Question : If $x_{1}x_{2}x_{3}=4(4+x_{1}+x_{2}+x_{3})$, then what is the value of $\left [ \frac{1}{(2+x_{1})} \right ]+\left [ \frac{1}{(2+x_{2})} \right ]+\left [ \frac{1}{(2+x_{3})} \right ]?$
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $2$
Option 4: $\frac{1}{3}$
Correct Answer: $\frac{1}{2}$
Solution :
Given: $x_{1}x_{2}x_{3}=4(4+x_{1}+x_{2}+x_{3})$
Put $x_{2} = 1$ and $x_{3} = 1$,
⇒ $x_{1}= 4(4 + x_{1} + 1 + 1)$
⇒ $x_{1}= 16 + 4x_{1} + 8$
⇒ $x_{1}=-8$
Putting all these values;
$\left [ \frac{1}{(2+x_{1})} \right ]+\left [ \frac{1}{(2+x_{2})} \right ]+\left [\frac{1}{(2+x_{3})} \right ]$
= $-\frac{1}{6} + \frac{1}{3}+\frac{1}{3}$
= $\frac{1}{2}$
Hence, the correct answer is $\frac{1}{2}$.
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