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Question

Question: The range of the function \(f\left( x \right) = {}^{7 - x}{P_{x - 3}}\) is A) {1, 2, 3} B) {1, 2...

The range of the function f(x)=7xPx3f\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}

Explanation

Solution

The given function is f(x)=7xPx3f\left( x \right) = {}^{7 - x}{P_{x - 3}} .
Firstly, find the range of the variable x from the function f(x)f\left( x \right) .
Then, put the different values of x in function f(x)f\left( x \right) .
Finally, this will give the range of the given function f(x)=7xPx3f\left( x \right) = {}^{7 - x}{P_{x - 3}}

Complete step by step solution:
The given function is f(x)=7xPx3f\left( x \right) = {}^{7 - x}{P_{x - 3}} .
So, the given function is defined when x30x - 3 \geqslant 0 .
x3\therefore x \geqslant 3
Now, f(x)=7xPx3=(7x)!(7xx+3)!=(7x)!(102x)!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 102x010 - 2x \geqslant 0 .
2x10 x5  \therefore - 2x \geqslant - 10 \\\ \therefore x \leqslant 5 \\\
Thus, we get a range of variable x as 3x53 \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(3)=73P33=4P0=4!(40)!=4!4!=1f\left( 3 \right) = {}^{7 - 3}{P_{3 - 3}} = {}^4{P_0} = \dfrac{{4!}}{{\left( {4 - 0} \right)!}} = \dfrac{{4!}}{{4!}} = 1
f(4)=74P43=3P1=3!(31)!=3!2!=3f\left( 4 \right) = {}^{7 - 4}{P_{4 - 3}} = {}^3{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = 3
f(5)=75P53=2P2=2!(22)!=2!0!=2f\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(x)=7xPx3f\left( x \right) = {}^{7 - x}{P_{x - 3}} .
So, x3x-3 must be greater than or equal to 0.
x30x - 3 \geqslant 0
x3\therefore x \geqslant 3
Also, any permutation nPr{}^n{P_r} is defined when nrn \geqslant r .
So, in this question 7xx37 - x \geqslant x - 3
7+32x 2x10 x5  \therefore 7 + 3 \geqslant 2x \\\ \therefore 2x \leqslant 10 \\\ \therefore x \leqslant 5 \\\
Thus, we get the range of variable x as 3x53 \leqslant x \leqslant 5 .