Question

Let $f\left(x\right)=x^2-2x$ and $g\left(x\right)=f\left(f\right(x)-1)+f(5-f(x\left)\right),$ then

• A. $g\left(x\right)<0,\forall x\in R$
• B. $g\left(x\right)<0$ for some $x\in R$
• C. $g\left(x\right)\le 0$ for some $x\in R$
• D. $g\left(x\right)\ge 0,\forall x\in R$

Answer 1

The correct option is D $g\left(x\right)\ge 0,\forall x\in R$

Given, $f\left(x\right)=x^2-2x$

Now, $f\left(f\right(x)-1)=f(x^2-2x-1)=(x^2-2x-1{)}^2-2(x^2-2x-1)$

And, $f(5-f(x\left)\right)=f(5-x^2+2x)=(5-x^2+2x{)}^2-2(5-x^2+2x)$

Hence, $g\left(x\right)=f\left(f\right(x)-1)+f(5-f(x\left)\right)\therefore g\left(x\right)=(x^2-2x-1{)}^2-2(x^2-2x-1)+(5-x^2+2x{)}^2-2(5-x^2+2x)\Rightarrow g\left(x\right)=(x^2-2x-1{)}^2+(5-x^2+2x{)}^2-2(x^2-2x-1+5-x^2+2x)\therefore g\left(x\right)=(x^2+2x-1{)}^2+(5-2x^2+2x^2{)}^2-8$

Since the first two terms are in square,

$\therefore$ it can not be negative and if x = 9 then we also get positive value.

$\therefore g\left(x\right)\ge 0\forall x\in R$

$\therefore$ option D is correct