JEE MAIN - Mathematics (2004 - No. 2)
The range of the function f(x) = $${}^{7 - x}{P_{x - 3}}$$ is
{1, 2, 3, 4, 5}
{1, 2, 3, 4, 5, 6}
{1, 2, 3, 4}
{1, 2, 3}
Explanation
The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.
The domain of $f(x)$ is the set of all real numbers such that
(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)
(ii) 7 - x > 0 $$ \Rightarrow $$ x < 7
(iii) x - 3 $$ \le $$ 7 - x $$ \Rightarrow $$ 2x $$ \le $$ 10 $$ \Rightarrow $$ x $$ \le $$ 5.
To find the range of $f(x)$, we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.
For $x=3$, we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$.
For $x=4$, we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$.
For $x=5$, we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$.
Therefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$.
The domain of $f(x)$ is the set of all real numbers such that
(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)
(ii) 7 - x > 0 $$ \Rightarrow $$ x < 7
(iii) x - 3 $$ \le $$ 7 - x $$ \Rightarrow $$ 2x $$ \le $$ 10 $$ \Rightarrow $$ x $$ \le $$ 5.
To find the range of $f(x)$, we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.
For $x=3$, we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$.
For $x=4$, we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$.
For $x=5$, we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$.
Therefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$.
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