Lf f-R- R such that fx-y-Kxy - fx-2y2 for all x-y - R and f1-2-f2-8 then f20-f10

The correct option is A 600 f(x+y) kxy=f(x)+2y^2 Substitute x=0, y=1 f(1) k× 0=f(0)+2 f(1)=f(0)+2 ∴ f(0)=0 Now Substitute x= ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

lf f:R→ R s...

Question

lf $f : R \to R$ such that $f (x + y) - K x y = f (x) + 2 y^{2}$ for all $x, y \in R$ and $f (1) = 2, f (2) = 8$ then $f (20) - f (10)$

600

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300

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60

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200

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Solution

The correct option is A $600$
$f (x + y) - k x y = f (x) + 2 y^{2}$
Substitute $x = 0, y = 1$
$f (1) - k \times 0 = f (0) + 2$
$f (1) = f (0) + 2$
$∴ f (0) = 0$
Now Substitute $x = 0, y = x$
$f (x) - k \times 0 = f (0) + 2 x^{2}$
$f (x) = 2 x^{2}$
$f (10) = 200$
$f (20) = 800$
$f (20) - f (10) = 600$
Hence, option 'A' is correct.

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Similar questions

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lf f:R→R such that f(x+y)−Kxy=f(x)+2y2 for all x,y∈R and f(1)=2,f(2)=8 then f(20)−f(10)

The correct option is A 600f(x+y)−kxy=f(x)+2y2Substitute x=0,y=1f(1)−k×0=f(0)+2f(1)=f(0)+2∴f(0)=0Now Substitute x=0,y=xf(x)−k×0=f(0)+2x2f(x)=2x2f(10)=200f(20)=800f(20)−f(10)=600Hence, option 'A' is correct.

lf $f : R \to R$ such that $f (x + y) - K x y = f (x) + 2 y^{2}$ for all $x, y \in R$ and $f (1) = 2, f (2) = 8$ then $f (20) - f (10)$