Question

$K\left(x\right)$ is a function such that $K\left(f\right(x\left)\right)=a+b+c+d$,
Where,
$a=\begin{cases}
0 & \text{ if f(x) is even} \\
-1 & \text{ if f(x) is odd} \\
2 & \text{ if f(x) is neither even nor odd}
\end{cases}$
$b=\begin{cases}
3 & \text{ if f(x) is periodic} \\
4 & \text{ if f(x) is aperiodic}
\end{cases}$
$c=\begin{cases}
5 & \text{ if f(x) is one one} \\
6 & \text{ if f(x) is many one}
\end{cases}$
$d=\begin{cases}
7 & \text{ if f(x) is onto} \\
8 & \text{ if f(x) is into}
\end{cases}$
$h:R\to R,h\left(x\right)=\left(\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}\right)$
On the basis of above information, answer the following questions.$K\left(h\right(x\left)\right)=$

• A. $15$
• B. $16$
• C. $17$
• D. $18$

Answer 1

The correct option is D $18$

$h(-x)=\frac{e^{-2x}+e^{-x}+1}{e^{-2x}-e^{-x}+1}=\frac{e^{2x}(e^{-2x}+e^{-x}+1)}{e^{2x}(e^{-2x}-e^{-x}+1)}=h\left(x\right)$
Hence $h\left(x\right)$ is even, Since it is even , it is many-one in R
It is not onto because it cannot take all real values, for e.g. it cannot take 0 value
It is not aperiodic because no such t exists such that $h\left(x\right)=h(t+x)$
Therefor $a=0,b=4,c=6,d=8$
$k\left(h\right(x\left)\right)=18$