Question

$K\left(x\right)$ is a function such that $K\left(f\right(x\left)\right)=a+b+c+d$,
Where,

$a=\{\begin{array}{cc}0& \text{if f(x) is even}\\ -1& \text{if f(x) is odd}\\ 2& \text{if f(x) is neither even nor odd}\end{array}$

$b=\{\begin{array}{cc}3& \text{if f(x) is periodic}\\ 4& \text{if f(x) is aperiodic}\end{array}$

$c=\{\begin{array}{cc}5& \text{if f(x) is one one}\\ 6& \text{if f(x) is many one}\end{array}$

$d=\{\begin{array}{cc}7& \text{if f(x) is onto}\\ 8& \text{if f(x) is into}\end{array}$

$h:R\to R,h\left(x\right)=\left(\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}\right)$

On the basis of above information, answer the following questions.$g\left(x\right)=sin\left(x\right)$
$K\left(g\right(x\left)\right)=$

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

Answer 1

The correct option is B $16$

$g\left(x\right)=\sin \left(x\right)$
Hence $g(-x)=-g\left(x\right)$, therefore an odd function.
It is periodic, with a period of $\pi$
$sin\left(x\right)$ is an onto function only for a co-domain of $[\frac{-\pi }{2},\frac{\pi }{2}]$ since its range is $[-1,1]$
Thus for $R\to R$ or a co-domain, R, it is an into function.
Also $\sin \left(x\right)$ is a many on one function since
$\sin \left(x_1\right)=\sin \left(x_2\right)$
Implies
$x_1=(2n+1)\pi -x_2$ where $n\in W$.
Hence
$k\left(g\right(x\left)\right)=-1+3+6+8$
$=16$