Question

The solution set of the equation $\raisebox{1ex}{$\left[x-a\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-b\right]$}\right.+\raisebox{1ex}{$\left[x-b\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-a\right]$}\right.=\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\right)+\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right),\left(a\ne b\right)$ is

• A. $\left\{a-b,0\right\}$
• B. $\left\{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.,0\right\}$
• C. $\left\{a+b,0\right\}$
• D. $\left\{ab,0\right\}$

Answer 1

The correct option is C

$\left\{a+b,0\right\}$

Explanation for the correct option:

Given, $\raisebox{1ex}{$\left[x-a\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-b\right]$}\right.+\raisebox{1ex}{$\left[x-b\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-a\right]$}\right.=\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\right)+\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right),\left(a\ne b\right)$

Let, $\raisebox{1ex}{$\left[x-a\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-b\right]$}\right.=m$

$$\begin{gathered} \Rightarrow m+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\right)=\raisebox{1ex}{$\left[a^2+b^2\right]$}\!\left/ \!\raisebox{-1ex}{$ab$}\right. \\ \Rightarrow ab{m}^2-\left(a^2+b^2\right)m+ab=0 \\ \Rightarrow (bm-a)(am-b)=0 \\ \Rightarrow m=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.,m=\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right. \end{gathered}$$

If $m=\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\right),\raisebox{1ex}{$\left[x-a\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-b\right]$}\right.=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.$

$$\begin{gathered} \Rightarrow bx-ab=ax-ab \\ \Rightarrow ax-bx=0 \\ \Rightarrow x(a-b)=0 \\ \Rightarrow x=0 \end{gathered}$$

If $m=\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right),\raisebox{1ex}{$\left[x-a\right]$}\!\left/ \!\raisebox{-1ex}{$\left[x-b\right]$}\right.=\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.$

$$\begin{gathered} \Rightarrow ax-a^2=bx-b^2 \\ \Rightarrow ax-bx=a^2-b^2 \\ \Rightarrow x(a-b)=(a-b)(a+b) \\ \Rightarrow x=a+b \end{gathered}$$

Hence the correct answer is option (C), $\left\{a+b,0\right\}$