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

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Sets Relations and Functions

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) - xy

, for all

x, y \in R

, then the value of

∣2 (f (1) + f (2) + f (3) + \dots\dots + f (20)) ∣

is equal to ____.

Answer: 3395

From the functional equation

f (x + y) = f (x) + f (y) - x y

, differentiating with respect to

y

and setting

y = 0

f^{'} (x) = f^{'} (0) - x

Integrating:

f (x) = - \frac{x ^{2}}{2} + f^{'} (0) \cdot x + λ

.

Since

f (0) = 2 k

from the given form, and

f (0) = λ

, we need

λ = 0

(from the functional equation with

x = y = 0

).

Comparing with

f (x) = (c + 1) x^{2} + (1 - c^{2}) x + 2 k

c + 1 = - \frac{1}{2} \Rightarrow c = - \frac{3}{2}

1 - c^{2} = f^{'} (0) = - \frac{5}{4}

2 k = 0

f (x) = - \frac{1}{2} x^{2} - \frac{5}{4} x

.

Then:

∣2 x = 1 \sum 20 f (x) ∣ = x = 1 \sum 20 x^{2} + \frac{5}{2} x = 1 \sum 20 x = ∣ 2870 + 525 ∣ = 3395

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer