Set, Relation and FunctionHard
Question
A function f(x) satisfies the relation f(x + y) = f(x) + f(y) + xy (x + y) ∀, x, y, ∈ R. If f′(0) = - 1, then
Options
A.f(x) is an exponential function
B.f(x) is a polynomial function
C.f(x) is twice differentiable for all x ∈ R
D.f′(3) = 8
Solution
f(x + y) = f(x) + f(y) + xy (x + y)
f(0) = 0
∴
= - 1
∴
= 
= 
+ 
x (x + h) = - 1 + x2
∴ f′(x) = - 1 + x2
∴ f(x) =
- x + c.
∴ f(x) is a polynomial function, f(x) is twice differentiable for all x ∈ R and f′(3) = 32 - 1 = 8
f(0) = 0
∴
= - 1∴
= = + x (x + h) = - 1 + x2∴ f′(x) = - 1 + x2
∴ f(x) =
- x + c.∴ f(x) is a polynomial function, f(x) is twice differentiable for all x ∈ R and f′(3) = 32 - 1 = 8
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