Question

Let $f\left(x\right)$ be a continuous function which satisfies $f(x^2+1)=\frac{2}{f\left(2^x\right)-1}$ & $f\left(x\right)>0\forall x\epsilon R$ Then $\underset{x\to 1}{lim}f\left(x\right)$ is-

• A. 4
• B. 2
• C. 1
• D. does not exist

Answer 1

The correct option is B 2

$f(x^2+1)=\frac{2}{f\left(2^x\right)-1}$
$\Rightarrow \underset{x\to 0}{lim}f(x^2+1)=\underset{x\to 0}{lim}\frac{2}{f\left(2^x\right)-1}$
$\Rightarrow L=\frac{2}{L-1}\{whereL=\underset{x\to 1}{lim}f\left(x\right)\}$
$\Rightarrow L^2-L-2=0\Rightarrow (L-2)(L+1)=0$
$\therefore L=2$ $\{\because f\left(x\right)>0\}$