Let $a_{i}, i = 1, 2, 3, ⟶ n$ denote integers in the domain of the function $f (x) = \sqrt{\log_{1 / 2} (\frac{4 x - 25}{x - 21})}$ where $a_{i} < a_{i + 1} \forall i \in N$ If the line $L : \frac{2 x - a_{1}}{4} = \frac{y + a_{1}}{a_{2}} = \frac{z - a_{3}}{a_{5}}$ meets the xy, yz and zx planes at A, B and Cn respectively. If volume of the tetrahedron $O A B C^{1}$ is V, where O is the origin and $C^{1}$ is image of C in the X-axis then the value of [V] , (when [.] is greatest integer function) is_____

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answer is 3.

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Detailed Solution

For domain of f(x)
we must have $\log_{\frac{1}{2}} (\frac{4 x - 25}{x - 21}) \geq 0$

$\Rightarrow 0 < \frac{4 x - 25}{x - 21} \leq 1 \Rightarrow x \in [\frac{4}{3}, \frac{25}{4})$

$∴$ Integers in the domain are 2,3,…….,6.

$\Rightarrow a_{1} = 2, a_{2} = 3, \dots \dots, a_{5} = 6$

$∴ The line l : \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 4}{6} = r$

At xy plane, put z = 0

$\Rightarrow 6 r + 4 = 0 \Rightarrow r = \frac{- 2}{3}$

$∴ A (2 r + 1, 3 r, 0) = (- \frac{1}{3} - 4, 0)$

At yz plane, put x = 0

$\Rightarrow 2 r + 1 = 0 \Rightarrow r = \frac{- 1}{2}$

$∴ B (0, 3 r - 2, 6 r + 4) = (0, \frac{- 7}{2}, 1)$

At zx plane, put y = 0

$\Rightarrow 3 r - 2 = 0 \Rightarrow r = \frac{2}{3}$

$∴ C (\frac{7}{3}, 0, 8)$

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