Let (7 + 4sqrt{3})^n = p + beta,where n and p | vedclass

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(D) Let $x = (7 + 4\sqrt{3})^n = p + \beta$. Since $7 + 4\sqrt{3} > 1$,$p + \beta$ is a large number. Consider $f = (7 - 4\sqrt{3})^n$. Since $0 < 7 -

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none of these

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(D) Let $x = (7 + 4\sqrt{3})^n = p + \beta$. Since $7 + 4\sqrt{3} > 1$,$p + \beta$ is a large number. Consider $f = (7 - 4\sqrt{3})^n$. Since $0 Expanding $(7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n$ using the Binomial Theorem,all irrational terms cancel out,resulting in an even integer,say $2k$. Thus,$(p + \beta) + f = 2k$. Since $p$ is an integer and $0 Therefore,$(1 - \beta)(p + \beta) = f(p + \beta) = (7 - 4\sqrt{3})^n (7 + 4\sqrt{3})^n = (49 - 48)^n = 1^n = 1$. Since $1$ is neither prime nor composite,the correct option is $D$.

Let $(7 + 4\sqrt{3})^n = p + \beta$,where $n$ and $p$ are positive integers and $\beta \in (0, 1)$. Then $(1 - \beta)(p + \beta)$ is

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