Mathematics - Functions Question with Solution | TestHub

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

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

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

Now $\underset{y\to 0}{\lim }\frac{f\left(x+y\right)-f\left(x\right)}{y}=\underset{y\to 0}{\lim }\frac{f\left(y\right)-xy}{y}=\underset{y\to 0}{\lim }\frac{f\left(y\right)}{y}-x$

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

Integrating, we get

$f\left(x\right)=-\frac{x^2}{2}+f^{\text{'}}\left(0\right)x+\lambda$ but $f\left(0\right)=0\Rightarrow \lambda =0$

i.e. $f\left(x\right)=-\frac{x^2}{2}+\left(1-c^2\right)x \cdots \left(ii\right)$

$\therefore$  as $f^{\text{'}}\left(0\right)=1-c^2$

Now comparing equation $\left(i\right)$ and $\left(ii\right)$, we get

$c=-\frac{3}{2}$

$\therefore f\left(x\right)=-\frac{1}{2}{x}^2-\frac{5}{4}x$

Now $\left|2\sum _{x=1}^{20}f\left(x\right)\right|=\sum _{x=1}^{20}{x}^2+\frac{5}{2}·\sum _{x=1}^{20}x=\frac{20\left(21\right)\left(41\right)}{6}+\frac{5}{2}\left(\frac{20\left(21\right)}{2}\right)$

$=2870+525$

$=3395$