Let f(x) be a real-valued function such that ∣∣f(x)+x2+1∣∣≥| – InterviewSolution

1.	Let f(x) be a real-valued function such that ∣∣f(x)+x2+1∣∣≥\|f(x)\|+∣∣x2+1∣∣ and f(x)≤0 for all real values of x. Then the absolute value of 5∑r=1(1+f(r)) is
Answer» Let $f (x)$ be a real-valued function such that $∣ ∣ f (x) + x^{2} + 1 ∣ ∣ \geq \| f (x) \| + ∣ ∣ x^{2} + 1 ∣ ∣$ and $f (x) \leq 0$ for all real values of $x$ . Then the absolute value of $5 \sum r = 1 (1 + f (r))$ is

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