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Question: If \[f:\mathbb{R} \to \mathbb{R}\] and \[g:\mathbb{R} \to \mathbb{R}\] defined by,\[f\left( x \right...

If f:R→Rf:\mathbb{R} \to \mathbb{R} and g:R→Rg:\mathbb{R} \to \mathbb{R} defined by,f(x)=∣x∣f\left( x \right) = \left| x \right|and g(x)=[x−3]g\left( x \right) = \left[ {x - 3} \right] for x∈Rx \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\\}

Explanation

Solution

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 [x]\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:R→Rf:\mathbb{R} \to \mathbb{R} and g:R→Rg:\mathbb{R} \to \mathbb{R} defined by, f(x)=∣x∣f\left( x \right) = \left| x \right|and g(x)=[x−3]g\left( x \right) = \left[ {x - 3} \right] for x∈Rx \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(f(x))=g(∣x∣)g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right), since, f(x)=∣x∣f\left( x \right) = \left| x \right| is given.
Again, g(f(x))=g(∣x∣)=[∣x∣−3]g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left[ {\left| x \right| - 3} \right] since, g(x)=[x−3]g\left( x \right) = \left[ {x - 3} \right] for x∈Rx \in \mathbb{R}.
Now we need to consider the region, −85<x<85\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}
If we take, x=0x = 0 then, g(f(x))=[∣0∣−3]=−3g\left( {f\left( x \right)} \right) = \left[ {\left| 0 \right| - 3} \right] = - 3.
If we take, x=12x = \dfrac{1}{2} then g(f(x))=[∣12∣−3]=[12−3]=[−52]=−3g\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<\-52<\-2 - 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=−12x = - \dfrac{1}{2} then g(f(x))=[∣−12∣−3]=[12−3]=[−52]=−3g\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=1x = 1 then, g(f(x))=[∣1∣−3]=[1−3]=−2g\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, ∣2∣=2&\left| 2 \right| = 2\& also ∣−2∣=2\left| { - 2} \right| = 2.
The box function is the greatest integer function, returning the integer just below the value entered, denoted by [x]\left[ x \right] .
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.