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ધારો કે $c , k \in R$ ને પ્રત્યેક $x, y \in R$ માટે $f(x)=( c +1) x^{2}+\left(1- c ^{2}\right) x+2 k$ અને $f(x+y)=f(x)+f(y)-x y$ હોય,તો $|2(f(1)+f(2)+f(3)+\ldots \ldots . .+f(20))|$નું મૂલ્ય $\dots\dots$ છે.
$3365$
$3375$
$3385$
$3395$
Solution
$f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$ $\dots(1)$
$f(x+y)=f(x)+f(y)-x y \quad \forall x y \in R$
$\lim \limits_{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim \limits_{y \rightarrow 0} \frac{f(y)-x y}{y}\Rightarrow f^{\prime}(x)=f^{\prime}(0)-x$
$f(x)=-\frac{1}{2} x^{2}+f^{\prime}(0) \cdot x+\lambda \quad \text { but } f(0)=0 \Rightarrow \lambda=0$
$f(x)=-\frac{1}{2} x^{2}+\left(1-c^{2}\right) \cdot x$ $\dots(2)$
$\therefore \quad$ as $f ^{\prime}(0)=1- c ^{2}$
Comparing equation $(1)$ and $(2)$
We obtain, $c =-\frac{3}{2}$
$\therefore \quad f ( x )=-\frac{1}{2} x ^{2}-\frac{5}{4} x$
Now $\left|2 \sum \limits_{x=1}^{20} f(x)\right|=\sum \limits_{x=1}^{20} x^{2}+\frac{5}{2} \cdot \sum \limits_{x=1}^{20} x$
$=2870+525$
$=3395$
