Question
The domain of the function $$f\left( x \right) = {}^{24 - x}{C_{3x - 1}} + {}^{40 - 6x}{C_{8x - 10}}$$ is :
A.
$$\left\{ {2,\,3} \right\}$$
B.
$$\left\{ {1,\,2,\,3} \right\}$$
C.
$$\left\{ {1,\,2,\,3,\,4} \right\}$$
D.
None of these
Answer :
$$\left\{ {2,\,3} \right\}$$
Solution :
$$\eqalign{
& {}^{24 - x}{C_{3x - 1}}{\text{ is defined if,}} \cr
& 24 - x > 0,\,3x - 1 \geqslant 0{\text{ and }}24 - x \geqslant 3x - 1 \cr
& \Rightarrow x < 24,\,x \geqslant \frac{1}{3}{\text{ and }}x \leqslant \frac{{25}}{4} \cr
& \Rightarrow \frac{1}{3} \leqslant x \leqslant \frac{{25}}{4} \cr
& {}^{40 - 6x}{C_{8x - 10}}{\text{ is defined if,}} \cr
& 40 - 6x > 0,\,8x - 10 \geqslant 0{\text{ and }}40 - 6x \geqslant 8x - 10 \cr
& \Rightarrow x < \frac{{20}}{3},\,x \geqslant \frac{5}{4}{\text{ and }}x \leqslant \frac{{25}}{7} \cr
& \Rightarrow \frac{5}{4} \leqslant x \leqslant \frac{{25}}{7} \cr
& {\text{From (1) and (2), we get }}\frac{5}{4} \leqslant x \leqslant \frac{{25}}{7} \cr
& {\text{But }}24 - x\, \in \,N, \cr
& \therefore \,\,x{\text{ must be an integer, }}x = 2,\,3 \cr
& {\text{Hence domain }}\left( f \right) = \left\{ {2,\,3} \right\} \cr} $$