Let c, k in R. If f(x)=(c+1) x^{2}+(1-c^{2}) | vedclass

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(D) Given $f(x)=(c+1) x^{2}+(1-c^{2}) x+2 k$  $(1)$Given $f(x+y)=f(x)+f(y)-x y$ for all $x, y \in R$.Putting $x=0, y=0$,we get $f(0)=f(0)+f(0)-0 \Righ

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$3395$

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(D) Given $f(x)=(c+1) x^{2}+(1-c^{2}) x+2 k$  $(1)$Given $f(x+y)=f(x)+f(y)-x y$ for all $x, y \in R$.Putting $x=0, y=0$,we get $f(0)=f(0)+f(0)-0 \Rightarrow f(0)=0$.Since $f(0)=2k$,we have $2k=0 \Rightarrow k=0$.Now,$f(x+y)=f(x)+f(y)-x y$.Differentiating with respect to $y$,we get $f'(x+y)=f'(y)-x$.Putting $y=0$,$f'(x)=f'(0)-x$.Integrating,$f(x)=-\frac{1}{2} x^{2}+f'(0) x+C$.Since $f(0)=0$,$C=0$.Thus,$f(x)=-\frac{1}{2} x^{2}+f'(0) x$.Comparing with $(1)$,$c+1=-\frac{1}{2} \Rightarrow c=-\frac{3}{2}$.Also,$f'(0)=1-c^{2}=1-(-\frac{3}{2})^{2}=1-\frac{9}{4}=-\frac{5}{4}$.So,$f(x)=-\frac{1}{2} x^{2}-\frac{5}{4} x$.We need to find $|2 \sum_{x=1}^{20} f(x)| = |2 \sum_{x=1}^{20} (-\frac{1}{2} x^{2}-\frac{5}{4} x)| = |-\sum_{x=1}^{20} x^{2} - \frac{5}{2} \sum_{x=1}^{20} x|$.Using $\sum_{x=1}^{n} x^{2} = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{x=1}^{n} x = \frac{n(n+1)}{2}$ for $n=20$:$\sum_{x=1}^{20} x^{2} = \frac{20 	imes 21 	imes 41}{6} = 2870$.$\sum_{x=1}^{20} x = \frac{20 	imes 21}{2} = 210$.Value $= |-(2870) - \frac{5}{2}(210)| = |-(2870 + 525)| = |-3395| = 3395$.

Let $c, k \in R$. If $f(x)=(c+1) x^{2}+(1-c^{2}) x+2 k$ and $f(x+y)=f(x)+f(y)-x y$,for all $x, y \in R$,then the value of $|2(f(1)+f(2)+f(3)+\ldots+f(20))|$ is equal to

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