Question

Match the two column.

Column 1	Column 2
If $x=2+\sqrt{3}$, then $x^2+\frac{1}{x^2}$	$4$
If $x=9+4\sqrt{5}$, then $\sqrt{x}-\frac{1}{\sqrt{x}}$	$1$
$\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}$	$2$
If ${\left(\frac{a}{b}\right)}^{x-1}={\left(\frac{b}{a}\right)}^{x-3}$, then $x=$	$14$

Answer 1

For (A)-(iv)

$x=2+\sqrt{3}$ then $\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-3}=2-\sqrt{3}$ [$\frac{1}{x}$is rationalized]

$$\begin{gathered} \left(x^2+\frac{1}{x^2}\right)={\left(2+\sqrt{3}\right)}^2+{\left(2-\sqrt{3}\right)}^2 \\ =2^2+4\sqrt{3}+{\left(\sqrt{3}\right)}^2+2^2-4\sqrt{3}+{\left(\sqrt{3}\right)}^2 \\ =4+4+3+3=14 \end{gathered}$$ [$$\begin{gathered} {\left(a+b\right)}^2=a^2+2ab+b^2 \\ {\left(a-b\right)}^2=a^2-2ab+b^2 \end{gathered}$$is used]

Hence, $14$ is the correct answer.

For (B)-(i)

$x=9+4\sqrt{5}$ then

$\frac{1}{x}=\frac{1}{9+4\sqrt{5}}=\frac{1}{9+4\sqrt{5}}\times \frac{9-4\sqrt{5}}{9-4\sqrt{5}}=\frac{9-4\sqrt{5}}{{\left(9\right)}^2-{\left(4\sqrt{5}\right)}^2}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}$ [$\frac{1}{x}$is rationalized]

We will find the value of ${\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)}^2$and find the square root at last.

$$\begin{gathered} {\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)}^2={\left(\sqrt{x}\right)}^2+{\left(\frac{1}{\sqrt{x}}\right)}^2-2\times \sqrt{x}\times \frac{1}{\sqrt{x}} \\ =x+\frac{1}{x}-2 \\ =9+4\sqrt{5}+9-4\sqrt{5}-2 \\ =16 \\ \Rightarrow \sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{16}=4 \end{gathered}$$ [${\left(a-b\right)}^2=a^2-2ab+b^2$is used]

Hence, $4$ is the correct answer.

For (C)-(ii)

Simplification

$\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}$

$\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}=\frac{1}{1+{\displaystyle \frac{x^b}{x^a}}+{\displaystyle \frac{x^c}{x^a}}}+\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}+{\displaystyle \frac{x^c}{x^b}}}+\frac{1}{1+{\displaystyle \frac{x^a}{x^c}}+{\displaystyle \frac{x^b}{x^c}}}$ $\left[a^{x-y}=\frac{a^x}{a^y}\right]$

$$\begin{gathered} =\frac{1}{{\displaystyle \frac{x^a+x^b+x^c}{x^a}}}+\frac{1}{{\displaystyle \frac{x^b+x^c+x^a}{x^b}}}+\frac{1}{{\displaystyle \frac{x^c+x^a+x^b}{x^c}}} \\ =\frac{x^a}{{\displaystyle x^a+x^b+x^c}}+\frac{x^b}{{\displaystyle x^a+x^b+x^c}}+\frac{x^c}{{\displaystyle x^a+x^b+x^c}} \end{gathered}$$

$=\frac{x^a+x^b+x^c}{{\displaystyle x^a+x^b+x^c}}=1$

Hence, $1$ is the correct answer.

For (D)-(iii)

Simplification

${\left(\frac{a}{b}\right)}^{x-1}={\left(\frac{b}{a}\right)}^{x-3}$

$$\begin{gathered} \Rightarrow {\left(\frac{a}{b}\right)}^{x-1}={\left(\frac{a}{b}\right)}^{-\left(x-3\right)} \\ \Rightarrow x-1=-x+3 \\ \Rightarrow x+x=3+1 \\ \Rightarrow 2x=4 \\ \Rightarrow x=2 \end{gathered}$$ $\left[\frac{x}{y}={\left(\frac{y}{x}\right)}^{-1}\right]$

Hence, $2$ is the correct answer.

Hence, the final answer is (A)-(iv),(B)-(i),(C)-(ii),(D)-(iii)