JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 18)
Let f : R $$\to$$ R be a function such that $$f(x + y) = f(x) + f(y),\,\forall x,y \in R$$. If f(x) is differentiable at x = 0, then
f(x) is differentiable only in a finite interval containing zero.
f(x) is continuous $$\forall x \in R$$.
f'(x) is constant $$\forall x \in R$$.
f(x) is differentiable except at finitely many points.
Explanation
Set x = 0 in the functional equation to obtain
$$f(0) = f(0) + f(0)$$ $$\therefore$$ $$f(0) = 0$$
$$f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x + 0)} \over h} = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) - f(x) - f(0)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over h} = f'(0)$$
Thus, $$f'(x) = \lambda $$ (say). Also $$f(x) = \lambda x + \mu $$
As $$f(x) = 0$$ we have $$\mu = 0$$ $$\therefore$$ $$f(x) = \lambda x$$.
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