JEE Main 2022 — Mathematics Algebra

Given,

$f(x)=2{x}^{2}-x-1$f(x)=2x2−x−1 and $|f(n)|\leq 800$∣f(n)∣≤800

So, $2{n}^{2}-n-801\leq 0$2n2−n−801≤0

$$\Rightarrow {n}^{2}-\frac{1}{2}n-\frac{801}{2}\leq 0$$⇒n2−21​n−2801​≤0

$$\Rightarrow {(n-\frac{1}{4})}^{2}-\frac{801}{2}-\frac{1}{16}\leq 0$$⇒(n−41​)2−2801​−161​≤0

$$\Rightarrow {(n-\frac{1}{4})}^{2}-\frac{6409}{16}\leq 0$$⇒(n−41​)2−166409​≤0

$$\Rightarrow (n-\frac{1}{4}-\frac{\sqrt{6409}}{4})(n-\frac{1}{4}+\frac{\sqrt{6409}}{16})\leq 0$$⇒(n−41​−46409​​)(n−41​+166409​​)≤0

$$\Rightarrow \frac{1-\sqrt{6409}}{4}\leq n\leq \frac{1+\sqrt{6409}}{4}$$⇒41−6409​​≤n≤41+6409​​

Since given $n$n is integer,

So, $n={-19,-18-17,\ldots \ldots ..0,1,2,\ldots \ldots ,20}$n=−19,−18−17,……..0,1,2,……,20

Now $\displaystyle \underset{n\in S}{\sum }f(x)=\sum (2{x}^{2}-x-1)$n∈S∑​f(x)=∑(2x2−x−1)

$$=(2\sum {x}^{2}-\sum x-\sum 1)$$=(2∑x2−∑x−∑1)

$=2[{19}^{2}+{18}^{2}+\ldots .+{1}^{2}+{1}^{2}+{2}^{2}+\ldots .+{19}^{2}+{20}^{2}]-20-40$=2[192+182+….+12+12+22+….+192+202]−20−40 \displaystyle (\text{as }\sum x=20&\sum 1=40)

$$=4[{1}^{2}+{2}^{2}+\ldots ..+{19}^{2}]+2[{20}^{2}]-20-40$$=4[12+22+…..+192]+2[202]−20−40

$$=\frac{4\times 19\times 20\times (2\times 19+1)}{6}+2\times 400-60$$=64×19×20×(2×19+1)​+2×400−60

$$=\frac{4\times 19\times 20\times 39}{6}+800-60-9880+800-60$$=64×19×20×39​+800−60−9880+800−60

$$=10620$$=10620