
The range of function $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$ is:
A. $\left\{ {1,2,3} \right\}$
B. $\left\{ {1,2,3,4,5} \right\}$
C. $\left\{ {1,2,3,4} \right\}$
D. $\left\{ {1,2,3,4,5,6} \right\}$
Answer
575.4k+ views
Hint: In the given question first we will put both powers in the given expression greater than zero. We will make all three possible conditions for it. By doing this we will get the value of$x$. Thus we get three values of$x$. Then we will put these values of $x$ in the given function and find the range of the expression. Hence we will get the correct answer.
Complete step by step Answer:
In the Question, The function is given that:
$f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$
The given function $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$can be defined as three conditions:
Condition1:
$7 - x > 0$
It means, we can write that
$x < 7$
Condition 2:
$x - 3 > 0$ In the given question
It means, we can write that
$x < 3$
Condition 3:
$\left( {x - 3} \right) < \left( {x - 7} \right)$
It means, we can write that
Assume that:
$
\left( {x - 3} \right) + \left( {x - 7} \right) \\
\Rightarrow x - 3 + x - 7 \\
\Rightarrow 2x - 10 \\
\Rightarrow 2x = 10 \\
$
Or we can write as
$
\Rightarrow 2x < 10 \\
\Rightarrow x < 5 \\
$
It means, we can write that, the value of $x$ will be
$x = 3,4,5$
Because $x < 5$
Hence the range of $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$can be written as:
Put the value of $x$ in the given expression.
$
\left\{ {^{7 - 3}{P_{3 - 3}}{,^{7 - 4}}{P_{4 - 3}}{,^{7 - 5}}{P_{5 - 3}}} \right\} \\
\Rightarrow \left\{ {^4{P_0}{,^3}{P_1}{,^2}{P_2}} \right\} \\
$
Hence the range of the given expression $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$ is $\left\{ {1,2,3} \right\}$.
And this is our answer.
Hence the correct answer is option A.
Note: In the given question remember that we have to put both powers in the given expression greater than zero. Make all the required conditions for the expression. Thus we get the value of $x$ after that we have to put the value of $x$in the given expression, thus we get the range of the given function.
Complete step by step Answer:
In the Question, The function is given that:
$f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$
The given function $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$can be defined as three conditions:
Condition1:
$7 - x > 0$
It means, we can write that
$x < 7$
Condition 2:
$x - 3 > 0$ In the given question
It means, we can write that
$x < 3$
Condition 3:
$\left( {x - 3} \right) < \left( {x - 7} \right)$
It means, we can write that
Assume that:
$
\left( {x - 3} \right) + \left( {x - 7} \right) \\
\Rightarrow x - 3 + x - 7 \\
\Rightarrow 2x - 10 \\
\Rightarrow 2x = 10 \\
$
Or we can write as
$
\Rightarrow 2x < 10 \\
\Rightarrow x < 5 \\
$
It means, we can write that, the value of $x$ will be
$x = 3,4,5$
Because $x < 5$
Hence the range of $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$can be written as:
Put the value of $x$ in the given expression.
$
\left\{ {^{7 - 3}{P_{3 - 3}}{,^{7 - 4}}{P_{4 - 3}}{,^{7 - 5}}{P_{5 - 3}}} \right\} \\
\Rightarrow \left\{ {^4{P_0}{,^3}{P_1}{,^2}{P_2}} \right\} \\
$
Hence the range of the given expression $f\left( x \right){ = ^{7 - x}}{P_{x - 3}}$ is $\left\{ {1,2,3} \right\}$.
And this is our answer.
Hence the correct answer is option A.
Note: In the given question remember that we have to put both powers in the given expression greater than zero. Make all the required conditions for the expression. Thus we get the value of $x$ after that we have to put the value of $x$in the given expression, thus we get the range of the given function.
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