Let f(x) = 2 x ^ 2 - x - 1 and S = \ n ∈ ℤ: |f(n)| ≤ 800\ . Then,...

Given,

$f\left(x\right) = 2{x}^{2} - x - 1$ and \left|f\left(n\right)\right| \le 800$</p><p>So,$2{n}^{2} - n - 801 \le 0$</p><p>$⇒ {n}^{2} - \dfrac{1}{2}n - \dfrac{801}{2} \le 0$</p><p>$⇒ {\left(n - \dfrac{1}{4}\right)}^{2} - \dfrac{801}{2} - \dfrac{1}{16} \le 0$</p><p>$⇒ {\left(n - \dfrac{1}{4}\right)}^{2} - \dfrac{6409}{16} \le 0$</p><p>$⇒ \left(n - \dfrac{1}{4} - \dfrac{\sqrt{6409}}{4}\right)\left(n - \dfrac{1}{4} + \dfrac{\sqrt{6409}}{16}\right) \le 0$</p><p>$⇒ \dfrac{1 - \sqrt{6409}}{4} \le n \le \dfrac{1 + \sqrt{6409}}{4}$</p><p>Since given$n$is integer,</p><p>So,$n = \left\{- 19, - 18 - 17, \ldots \ldots . . 0, 1, 2, \ldots \ldots , 20\right\}$</p><p>Now$\mathop{∑}_{n \in S}f\left(x\right) = ∑ \left(2{x}^{2} - x - 1\right)$</p><p>$= \left(2 ∑ {x}^{2} - ∑ x - ∑ 1\right)$</p><p>$= 2\left[{19}^{2} + {18}^{2} + \ldots . + {1}^{2} + {1}^{2} + {2}^{2} + \ldots . + {19}^{2} + {20}^{2}\right] - 20 - 40$\left(\text{as} ∑ x = 20 \text{ and } ∑ 1 = 40\right)$

$= 4\left[{1}^{2} + {2}^{2} + \ldots . . + {19}^{2}\right] + 2\left[{20}^{2}\right] - 20 - 40$

$= \dfrac{4 \times 19 \times 20 \times \left(2 \times 19 + 1\right)}{6} + 2 \times 400 - 60$

$= \dfrac{4 \times 19 \times 20 \times 39}{6} + 800 - 60 - 9880 + 800 - 60$

$= 10620$