Question

Consider the function $f:R\to (0,\infty )$ defined by $f\left(x\right)=2^x+2^{|x|}$
f(x) is

• A. One–One Onto
• B. One–One Into
• C. Many–One Onto
• D. Many–One Into

Answer 1

The correct option is D Many–One Into

Consider the function $f:R\to (0,\infty )$ defined by $f\left(x\right)=2^x+2^{|x|}$
$f\left(x\right)=2^x+2^{|x|}$ For $x\ge 0,f\left(x\right)={2.2}^x$ And for $x<0;f\left(x\right)=2^x+2^{-x}$
and in this case $f^{\prime }\left(x\right)=2^{x}ln2-2^{-x}ln2=ln2\left(\frac{2^{2x}-1}{2^x}\right)=\frac{ln2}{2^x}.(2^x+1)(2^x-1)<0$
So; is decreasing in $x\in (-\infty ,0)$ and increasing in $x\in (0,\infty )$
So, f(x) is many one Also, Range of f(x) is $[2,\infty )$
f is many –one Into