Question

What is the domain of the function \[f\left( x \right) = {}^{24 - x}{C_{3x - 1}} + {}^{40 - 6x}{C_{8x - 10}}\] ?
A. \[\left\{ {2,3} \right\}\]
B. \[\left\{ {1,2,3} \right\}\]
C. \[\left\{ {1,2,3,4} \right\}\]
D. None of these

Answer 1

Hint: Here, the function of combination terms is given. First, find the term of the given function \[{}^{24 - x}{C_{3x - 1}}\] is defined in which interval by solving the conditions for \[24 - x\], and \[3x - 1\]. Similarly, find the term of the given function \[{}^{40 - 6x}{C_{8x - 10}}\] is defined in which interval by solving the conditions for \[40 - 6x\], and \[8x - 10\]. In the end, compare both intervals and calculate the domain of the given function to get the required answer.

Formula Used: In a combination term \[{}^n{C_r}\]we have, \[n > 0\] and \[r \ge 0\]


Complete step by step solution: The given function is \[f\left( x \right) = {}^{24 - x}{C_{3x - 1}} + {}^{40 - 6x}{C_{8x - 10}}\].

Let’s find out in which domain \[{}^{24 - x}{C_{3x - 1}}\] is defined.
We know that, in a combination term \[{}^n{C_r}\], \[n > 0\] and \[r \ge 0\]
So, we get
\[24 - x > 0\] and \[3x - 1 \ge 0\]
Solve the above inequalities.
\[24 > x\] and \[3x \ge 1\]
\[ \Rightarrow 24 > x\] and \[x \ge \dfrac{1}{3}\]

Also, from the above inequalities, we get
\[24 - x \ge 3x - 1\]
\[ \Rightarrow 25 - x \ge 3x\]
\[ \Rightarrow 25 \ge 4x\]
\[ \Rightarrow \dfrac{{25}}{4} \ge x\]
So, the term \[{}^{24 - x}{C_{3x - 1}}\] is defined in the interval \[\dfrac{1}{3} \le x \le \dfrac{{25}}{4}\].

Similarly, find out in which domain \[{}^{40 - 6x}{C_{8x - 10}}\] is defined.
We know that, in a combination term \[{}^n{C_r}\], \[n > 0\] and \[r \ge 0\]
So, we get
\[40 - 6x > 0\] and \[8x - 10 \ge 0\]
Solve the above inequalities.
\[40 > 6x\] and \[8x \ge 10\]
\[ \Rightarrow \dfrac{{20}}{3} > x\] and \[x \ge \dfrac{5}{4}\]

Also, from the above inequalities, we get
\[40 - 6x \ge 8x - 10\]
\[ \Rightarrow 50 - 6x \ge 8x\]
\[ \Rightarrow 50 \ge 14x\]
\[ \Rightarrow 25 \ge 7x\]
\[ \Rightarrow \dfrac{{25}}{7} \ge x\]
So, the term \[{}^{40 - 6x}{C_{8x - 10}}\] is defined in the interval \[\dfrac{5}{4} \le x \le \dfrac{{25}}{7}\].

From the intervals of both terms, we get
\[\dfrac{5}{4} \le x \le \dfrac{{25}}{7}\]

But we know that \[24 - x \in N\].
So, \[x\] is a natural number.
And the natural numbers present in the given interval \[\dfrac{5}{4} \le x \le \dfrac{{25}}{7}\] are 2 and 3.
Hence, the domain of the given function is \[\left\{ {2,3} \right\}\].

Option ‘A’ is correct

Note: Students often make mistakes while comparing or solving the inequality equation. Remember the following rules.
If we multiply or divide any inequality equation by a positive number, then the inequality sign does not change.
If we multiply or divide any inequality equation by a negative number, then the inequality sign changes its direction.