Question

If $a,b\in R$ and satisfy $a=b+\frac{1}{a+\frac{1}{b+\frac{1}{a+\dots \dots \infty }}},b=a-\frac{1}{b+\frac{1}{a-\frac{1}{b+\dots \dots \infty }}}$ then $a^2-b^2$ is equal to

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (B)

We have $a=b+\frac{1}{a+\frac{1}{a}}\Rightarrow (a-b)=\frac{1}{a+\frac{1}{a}}$ .....(1)
Also, $b=a-\frac{1}{b+\frac{1}{b}}\Rightarrow (a-b)=\frac{1}{b+\frac{1}{b}}$ ....(2)
$\therefore$ From (1) and (2), we get
$a+\frac{1}{a}=b+\frac{1}{b}\Rightarrow (a-b)+\left(\frac{1}{a}-\frac{1}{b}\right)=0\Rightarrow (a-b)\left(1-\frac{1}{ab}\right)=0$
But reject $a=b$ (Think?).
$\therefore ab=1.$ So, from $(1)$, we get $(a-b)(a+b)=1\Rightarrow a^2-b^2=1$