Question

.If \[f:R \to R\]and \[g:R \to R\]are given by \[f(x) = \left| x \right|\]and \[g(x) = \left[ x \right]\]for each \[x \in R\], then \[\{ x \in R:g(f(x)) \le f(g(x))\} \]=
A) \[Z \cup ( - \infty ,0)\]
B) \[x = \dfrac{9}{2}\]
C) Z
D) R

Answer 1

Hint: Here are two function f which is mapping from real number to real number and g which is also mapping from real number to real number. We have to relate the function f to g and then g to f. then we will get the relation of \[g[f(x)]\] to \[f[g(x)]\].



Formula Used:For every \[x \in R\]
               If \[f(x) = x\], \[g(x) = x'\]
                \[f[g(x)] = f(x')\]
                                =\[x'\]
Also \[\forall x \in R\], \[\left[ { - x} \right]\]=\[ - x\]
                                           \[\left| x \right| = x\]



Complete step by step solution:Given \[f:R \to R\]and \[g:R \to R\]
\[f(x) = \left| x \right|\]and \[g(x) = \left[ x \right]\], \[\forall x \in R\]
Now given \[g[f(x)] \le f[g(x)]\]
                        \[g(\left| x \right|) \le f(\left[ x \right])\]
                         \[\left[ {\left| x \right|} \right] \le \left| {\left[ x \right]} \right|\]
Here now two cases arises,
Case (1): \[x = ( - \infty ,0)\]
 Put \[x = \dfrac{{ - 9}}{2}\]
                      \[\left[ {\left| x \right|} \right] = \left[ {\dfrac{9}{2}} \right]\]
                     \[\left[ {\left| x \right|} \right] = \left[ {4.5} \right]\]
                              =4
         Also, \[\left| {\left[ x \right]} \right| = \left[ {\dfrac{{ - 9}}{2}} \right]\]
                    \[\left| {\left[ x \right]} \right| = \left| { - 5} \right|\]
                            =5
Thus \[\left[ {\left| x \right|} \right] \le \left| {\left[ x \right]} \right|\]
Case (2): \[x = (0, + \infty )\]
                      \[x = \dfrac{9}{2}\]
                \[\left[ {\left| x \right|} \right] = \left[ {\left| {\dfrac{9}{2}} \right|} \right]\]
               \[\left[ {\left| x \right|} \right] = \left[ {4.5} \right]\]
                         =4
               \[\left| {\left[ x \right]} \right| = \left| {\dfrac{9}{2}} \right|\]
                       =4
Thus \[\left[ {\left| x \right|} \right] = \left| {\left[ x \right]} \right|\]
Hence we can see that this function\[g[f(x)] \le f[g(x)]\]is true for all the value of real number.



Option ‘D’ is correct



Note:Student must remember when the closed interval function of \[\left[ x \right]\]is negative then its value is negative and when mod of function \[\left| x \right|\]is negative then its value is positive.