ધારોકે f(x)=2 x^{2}-x-1 અને S ={n in Z :|f(n) | vedclass

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Question

$f ( x )=2 x ^{2}- x -1$

$| f ( x )| \leq 800$

$2 n ^{2}- n -801 \leq 0$

$n^{2}-\frac{1}{2} n-\frac{801}{2} \leq 0$

$\left(n-\fr

Vedclass · Accepted Answer

$10620$

Vedclass · Answer

$f ( x )=2 x ^{2}- x -1$

$| f ( x )| \leq 800$

$2 n ^{2}- n -801 \leq 0$

$n^{2}-\frac{1}{2} n-\frac{801}{2} \leq 0$

$\left(n-\frac{1}{4}ight)^{2}-\frac{801}{2}-\frac{1}{16} \leq 0$

$\left(n-\frac{1}{4}ight)^{2}-\frac{6409}{16} \leq 0$

$\left(n-\frac{1}{4}-\frac{\sqrt{6409}}{4}ight)\left(n-\frac{1}{4}+\frac{\sqrt{6409}}{16}ight) \leq 0$

$\frac{1-\sqrt{6409}}{4} \leq n \leq \frac{1+\sqrt{6409}}{4}$

$n=\{-19,-18-17, \ldots \ldots 0,1,2, \ldots \ldots, 20\}$

$\sum_{n \in s} f(x)=\sum\left(2 x^{2}-x-1ight)$

$=2\left[19^{2}+18^{2}+\ldots . .+1^{2}+1^{2}+2^{2}+\ldots .+19^{2}+20^{2}ight]$

$=4\left[1^{2}+2^{2}+\ldots . .+19^{2}ight]+2\left[20^{2}ight]-20-40$

$=\frac{4 	imes 19 	imes 20 	imes(2 	imes 19+1)}{6}+2 	imes 400-60$

$=\frac{4 	imes 19 	imes 20 	imes 39}{6}+800-60-9880+800-60$

$=10620$

ધારોકે $f(x)=2 x^{2}-x-1$ અને $S =\{n \in Z :|f(n)| \leq 800\}$ છે, તો $\sum_{n \in S} f(n)$ નું મૂલ્ય ............ છે.

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