Question

The range of the function \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}}\]
(1) \[\left\{ {1,2} \right\}\]
(2) \[\left\{ {1,2,3} \right\}\]
(3) \[\left\{ {1,2,3,4,5} \right\}\]
(4) \[\left\{ 1 \right\}\]

Answer 1

Hint: The given function is \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}}\] . Firstly, we will find the range of the variable \[x\] from the function \[f\left( x \right)\] and then from the formula of permutation i.e., \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .After that we will put the different values of \[x\] in the function \[f\left( x \right)\] .And finally, we will get the range of the given function.

Complete step-by-step answer:
The given function is, \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}}\]
First of all, we will find the range of the variable \[x\] from the function \[f\left( x \right)\]
Now, we know that the given function is defined when
\[x - 1 \geqslant 0\]
\[ \Rightarrow x \geqslant 1{\text{ }} - - - \left( 1 \right)\]
Now, we will find the range of the variable \[x\] by applying the formula of permutation i.e., \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
So, \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}}\]
Here, \[n = 5 - x\] and \[r = x - 1\]
Therefore, \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}} = {\text{ }}\dfrac{{\left( {5 - x} \right)!}}{{\left( {5 - x - \left( {x - 1} \right)} \right)!}}\]
\[ \Rightarrow f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}} = {\text{ }}\dfrac{{\left( {5 - x} \right)!}}{{\left( {5 - x - x + 1} \right)!}}\]
On simplifying, we get
\[ \Rightarrow f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}} = {\text{ }}\dfrac{{\left( {5 - x} \right)!}}{{\left( {6 - 2x} \right)!}}\]
So, the given function is also defined when
\[6 - 2x \geqslant 0\]
\[ \Rightarrow - 2x \geqslant - 6\]
Now, we know that whenever any term is multiplied or divided by a negative number then the sign of the inequality changes.
Therefore, on multiplying the above inequality by \[ - 1\] , we get
\[x \leqslant 3{\text{ }} - - - \left( 2 \right)\]
So, from \[\left( 1 \right)\] and \[\left( 2 \right)\] we get arrange of variable \[x\] as \[1 \leqslant x \leqslant 3\]
Thus, the range of \[x\] will be \[\left\{ {f\left( 1 \right),{\text{ }}f\left( 2 \right),{\text{ }}f\left( 3 \right)} \right\}\]
Now,
\[f\left( 1 \right) = {\text{ }}{}^{5 - 1}{P_{1 - 1}} = {\text{ }}{}^4{P_0} = {\text{ }}\dfrac{{4!}}{{\left( {4 - 0} \right)!}}{\text{ }} = \dfrac{{4!}}{{4!}} = 1\]
\[f\left( 2 \right) = {\text{ }}{}^{5 - 2}{P_{2 - 1}} = {\text{ }}{}^3{P_1}{\text{ }} = {\text{ }}\dfrac{{3!}}{{\left( {3 - 1} \right)!}}{\text{ }} = \dfrac{{3!}}{{2!}}{\text{ }} = \dfrac{{3 \cdot 2!}}{{2!}}{\text{ }} = 3\]
\[f\left( 3 \right) = {\text{ }}{}^{5 - 3}{P_{3 - 1}} = {\text{ }}{}^2{P_2}{\text{ }} = {\text{ }}\dfrac{{2!}}{{\left( {2 - 2} \right)!}}{\text{ }} = \dfrac{{2!}}{{0!}}{\text{ }} = \dfrac{{2!}}{1}{\text{ }} = 2\]
Thus, we get the range of the given function as \[\left\{ {1,3,2} \right\}\]
Hence, the range of the function is \[\left\{ {1,2,3} \right\}\]
So, the correct answer is “Option 2”.

Note: There is an alternative way to find the range of the variable \[x\]
Here, the given function is \[f\left( x \right) = {}^{\left( {5 - x} \right)}{P_{\left( {x - 1} \right)}}\]
Now, we know that the given function is defined when
\[x - 1 \geqslant 0\]
\[ \Rightarrow x \geqslant 1{\text{ }} - - - \left( 1 \right)\]
Also, we know that any permutation \[{}^n{P_r}\] is defined when
\[n \geqslant r\]
So, according to this question
\[5 - x \geqslant x - 1\]
Taking \[x\] terms on the right-hand side and constant terms on the left-hand side, we get
\[5 + 1 \geqslant x + x\]
\[ \Rightarrow 6 \geqslant 2x\]
\[ \Rightarrow x \leqslant 3{\text{ }} - - - \left( 2 \right)\]
Thus from \[\left( 1 \right)\] and \[\left( 2 \right)\] we get the range of the variable \[x\] as
\[1 \leqslant x \leqslant 3\]