is equal to $1 + x - [x]$, which is given below:
$\Rightarrow g(x) = 1 + x - [x]$
Consider the expression $x - [x]$, as given below:
We know that $x$ is a real number and $[x]$ is definitely an integer, as it is the greatest integer function.
So the difference between a real number and its integer, gives only the fractional part.
\Rightarrow x - [x] = \left\\{ x \right\\}
Where \left\\{ x \right\\} is the fractional part where it lies in between only 0 and 1.
Here \left\\{ x \right\\} \in \left( {0,1} \right)
So the function $g(x)$ is equal to , as given below:
$\Rightarrow g(x) = 1 + x - [x]$
\Rightarrow g(x) = 1 + \left\\{ x \right\\}
Given that the function $f(x)$ is equal to 1, whenever $x$ is greater than zero.
$\Rightarrow f(x) = 1;x > 0$
We know that the function $g(x)$ is 1 + \left\\{ x \right\\}, where \left\\{ x \right\\} \in \left( {0,1} \right)
Here $g(x) > 0\forall x \in R$,
$\therefore f\left( {g(x)} \right) = 1,\forall x$
Now consider the integral $\int\limits_0^4 {f\left( {g(x)} \right)}$, as given below:
$\Rightarrow \int\limits_0^4 {f\left( {g(x)} \right)} dx$
We know that $f\left( {g(x)} \right) = 1,\forall x$
$\Rightarrow \int\limits_0^4 1 dx = \left. x \right|_0^4$
$\Rightarrow 4 - 0$
Hence the integral $\int\limits_0^4 {f\left( {g(x)} \right)}$ is equal to:

$\therefore \int\limits_0^4 {f\left( {g(x)} \right)} = 4$

Note:
Please note that when the intervals are in the form of $\left[ {n,n + 1} \right)$, the value of greatest integer function is $n$, where $n$ is an integer. $0 \leqslant x < 1$ will always lie in the interval $\left[ {0,0.9} \right)$ so here the greatest integer function of $x$ will be 0. Whereas the greatest the negative integer is the first negative integer from zero. The greatest negative integer is the first negative integer from zero. The first negative integer from zero is one less than 0 and the number is -1.