Question

Let c, k ∈ R. If f(x) = (c + 1)x² + (1 – c²)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ... + f(20))| is equal to ______.

Options

• 3392

• 3393

• 3394

• 3395

Answer 1

Let c, k ∈ R. If f(x) = (c + 1)x² + (1 – c²)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ... + f(20))| is equal to 3395.

Explanation:

f(x) = (c + 1)x² + (1 – c²)x + 2k and f(x + y) = f(x) + f(y) – xy ∀ x, y ∈ R  ...(1)

`lim_(y → 0) (f(x + y) - f(x))/y` = `lim_(y → 0) (f(y) - xy)/y`

⇒ f(x) = `f(0) - x`

f(x) = `-1/2x^2 + f(0). x + λ` but f(0) = 0 ⇒ λ = 0

f(x) = `-1/2x^2 + (1 - c).x`  ...(2)

as f = 1 – c²

Comparing equation (1) and (2)

We obtain, c = `-3/2`

∴ f(x) = `-1/2x^2 - 5/4x`

`|2sum_(x = 1)^20 f(x)| = sum_(x = 1)^20 x^2 + 5/2.sum_(x = 1)^20x`

= 2870 + 525

= 3395