Question

The domain of the real valued function $f\left(x\right)$ for which $4^{f\left(x\right)+4^{1-f\left(x\right)}}=4^x$ is

• A. $(-1,1)$
• B. $(-\infty ,1)$
• C. $[1,\infty )$
• D. R

Answer 1

The correct option is C $[1,\infty )$

Given,

$4^{f\left(x\right)}+4^{(1-f(x)}=4^x$

$\Rightarrow 4^{f\left(x\right)}+\frac{4}{4^{f\left(x\right)}}=4^x$

$\Rightarrow \frac{4^{f\left(x\right)}\cdot 4^{f\left(x\right)}+4}{4^{f\left(x\right)}}=4^x$

$\Rightarrow y^{2f\left(x\right)}+4=4^{x+f\left(x\right)}$

Dividing above equation by 4 on both sides

$\begin{array}{cc}& \frac{4^{2f\left(x\right)}}{4}+1=\frac{y^{x+f\left(x\right)}}{4}\\ \Rightarrow & 4^{2f\left(x\right)-1}+1=4^{x+f\left(x\right)-1}-\left(i\right)\\ & 1+4^{-(1-2f(x)}=4^{x-f\left(x\right)}-\left(ii\right)\end{array}$

$\text{both equation are same,}$

$\text{Then,}$

$4^{2f\left(x\right)-1}=4^{1-2f\left(x\right)}$

$2f\left(x\right)-1=1-2f\left(x\right)$

$f\left(x\right)+3f\left(x\right)=2$

$\Rightarrow 4f\left(x\right)=2$

$7f\left(x\right)=\frac{1}{2}$

$4^{x+f\left(x\right)-1}=4^{x-f\left(x\right)}$

$\Rightarrow x+f\left(x\right)-1=x-f\left(x\right)$

$\Rightarrow 2f\left(x\right)=1$

$\Rightarrow f\left(x\right)=\frac{1}{2}$

$Therefore=f\left(x\right)$ is a constant function.

but from above equation

$4^x=4^{f\left(x\right)}+4^{1-f\left(x\right)}$

$\text{= Putting the value of}f\left(x\right)=\frac{1}{2}$

$\text{in above equation.}$

$\Rightarrow 4^x=4^{\frac{1}{2}}+4^{1-\frac{1}{2}}$

$\Rightarrow 4^x=4$

$x=1$

$x$ has one value in i.e. $x=1$

So, $x\in [1,\infty )$

Option C