
If \[f:\mathbb{R} \to \mathbb{R}\] and \[g:\mathbb{R} \to \mathbb{R}\] defined by,\[f\left( x \right) = \left| x \right|\]and \[g\left( x \right) = \left[ {x - 3} \right]\] for \[x \in \mathbb{R}\].
\[\left[ . \right]\] denotes greatest integer function, then \[\left\{ {g\left( {f\left( x \right)} \right):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}} \right\} = \]
(A) \[\left\{ {0,1} \right\}\]
(B) \[\left\{ { - 1, - 2} \right\}\]
(C) \[\left\{ { - 3, - 2} \right\}\]
(D) \[\left\{ {2,3} \right\}\]
Answer
476.1k+ views
Hint: In this question, we have to find out the region in the specific range.
We know that Modulus is the function which gives the absolute value of a real or complex number. It will always return the positive value.
The box function, more commonly known as the greatest integer function, returns the integer just below the value entered, denoted by \[\left[ x \right]\] .
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.
Complete step-by-step answer:
It is given that \[f:\mathbb{R} \to \mathbb{R}\] and \[g:\mathbb{R} \to \mathbb{R}\] defined by, \[f\left( x \right) = \left| x \right|\]and \[g\left( x \right) = \left[ {x - 3} \right]\] for \[x \in \mathbb{R}\].
\[\left[ . \right]\] denotes greatest integer function,
We have to find out the set \[\left\{ {g\left( {f\left( x \right)} \right):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}} \right\}\]
Now, \[g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right)\], since, \[f\left( x \right) = \left| x \right|\] is given.
Again, \[g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left[ {\left| x \right| - 3} \right]\] since, \[g\left( x \right) = \left[ {x - 3} \right]\] for \[x \in \mathbb{R}\].
Now we need to consider the region, \[\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}\]
If we take, \[x = 0\] then, \[g\left( {f\left( x \right)} \right) = \left[ {\left| 0 \right| - 3} \right] = - 3\].
If we take, \[x = \dfrac{1}{2}\] then \[g\left( {f\left( x \right)} \right) = \left[ {\left| {\dfrac{1}{2}} \right| - 3} \right] = \left[ {\dfrac{1}{2} - 3} \right] = \left[ {\dfrac{{ - 5}}{2}} \right] = - 3\]
As \[ - 3 < - \dfrac{5}{2} < - 2\] and for box function the greatest integer function, returns the integer just below the value entered.
If we take, \[x = - \dfrac{1}{2}\] then \[g\left( {f\left( x \right)} \right) = \left[ {\left| { - \dfrac{1}{2}} \right| - 3} \right] = \left[ {\dfrac{1}{2} - 3} \right] = \left[ {\dfrac{{ - 5}}{2}} \right] = - 3\]
If we take, \[x = 1\] then, \[g\left( {f\left( x \right)} \right) = \left[ {\left| 1 \right| - 3} \right] = \left[ {1 - 3} \right] = - 2\].
Hence, \[\left\{ {g\left( {f\left( x \right)} \right):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}} \right\} = \]\[\left\{ { - 3, - 2} \right\}\].
Thus (C) is the correct option.
Note: Modulus of a number will always return the positive value.
That is for example, \[\left| 2 \right| = 2\& \] also \[\left| { - 2} \right| = 2\].
The box function is the greatest integer function, returning the integer just below the value entered, denoted by \[\left[ x \right]\] .
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.
We know that Modulus is the function which gives the absolute value of a real or complex number. It will always return the positive value.
The box function, more commonly known as the greatest integer function, returns the integer just below the value entered, denoted by \[\left[ x \right]\] .
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.
Complete step-by-step answer:
It is given that \[f:\mathbb{R} \to \mathbb{R}\] and \[g:\mathbb{R} \to \mathbb{R}\] defined by, \[f\left( x \right) = \left| x \right|\]and \[g\left( x \right) = \left[ {x - 3} \right]\] for \[x \in \mathbb{R}\].
\[\left[ . \right]\] denotes greatest integer function,
We have to find out the set \[\left\{ {g\left( {f\left( x \right)} \right):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}} \right\}\]
Now, \[g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right)\], since, \[f\left( x \right) = \left| x \right|\] is given.
Again, \[g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left[ {\left| x \right| - 3} \right]\] since, \[g\left( x \right) = \left[ {x - 3} \right]\] for \[x \in \mathbb{R}\].
Now we need to consider the region, \[\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}\]
If we take, \[x = 0\] then, \[g\left( {f\left( x \right)} \right) = \left[ {\left| 0 \right| - 3} \right] = - 3\].
If we take, \[x = \dfrac{1}{2}\] then \[g\left( {f\left( x \right)} \right) = \left[ {\left| {\dfrac{1}{2}} \right| - 3} \right] = \left[ {\dfrac{1}{2} - 3} \right] = \left[ {\dfrac{{ - 5}}{2}} \right] = - 3\]
As \[ - 3 < - \dfrac{5}{2} < - 2\] and for box function the greatest integer function, returns the integer just below the value entered.
If we take, \[x = - \dfrac{1}{2}\] then \[g\left( {f\left( x \right)} \right) = \left[ {\left| { - \dfrac{1}{2}} \right| - 3} \right] = \left[ {\dfrac{1}{2} - 3} \right] = \left[ {\dfrac{{ - 5}}{2}} \right] = - 3\]
If we take, \[x = 1\] then, \[g\left( {f\left( x \right)} \right) = \left[ {\left| 1 \right| - 3} \right] = \left[ {1 - 3} \right] = - 2\].
Hence, \[\left\{ {g\left( {f\left( x \right)} \right):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}} \right\} = \]\[\left\{ { - 3, - 2} \right\}\].
Thus (C) is the correct option.
Note: Modulus of a number will always return the positive value.
That is for example, \[\left| 2 \right| = 2\& \] also \[\left| { - 2} \right| = 2\].
The box function is the greatest integer function, returning the integer just below the value entered, denoted by \[\left[ x \right]\] .
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
