JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 14)
The domain of the function $$f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$$ is : ( where $$[\mathrm{x}]$$ denotes the greatest integer less than or equal to $$x$$ )
$$(-\infty,-2) \cup[6, \infty)$$
$$(-\infty,-3] \cup[6, \infty)$$
$$(-\infty,-2) \cup(5, \infty)$$
$$(-\infty,-3] \cup(5, \infty)$$
Explanation
$$
f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}
$$
For Domain $[x]^2-3[x]-10>0$
$$ \begin{aligned} & \Rightarrow ([x]-5)([x]+2)>0 \\\\ & \Rightarrow [x] \in(-\infty,-2) \cup(5, \infty) \\\\ & \therefore x \in(-\infty,-2) \cup[6, \infty) \end{aligned} $$
For Domain $[x]^2-3[x]-10>0$
$$ \begin{aligned} & \Rightarrow ([x]-5)([x]+2)>0 \\\\ & \Rightarrow [x] \in(-\infty,-2) \cup(5, \infty) \\\\ & \therefore x \in(-\infty,-2) \cup[6, \infty) \end{aligned} $$
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