Question

The range of $\Psi \left(x\right)$ is given by

• A. $[e^{-1/2},e^{1/2}]$
• B. $[2^{-1/2},2^{1/2}]$
• C. $[-1,1]$
• D. $[-2,2]$

Answer 1

The correct option is B $[2^{-1/2},2^{1/2}]$

Given $\Psi \left(x\right)=f\left(g\left(x\right)\right)$ ...(1)

$f(mx+ny)={\left(f\left(x\right)\right)}^m{\left(f\left(x\right)\right)}^n\forall x,y,m,n\in R$ ...(2)

$f^{\prime }\left(0\right)=-\ln 2f\left(0\right)$ ...(3)

$g\left(x\right)$ is periodic function given by $g\left(x\right)=|x-1|-\frac{1}{2};0\le x\le 2$ ...(4)

$h\left(x\right)=g\left(x\right)+\sin \pi x$ is periodic with period $2$

$\Rightarrow h(x+2)=h\left(x\right)$$\Rightarrow g(x+2)+\sin \pi (x+2)=g\left(x\right)+\sin \pi x$

$\Rightarrow g(x+2)+\sin \pi x=g\left(x\right)+\sin \pi x$$\Rightarrow g(x+2)=g\left(x\right)$

$\Rightarrow g\left(x\right)$ must be a periodic function with period $2$

$\therefore g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}|x-1|-\frac{1}{2};& 0\le x\le 2\end{array}\\ \begin{array}{cc}|x-3|-\frac{1}{2};& 2\le x\le 4\end{array}\\ \begin{array}{cc}|x-5|-\frac{1}{2};& 4\le x\le 6\end{array}\\ andsoon...\end{array}$

Now, $f(mx+hy)={\left(f\left(x\right)\right)}^m{\left(f\left(y\right)\right)}^n$ $\forall x,y,m,n\in R$

for $m=n=0$

$f\left(0\right)=1;$ for $m=n=1;f(x+y)=f\left(x\right).f\left(y\right)$

$\therefore f^{\prime }\left(a\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}f\left(x\right)\left[\frac{f\left(h\right)-1}{h}\right]=\underset{h\to 0}{lim}f\left(x\right)\left[\frac{f\left(h\right)-f\left(0\right)}{h}\right]$

$(\because f\left(0\right)=1)$

$=f\left(x\right)f^{\prime }\left(0\right)=f\left(x\right)(\ln 2.f\left(0\right))=(-\ln 2)\left(f\left(x\right)\right)$

$\therefore \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=-\ln 2$$\Rightarrow \ln f\left(x\right)+C=-\left(\ln 2\right)x$$\Rightarrow \ln f\left(x\right)+x\ln 2=-C$$\Rightarrow \ln 2^x.f\left(x\right)=-C$

for $x=0,\ln 1=-C\Rightarrow C=0$

$\therefore \ln 2^{x}f\left(x\right)=0$$\Rightarrow 2^{x}f\left(x\right)=1\Rightarrow f\left(x\right)=2^{-x}$

$\Rightarrow \Psi \left(x\right)=\left(f\left(f\left(x\right)\right)\right)=2^{-g\left(x\right)}=2^{\frac{1}{2}-|x-1|}=\sqrt{2.}{2}^{-|x-1|}$

$\Psi \left(x\right)=2^{-g\left(x\right)}={\left(\frac{1}{2}\right)}^{g\left(x\right)}$

and $g\left(x\right)=|x-1|-\frac{1}{2};0\le x\le 2$ and $g\left(x\right)$ is periodic with period $2.$

Clearly $g\left(x\right)\in [-\frac{1}{2},\frac{1}{2}]$

Here $\Psi \left(x\right)$ is an exponential function with bass less than $1$

and hence is a decreasing function for

$g\left(x\right)\in [-\frac{1}{2},\frac{1}{2}]$ and also $\Psi \left(x\right)$ is periodic.

$\therefore$ Range of $\Psi \left(x\right)=[{\left(\frac{1}{2}\right)}^{\frac{1}{2}},{\left(\frac{1}{2}\right)}^{-\frac{1}{2}}]=\left[\frac{1}{\sqrt{2}}\right]=[2^{-\frac{1}{2}},2^{\frac{1}{2}}]$