A) \[\{2,3\}\]
B) \[\{1,2,3\}\]
C) \[\{1,2,3,4\}\]
D) None of these
Correct Answer: A
Solution :
| [a] \[^{24-x}{{C}_{3x-1}}\] is defined if, |
| \[24-x>0,3x-1\ge 0\] And \[24-x\ge 3x-1\] |
| \[\Rightarrow x<24,x\ge \frac{1}{3}\] And \[x\le \frac{25}{4}\Rightarrow \frac{1}{3}\le x\le \frac{25}{4}\] |
| \[^{40-6x}{{C}_{8x-10}}\] is defined if |
| \[40-6x>0,8x-10\ge 0\] And \[40-6x\ge 8x-10\] |
| \[\Rightarrow x<\frac{20}{3},x\ge \frac{5}{4}andx\le \frac{25}{7}\] |
| \[\Rightarrow \frac{5}{4}\le x\le \frac{25}{7}\] |
| From (1) and (2), we get \[\frac{5}{4}\le x\le \frac{25}{7}\] |
| But \[24-x\in N,\therefore x\]must be an integer, \[x=2,3.\] |
| Hence domain \[(f)=\{2,\,\,3\}.\] |
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