Question

$f\left(x\right)$ is a function defined on the set of Real Numbers $f\left(1\right)=1$.
Also, $f(x+5)\ge f\left(x\right)+5$ for all $xϵR$ and $f(x+1)\le f\left(x\right)+1$
If, $g\left(x\right)=\left(f\right(x){)}^2-f(x)$

• A. $g\left(2019\right)>(2019{)}^2$
• B. $g\left(2019\right)<(2019{)}^2$
• C. $g\left(2019\right)=(2019{)}^2$
• D. $g\left(2019\right)=2019\times 2018$

Answer 1

Answer: (B), (D)

The correct options are
B $g\left(2019\right)<(2019{)}^2$
D $g\left(2019\right)=2019\times 2018$

Let $f\left(x\right)+5\le f(x+5)$ be the first condition and $f(x+1)\le f\left(x\right)+1$ be the second.
$f\left(x\right)+5\le f(x+5)$ .... from $1^{st}$ condition
$f(x+5)=f(x+4+1)\le f(x+4)+1$ ..... from $2^{nd}$ condition
$f(x+4)=f(x+3+1)\le f(x+3)+1$ .... from $2^{nd}$ condition
$f(x+3)=f(x+2+1)\le f(x+2)+1$ .... from $2^{nd}$ condition
$f(x+2)=f(x+1+1)\le f(x+1)+1$ .... from $2^{nd}$ condition
$f(x+1)\le f\left(x\right)+1$ .... from $2^{nd}$ condition
$f\left(x\right)+5\le f(x+4)+1$
$f(x+4)+1\le f(x+3)+2$
$f(x+3)+2\le f(x+2)+3$
$f(x+2)+3\le f(x+1)+4$
$f(x+1)+4\le f\left(x\right)+5$
Hence, looking at all the above set of inequalities
we get that each of these expressions are infact equal.
So, $f(x+1)=f\left(x\right)+1$
$\Rightarrow f\left(2\right)=f\left(1\right)+1\Rightarrow f\left(2\right)=1+1=2$
$f\left(3\right)=f\left(2\right)+1\Rightarrow f\left(3\right)=2+1=3$
$f\left(4\right)=f\left(3\right)+1\Rightarrow f\left(4\right)=3+1=4$
$f\left(2019\right)=f\left(2018\right)+1\Rightarrow f\left(2019\right)=2018+1=2019$
$\therefore g\left(2019\right)=\left(f\right(2019){)}^2-f(2010)$
$g\left(2019\right)=(2019{)}^2-(2019)$
$g\left(2019\right)=2019\times 2018$