Let f - R - R be a function satisfying f x - y - f x -2 y 2- kxy for all x - y - R - If f 1-2 and f 2-8- then f x is equal to-A- 2 x2B- 6 x-4 C- x2-3 x-2D- -x2-9 x-6

The correct option is A 2x^2 We have f(x + y) = f(x) + 2y^2 + kxy for all x, y ϵ R ⇒f(x+y) f(x)/y = 2y + kx for all x ϵ R ⇒y→ 0limf(x+y) f(x)/y=y→ 0lim(2y+ ...

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Byju's Answer

Standard XII

Mathematics

Polynomial Functions

Let f : R → R...

Question

Let $f : R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + k x y$ for all $x, y ϵ R$ . If f(1) = 2 and f(2) = 8, then f(x) is equal to.

$2 x^{2}$

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$6 x - 4$

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$x^{2} + 3 x - 2$

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$- x^{2} + 9 x - 6$

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Solution

The correct option is A
$2 x^{2}$

We have
$f (x + y) = f (x) + 2 y^{2} + k x y$ for all $x, y ϵ R$
$\Rightarrow \frac{f (x + y) - f (x)}{y} = 2 y + k x$ for all $x ϵ R$
$\Rightarrow lim y \to 0 \frac{f (x + y) - f (x)}{y} = lim y \to 0 (2 y + k x)$
$\Rightarrow f^{'} (x) = k x$ for all $x ϵ R .$
$\Rightarrow f (x) = \frac{k x^{2}}{2} + C$ for all $x ϵ R$
But, f(1) = 2 and f(2) = 8
Therefore, $2 = \frac{k}{2} + C$ and $8 = 2 k + C$
$\Rightarrow k = 4$ and C = 0
Hence, $f (x) = 2 x^{2}$ for all $x ϵ R$ .

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Let f:R→R be a function satisfying f(x+y)=f(x)+2y2+kxy for all x,yϵR. If f(1) = 2 and f(2) = 8, then f(x) is equal to.

Let $f : R \to R$ be a function satisfying $f (x + y) = f (x) + 2 y^{2} + k x y$ for all $x, y ϵ R$ . If f(1) = 2 and f(2) = 8, then f(x) is equal to.