- Home
- Standard 12
- Mathematics
माना $f(x)=2 x^2-x-1$ तथा $S=\{n \in Z :|f(n)| \leq 800\} \quad$ हैं। तब $\sum \limits_{n \in S} f(n)$ का मान है $............$
$10620$
$10630$
$10640$
$10650$
Solution
$f ( x )=2 x ^{2}- x -1$
$| f ( x )| \leq 800$
$2 n ^{2}- n -801 \leq 0$
$n^{2}-\frac{1}{2} n-\frac{801}{2} \leq 0$
$\left(n-\frac{1}{4}\right)^{2}-\frac{801}{2}-\frac{1}{16} \leq 0$
$\left(n-\frac{1}{4}\right)^{2}-\frac{6409}{16} \leq 0$
$\left(n-\frac{1}{4}-\frac{\sqrt{6409}}{4}\right)\left(n-\frac{1}{4}+\frac{\sqrt{6409}}{16}\right) \leq 0$
$\frac{1-\sqrt{6409}}{4} \leq n \leq \frac{1+\sqrt{6409}}{4}$
$n=\{-19,-18-17, \ldots \ldots 0,1,2, \ldots \ldots, 20\}$
$\sum_{n \in s} f(x)=\sum\left(2 x^{2}-x-1\right)$
$=2\left[19^{2}+18^{2}+\ldots . .+1^{2}+1^{2}+2^{2}+\ldots .+19^{2}+20^{2}\right]$
$=4\left[1^{2}+2^{2}+\ldots . .+19^{2}\right]+2\left[20^{2}\right]-20-40$
$=\frac{4 \times 19 \times 20 \times(2 \times 19+1)}{6}+2 \times 400-60$
$=\frac{4 \times 19 \times 20 \times 39}{6}+800-60-9880+800-60$
$=10620$
