NIMCET 2007 — Mathematics PYQ

To find the range of the function, we must first determine its valid domain based on the mathematical properties of permutations.

Step 1: Understand the Constraints of ${}^n{P}_r$

For a permutation function ${}^n{P}_r$ to be defined, the following standard conditions must be met:

1. $n$ must be a positive integer: n > 0

2. $r$ must be a non-negative integer: $r\ge 0$

3. The upper index must be greater than or equal to the lower index: $n\ge r$

Step 2: Apply Constraints to Find the Domain

Let's apply these three conditions to our given function where $n=7-x$ and $r=x-3$:

• Condition 1: 7 - x > 0 \implies x < 7

• Condition 2: $x-3\ge 0⟹x\ge 3$

• Condition 3: $7-x\ge x-3$

  $$7+3\ge x+x$$

  $$10\ge 2x⟹x\le 5$$

Combining all three conditions for integer values of $x$, the valid values in the domain are:

$$3\le x\le 5$$

Thus, the domain of the function is the set of integers:

$$\text{Domain}=\{3,4,5\}$$

Step 3: Substitute Domain Values to Find the Range

Now, we calculate the functional value $f(x)$ for each element in the domain to discover the range set.

1. For $x=3$:

   $$f(3)={}^{7-3}{P}_{3-3}={}^4{P}_0$$

   Using the formula ${}^n{P}_r=\frac{n!}{(n-r)!}$:

   $${}^4{P}_0=\frac{4!}{(4-0)!}=\frac{4!}{4!}=1$$

2. For $x=4$:

   $$f(4)={}^{7-4}{P}_{4-3}={}^3{P}_1$$

   $${}^3{P}_1=\frac{3!}{(3-1)!}=\frac{3\times 2!}{2!}=3$$

3. For $x=5$:

   $$f(5)={}^{7-5}{P}_{5-3}={}^2{P}_2$$

   $${}^2{P}_2=\frac{2!}{(2-2)!}=\frac{2!}{0!}=\frac{2}{1}=2$$

Step 4: Form the Range Set

Collecting all the computed output values from our steps:

$$\text{Range}=\{1,2,3\}$$

Conclusion

The distinct elements that comprise the range of the given permutation function are $1,2,$ and $3$.

Correct Option: (c) $\{1,2,3\}$