JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Let

f: R \rightarrow R

be a function satisfying

f(x+y)=f(x)+2 y^2+k x y

for all

x, y \in R

. If

f(1)=2

and

f(2)=8

, then

f(x)

is equal to

View answer and explanation

Correct option: A

2 x^2

Option A is correct because using the given condition

f(x+y)=f(x)+2 y^2+k x y

and the values

f(1)=2

and

f(2)=8

, we find

f(x)=2 x^2

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