First slide

Functions (XII)

Question

The domain of the function $f (x) =^{24 - x} C_{3 x - 1} +^{40 - 6 x} C_{8 x - 10} is,$

Moderate

A

{2, 3}
B

{1, 2, 3}
C

{1, 2, 3, 4}
D

None of these

Solution

$^{24 - x} C_{3 x - 1}$ is defined if,
$24 - x > 0, 3 x - 1 \geq 0 and 24 - x \geq 3 x - 1 \Rightarrow x < 24, x \geq \frac{1}{3} and x \leq \frac{25}{4} \Rightarrow \frac{1}{3} \leq x \leq \frac{25}{4} . . . . . . . . . (1)$
$^{40 - 6 x} C_{8 x - 10}$ is defined if
$40 - 6 x > 0, 8 x - 10 \geq 0 and 40 - 6 x \geq 8 x - 10 \begin{array}{l} \Rightarrow x < \frac{20}{3}, x \geq \frac{5}{4} and x \leq \frac{25}{7} \\ \Rightarrow \frac{5}{4} \leq x \leq \frac{25}{7} . . . . . . . . . . . . . (2) \end{array}$
From Eq. (1) and (2), we get
$\frac{5}{4} \leq x \leq \frac{25}{7}$
$But, 24 - x \in N$
x must be an integer,
x = 2, 3.
Hence, domain (f) = {2, 3}.

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