Question

If the third term in the binomial expansion of ${{\left( 1+{{x}^{{{\log }_{2}}x}} \right)}^{5}}$equals 2560, then the possible value of x is?
(a) $2\sqrt{2}$
(b) $\dfrac{1}{8}$
(c) $4\sqrt{2}$
(d) $\dfrac{1}{4}$

Answer 1

First, before proceeding for this, we must know the formula for the expansion of the binomial terms to get the third term of the expansion. Then, in the question it is given with its value as 256 to get the value of x. Then, by taking log with base 2 on both sides, we get the final result.

Complete step-by-step answer:
In this question, we are supposed to find the third term in the binomial expansion of ${{\left( 1+{{x}^{{{\log }_{2}}x}} \right)}^{5}}$whose value is given as 2560 to get the value of x.
So, before proceeding for this, we must know the formula for the expansion of the binomial terms to get the third term of the expansion as:
${{T}_{3}}={{T}_{2+1}}={}^{5}{{C}_{2}}{{\left( {{x}^{{{\log }_{2}}x}} \right)}^{2}}$
Then, in the question we are given with its value as 2560, so we get:
${}^{5}{{C}_{2}}{{\left( {{x}^{{{\log }_{2}}x}} \right)}^{2}}=2560$
Then, by solving the above expression to get the value of x as:

& \dfrac{5!}{\left( 5-2 \right)!2!}\times {{x}^{2{{\log }_{2}}x}}=2560 \\\ & \Rightarrow \dfrac{5\times 4\times 3!}{3!2!}\times {{x}^{2{{\log }_{2}}x}}=2560 \\\ & \\\ & \Rightarrow \dfrac{5\times 4}{2}\times {{x}^{2{{\log }_{2}}x}}=2560 \\\ & \Rightarrow 10\times {{x}^{2{{\log }_{2}}x}}=2560 \\\ & \Rightarrow {{x}^{2{{\log }_{2}}x}}=256 \\\ \end{aligned}$$ Then, by taking log with 2 on both sides, we get: $\begin{aligned} & 2{{\left( {{\log }_{2}}x \right)}^{2}}={{\log }_{2}}256 \\\ & \Rightarrow 2{{\left( {{\log }_{2}}x \right)}^{2}}=8 \\\ & \Rightarrow {{\left( {{\log }_{2}}x \right)}^{2}}=4 \\\ & \Rightarrow \left( {{\log }_{2}}x \right)=\pm 2 \\\ & \Rightarrow x={{2}^{\pm 2}} \\\ \end{aligned}$ Then, we get two values of x as: $\begin{aligned} & x={{2}^{2}} \\\ & \Rightarrow x=4 \\\ \end{aligned}$or $\begin{aligned} & x={{2}^{-2}} \\\ & \Rightarrow x=\dfrac{1}{4} \\\ \end{aligned}$ So, we get the value of x as 4 or $\dfrac{1}{4}$ **So, the correct answer is “Option d”.** **Note:** Now, to solve these types of the questions we need to know some of the basic formulas of factorial and combination. Now, to get the value of ${}^{n}{{C}_{r}}$ is given by the formula of the combination: $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4. $\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$