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

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Question

$f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$        $\dots(1)$

$f(x+y)=f(x)+f(y)-x y \quad \forall x y \in R$

$\lim \limits

Vedclass · Accepted Answer

$3395$

Vedclass · Answer

$f(x)=(c+1) x^{2}+\left(1-c^{2}ight) x+2 k$        $\dots(1)$

$f(x+y)=f(x)+f(y)-x y \quad \forall x y \in R$

$\lim \limits_{y ightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim \limits_{y ightarrow 0} \frac{f(y)-x y}{y}\Rightarrow f^{\prime}(x)=f^{\prime}(0)-x$

$f(x)=-\frac{1}{2} x^{2}+f^{\prime}(0) \cdot x+\lambda \quad 	ext { but } f(0)=0 \Rightarrow \lambda=0$

$f(x)=-\frac{1}{2} x^{2}+\left(1-c^{2}ight) \cdot x$        $\dots(2)$

$	herefore \quad$ as $f ^{\prime}(0)=1- c ^{2}$

Comparing equation $(1)$ and $(2)$

We obtain, $c =-\frac{3}{2}$

$	herefore \quad f ( x )=-\frac{1}{2} x ^{2}-\frac{5}{4} x$

Now $\left|2 \sum \limits_{x=1}^{20} f(x)ight|=\sum \limits_{x=1}^{20} x^{2}+\frac{5}{2} \cdot \sum \limits_{x=1}^{20} x$

$=2870+525$

$=3395$

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

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