A) f is periodic
B) f is differentiable for all \[x\in R\]
C) Range of f is singleton
D) f is neither odd nor even
Correct Answer: D
Solution :
\[\because \,\,f(x)=\int\limits_{0}^{x}{f(t)dt}\] Differentiating w.r.t, \[x,\,f'(x)=f(x)\] \[\Rightarrow \,\frac{f'(x)}{f(x)}=1\] Integrating, we get \[(f(x))=x+In\,C\]. \[\therefore \,\,f(x)=k.{{e}^{x}}\] But \[f(0)=0\Rightarrow \,k=0\] \[\therefore \,\,f(x)=0\Rightarrow \,f(x)\] is both odd as well as even.
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