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Q.
Let [x] denote the greatest integer less than or equal to x. If $x =(\sqrt{3}+1)^5$ then [x] is equal to
Binomial Theorem
Solution:
$x = \left(\sqrt{3}+1\right)^{5} = I + f$
where $0 \le f < 1$ and $I \in Z$
$f = \left(\sqrt{3}+1\right)^{5} ; 0 \le f < 1$
$I + f-f = 2\left[5\left(\sqrt{3}\right)^{4}+10\left(\sqrt{3}\right)^{2}+1\right]$
which is an integer.
$\therefore I = 152$