Question

If $a , b \in R$ and satisfy $a = b +\frac{1}{ a +\frac{1}{ b +\frac{1}{a+\ldots \ldots \infty}}}, b = a -\frac{1}{ b +\frac{1}{ a -\frac{1}{ b +\ldots \ldots \infty}}}$ 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
$a+\frac{1}{a}=b+\frac{1}{b} \Rightarrow(a-b)+\left(\frac{1}{a}-\frac{1}{b}\right)=0 \Rightarrow(a-b)\left(1-\frac{1}{a b}\right)=0$
But reject $a = b$ (Think?).
$\therefore a b=1 . $ So, from $(1)$, we get $(a-b)(a+b)=1 \Rightarrow a^2-b^2=1$