Question

If $f$ is a positive real valued differentiable function satisfying the functional rule $f^2(x+y)=f^2(x)+f^2(y)-2\sqrt{x}-2\sqrt{y}+2\sqrt{x+y}-1$ for all $x,y\ge 0$ such that $\underset{h\to 1}{\mathrm{Lim}}\frac{f(h)-2}{h-1}=\frac{1}{2}$, then find the number of solution(s) of the equation $\frac{f(|x|)}{\lambda |x|+1}=1(\lambda >0)$.

Answer 1

Answer: 3

$f^2(x+y)=f^2(x)+f^2(y)-2\sqrt{x}-2\sqrt{y}+2\sqrt{x+y}-1$
Partially differentiating w.r.t. $x$

$2f(x+y)f^{\prime }(x+y)=2f(x)f^{\prime }(x)-\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+y}}$
Putting $x=1$
$2f(1+y)f^{\prime }(1+y)=2f(1)f^{\prime }(1)-1+\frac{1}{\sqrt{1+y}}$
$2f(1+y)f^{\prime }(1+y)=2\cdot 2\cdot \frac{1}{2}-1+\frac{1}{\sqrt{1+y}}$
$2f(1+y)f^{\prime }(1+y)=1+\frac{1}{\sqrt{1+y}}$
$2f(x)f^{\prime }(x)=1+\frac{1}{\sqrt{x}}\Rightarrow (f(x){)}^2=x+2\sqrt{x}+C$
$\text{ Putting }x=1,C=1$
$\Rightarrow f(x)=\sqrt{x}+1.$