(10)

Sol. Given f(x + y 3 ) = f(x) + [f(y)] 3 and f ’ (0) > 0

putting x = y = 0, we get

f(0) = f(0) + (f(0)) 3 ⇒ f(0) = 0

also f ′ (0) = =

Let L = f ’ (0) = = = L 3

or L = L 3 or L = 0 , 1, –1 as f ′ (0) > 0 ⇒ f ′ (0) = 0, 1

Thus f ’ (x) = =

f ’ (x) = ⇒ f ’ (x) = 0 , 1

Integrating both sides, we get

f(x) = 0 or f(x) = x + c

As f(0) = 0 , we have f(x) = 0 or f(x) = x

Now f(x) = 0 is imposible as f(x) is not identically zero

∴ f(x) = x and f(10) = 10

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