Question

Let $\left[x\right]$ denotes the greatest integer less than or equal of $x$. If $x={(\sqrt{3}+1)}^5,$ then $\left[x\right]$ is equal to

• A. $75$
• B. $50$
• C. $76$
• D. $51$
• E. $152$

Answer 1

The correct option is E

$152$

Explanation for the correct option:

Finding the value of $x$ using binomial theorem:

Given $x={\left(\sqrt{3}+1\right)}^5$

We know that ${\left(a+b\right)}^n={}^{n}C_0{a}^n+{}^{n}C_1{a}^{n-1}b+{}^{n}C_2{a}^{n-2}{b}^2+\dots \dots \dots \dots +{}^{n}C_n{b}^n$

Applying binomial theorem in the given equation we get

$$\begin{gathered} x=5C_0{\left(\sqrt{3}\right)}^5+5C_1{\left(\sqrt{3}\right)}^4+5C_2{\left(\sqrt{3}\right)}^3+5C_3{\left(\sqrt{3}\right)}^2+5C_4{\left(\sqrt{3}\right)}^1+5C_5 \\ =9\sqrt{3}+5\times 9+10\times 3\sqrt{3}+10\times 3+5\times \sqrt{3}+1 \\ =9\sqrt{3}+45+30\sqrt{3}+30+5\sqrt{3}+1 \\ =76+44\sqrt{3} \end{gathered}$$

For the greatest integer

$\begin{array}{rcl}\left[x\right]& =& 76+44\sqrt{3}\\ & =& 76+44(1.732)\\ & =& 152\end{array}$

Hence, the correct answer is option (E).