Let c, k ∈ R . If f(x) = (c + 1) x ^ 2 + (1 - c ^ 2 )x + 2k and f(x...

Given $f\left(x\right) = \left(c + 1\right){x}^{2} + \left(1 - {c}^{2}\right)x + 2k ⋯ \left(i\right)$

& $f\left(x + y\right) = f\left(x\right) + f\left(y\right) - xy \forall xy \in R$

So ${f}^{'}\left(x\right) = 2\left(c + 1\right)x + 1 - {c}^{2}$

Now $\lim_{y \to 0}\dfrac{f\left(x + y\right) - f\left(x\right)}{y} = \lim_{y \to 0}\dfrac{f\left(y\right) - xy}{y} = \lim_{y \to 0}\dfrac{f\left(y\right)}{y} - x$

$⇒ {f}^{'}\left(x\right) = {f}^{'}\left(0\right) - x$

Integrating, we get

$f\left(x\right) = - \dfrac{{x}^{2}}{2} + {f}^{'}\left(0\right)x + \lambda$ but $f$\left(0\right) = 0 ⇒ \lambda = 0$</p><p>i.e.$f\left(x\right) = - \dfrac{{x}^{2}}{2} + \left(1 - {c}^{2}\right)x ⋯ \left(ii\right)$</p><p>$∴$as${f}^{'}\left(0\right) = 1 - {c}^{2}$</p><p>Now comparing equation \left(i\right)$ and \left(ii\right)$, we get</p><p>$c = - \dfrac{3}{2}$</p><p>$∴ f\left(x\right) = - \dfrac{1}{2}{x}^{2} - \dfrac{5}{4}x$</p><p>Now \left|2\mathop{∑}_{x = 1}^{20}f\left(x\right)\right| = \mathop{∑}_{x = 1}^{20}{x}^{2} + \dfrac{5}{2} \cdot \mathop{∑}_{x = 1}^{20}x = \dfrac{20\left(21\right)\left(41\right)}{6} + \dfrac{5}{2}\left(\dfrac{20\left(21\right)}{2}\right)$

$= 2870 + 525$

$= 3395$