Question

If $f:R\to R$ and $g:R\to R$, and $f\left(x\right)+2g\left(x\right)\ge f^2\left(x\right)+g^2\left(X\right)+\frac{5}{4}$
Then

• A. $f\left(x\right)$ and $g\left(x\right)$ are many-to-one.
• B. $f\left(x\right)$ and $g\left(x\right)$ are both one-to-one.
• C. $f\left(x\right)$ and $g\left(x\right)$ are strictly decreasing functions
• D. $f\left(x\right)$ and $g\left(x\right)$ are constant functions

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)$ and $g\left(x\right)$ are many-to-one.
D $f\left(x\right)$ and $g\left(x\right)$ are constant functions

$f^2\left(x\right)-f\left(x\right)+g^2\left(x\right)-2g\left(x\right)+\frac{5}{4}\le 0$
$\Rightarrow {\left(f\right(x)-\frac{1}{2})}^2+{\left(g\right(x)-1)}^2\le 0$
$\Rightarrow f\left(x\right)=\frac{1}{2}$ and $g\left(x\right)=1$
Thus $f\left(x\right)$ and $g\left(x\right)$ are both constant functions
$\Rightarrow f\left(x\right)$ and $g\left(x\right)$ are both many-to-one functions.
(Since, for more than are values of $x$, the functions take the same values, ${}^{\prime }{\frac{1}{2}}^{\prime }$ and ${}^{\prime }{1}^{\prime }$ respectively)
So select Option (A) and (D).