Question

Let $f\left(x\right)$ be a real-valued function such that $\left|f\right(x)+x^2+1|\ge |f\left(x\right)|+|x^2+1|$ and $f\left(x\right)\le 0$ for all real values of $x$. Then the absolute value of $\sum _{r=1}^5(1+f\left(r\right))$ is

Answer 1

We know that $|x+y|\le |x|+|y|,\forall x,y\in R$
But given that $|f\left(x\right)+(x^2+1)|\ge |f\left(x\right)|+|x^2+1|$
$\therefore |f\left(x\right)+(x^2+1)|\ge |f\left(x\right)|+|x^2+1|$ will be true iff $f\left(x\right)\cdot (x^2+1)\ge 0$
$\Rightarrow f\left(x\right)\ge 0$
But $f\left(x\right)\le 0,\therefore f\left(x\right)=0$
$\sum _{r=1}^5(1+f\left(r\right))=\sum _{r=1}^{5}1=5$