If f(x + y) = f (x) + 2y2 + kxy and f(a) = 2, f(b) = 8, then f(x) is of the form

Question

If f(x + y) = f (x) + 2y2 + kxy and f(a) = 2, f(b) = 8, then f(x) is of the form (A) 2x2 (B) 2x2 + 1 (C) 2x2 - 1 (D) x2

Tardigrade Admin · Accepted Answer

f (x + y) = f(x) + 2y2 + kxy f(1 + y) = 2 + 2y2 + ky, putting x = 1 putting y = 1 f(2) = 8 = 2 + 2 + k ⇒ k = 4 ∴ f (1 + y) = 2 + 2y2 + 4y = 2(y + 1)2 ∴ f (x) = 2x2

Q. If $f (x + y) = f (x) + 2 y^{2} + k x y$ and $f (a) = 2, f (b) = 8$ , then f(x) is of the form

$2 x^{2}$

$2 x^{2} + 1$

$2 x^{2} - 1$

$x^{2}$

$f (x + y) = f (x) + 2 y^{2} + k x y$
$f (1 + y) = 2 + 2 y^{2} + k y$ , putting $x = 1$
putting $y = 1$
$f (2) = 8 = 2 + 2 + k \Rightarrow k = 4$
$∴ f (1 + y) = 2 + 2 y^{2} + 4 y = 2 (y + 1)^{2}$
$∴ f (x) = 2 x^{2}$

Q. If f(x+y)=f(x)+2y2+kxy and f(a)=2,f(b)=8, then f(x) is of the form

2x2

2x2+1

2x2−1

x2