Question

The range of the function $f\left( x \right) = {}^{7 - x}{P_{x - 3}}$ is
A) {1, 2, 3}
B) {1, 2, 3, 4, 5}
C) {1, 2, 3, 4}
D) {1, 2, 3, 4, 5, 6}

Answer 1

Hint:
The given function is $f\left( x \right) = {}^{7 - x}{P_{x - 3}}$ .
Firstly, find the range of the variable x from the function \[f\left( x \right)\] .
Then, put the different values of x in function \[f\left( x \right)\] .
Finally, this will give the range of the given function $f\left( x \right) = {}^{7 - x}{P_{x - 3}}$

Complete step by step solution:
The given function is $f\left( x \right) = {}^{7 - x}{P_{x - 3}}$ .
So, the given function is defined when $x - 3 \geqslant 0$ .
 $\therefore x \geqslant 3$
Now, $f\left( x \right) = {}^{7 - x}{P_{x - 3}} = \dfrac{{\left( {7 - x} \right)!}}{{\left( {7 - x - x + 3} \right)!}} = \dfrac{{\left( {7 - x} \right)!}}{{\left( {10 - 2x} \right)!}}$
So, the given function is also defined when $10 - 2x \geqslant 0$ .
 $
  \therefore - 2x \geqslant - 10 \\
  \therefore x \leqslant 5 \\
 $
Thus, we get a range of variable x as $3 \leqslant x \leqslant 5$ .
So, the range of x will be \[\left\{ {f\left( 3 \right),f\left( 4 \right),f\left( 5 \right)} \right\}\]
Now,
 $f\left( 3 \right) = {}^{7 - 3}{P_{3 - 3}} = {}^4{P_0} = \dfrac{{4!}}{{\left( {4 - 0} \right)!}} = \dfrac{{4!}}{{4!}} = 1$
 $f\left( 4 \right) = {}^{7 - 4}{P_{4 - 3}} = {}^3{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = 3$
 $f\left( 5 \right) = {}^{7 - 5}{P_{5 - 3}} = {}^2{P_2} = \dfrac{{2!}}{{\left( {2 - 2} \right)!}} = \dfrac{{2!}}{{0!}} = 2$
Thus, we get the range of the given function as {1, 3, 2} = {1, 2, 3}

So, option (A) is correct.

Note:
Alternate method to find the range of the variable x:
Here, the given function is $f\left( x \right) = {}^{7 - x}{P_{x - 3}}$ .
So, \[x-3\] must be greater than or equal to 0.
 $x - 3 \geqslant 0$
 $\therefore x \geqslant 3$
Also, any permutation ${}^n{P_r}$ is defined when $n \geqslant r$ .
So, in this question $7 - x \geqslant x - 3$
 $
  \therefore 7 + 3 \geqslant 2x \\
  \therefore 2x \leqslant 10 \\
  \therefore x \leqslant 5 \\
 $
Thus, we get the range of variable x as $3 \leqslant x \leqslant 5$ .