Question

If $f:R \to R{\text{ and }}g:R \to R{\text{ defined by }}f(x) = \left| x \right|{\text{ and }}g(x) = \left[ {x - 3} \right]{\text{ for }}x \in R.$
[.] denotes greatest integer function, then $\\{ g(f(x)):\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}\\} =$
A. $\\{ 0,1\\}$
B. $\\{ - 1, - 2\\}$
C. $\\{ - 3, - 2\\}$
D. $\\{ 2,3\\}$

Answer 1

As $f(x) = \left| x \right|$ and $g(x) = \left[ {x - 3} \right]$
So we can take the two cases
Case (1) for $x > 0$
$\Rightarrow$ $f(x) = x{\text{ and }}g(x) = \left[ {x - 3} \right]$
$\Rightarrow$ $g(f(x)) = g(x) = \left[ {x - 3} \right]$
And for the case (2) for $x < 0$
$\Rightarrow$ $f(x) = - x{\text{ and }}g(x) = \left[ {x - 3} \right]$
$\Rightarrow$ $g(f(x)) = g( - x) = \left[ { - x - 3} \right]$
Where [.] denotes the greatest integer function.

Complete step-by-step answer:
Here we are given that $f:R \to R{\text{ and }}g:R \to R{\text{ }}$
This means that range and domain of both the functions can be real number and also we are given that $f(x) = \left| x \right|$ and $g(x) = \left[ {x - 3} \right]$
Where [.] denotes the greatest integer function.
So as we know that GIF of any number would be the integer just smaller to that of the given number
If the number is the integer then the GIF of that number will be integer itself.
Now we are given
$\Rightarrow$ $f(x) = \left| x \right|$
So we can take the two cases where $f(x) = \left| x \right|$ is defined.
Case (1) for $x > 0$
Here $x$ is positive
$\Rightarrow$ $f(x) = x{\text{ and }}g(x) = \left[ {x - 3} \right]$
In $g(f(x))$ we just need to replace $f(x)$ by $x$
$\Rightarrow$ $g(f(x)) = g(x) = \left[ {x - 3} \right]$
And for the case (2) for $x < 0$
Here $x$ is negative
$\Rightarrow$ $f(x) = - x{\text{ and }}g(x) = \left[ {x - 3} \right]$
$\Rightarrow$ $g(f(x)) = g( - x) = \left[ { - x - 3} \right]$
So for $x < 0$
$g(f(x)) = g( - x) = \left[ { - x - 3} \right]$
So we can write that $g(f(x))$ $= \left[ { - x - 3} \right]{\text{ for }}x < 0 \\\ = \left[ {x - 3} \right]{\text{ for }}x > 0 \\\$
And we need to find between $\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}$
So for $\dfrac{{ - 8}}{5} < x < \dfrac{8}{5}$ that means $- 1.6 < x < 1.6$
$\Rightarrow$ $g(f(x)) = \left[ { - x - 3} \right]$
So if we will multiply by -1, the equation will change
As we know that $- 1 > - 2$ and if we multiply $- 1$ on both the sides then $1 < 2$ sign of equality will change, so here we got$- \dfrac{8}{5} < x < 0$ and now if we multiply by$- 1$, we get
$\Rightarrow$ $0 < \- x < \dfrac{8}{5}$. Now if we subtract $- 3$ we will get
$\Rightarrow$ $0 - 3 < \- x - 3 < \dfrac{8}{5} - 3$
$\Rightarrow$ $- 3 < \- x - 3 < \dfrac{{ - 7}}{5}$
And we can write it in this way also
$\Rightarrow$ $- 3 < \- x - 3 < \- 2$ $\cup$ $- 2 < \- x - 3 < \dfrac{{ - 7}}{5}$
Which means that $- x - 3 \in [ - 3, - 2) \cup [ - 2, - \dfrac{7}{5})$
So if we take GIF of $\left[ { - x - 3} \right]$ we will get$\left[ { - x - 3} \right] = - 3{\text{ and }} - 2$
So for $\left[ { - x - 3} \right]$we get$\\{ - 3, - 2\\}$
For case (1)
$x > 0$
$\Rightarrow$ $g(f(x)) = [x - 3]$
And we have to find for $0 < x < \dfrac{8}{5}$
So subtracting $3$ we get that
$\Rightarrow$ $0 - 3 < x - 3 < \dfrac{8}{5} - 3$
$\Rightarrow$ $- 3 < x - 3 < \dfrac{{ - 7}}{5}$
So we got $(x - 3) = \left( { - 3,\dfrac{{ - 7}}{5}} \right)$
So upon taking GIF of $g(f(x)) = [x - 3]$ we get $\\{ - 3, - 2\\}$
So upon combining both the answers we get$\\{ - 3, - 2\\}$

Hence option C is correct.

Note: If any equality is given $a < x < b$ and if we multiply by $- 1$ then the in inequality will change which means $- b < \- x < \- a$
For example: As we know that $- 1 > - 2$ and if we multiply $- 1$ on both the sides then $1 < 2$ sign of equality gets changed.
Also now if we are given that $f$ is the inverse of $g$ then
$\Rightarrow$ $g(x) = {f^{ - 1}}(x)$ then $f(g(x)) = x$
Upon differentiating, we get
$\Rightarrow$ $f'(g(x)).g'(x) = 1$
$\Rightarrow$ $f'(g(x)) = \dfrac{1}{{g'(x)}}$