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  • Let <img alt="f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%3A%20%5Cmathbf%7BR%7D%20%5Crightarrow%20%5Cmath

Let f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R} be defined by \mathrm{g(x)=x f(x)} then

Option: 1

g is a differentiable function


Option: 2

g is differentiable at 0 if f is continuous at 0


Option: 3

g is one-one if f is one-one


Option: 4

none of these


Answers (1)

best_answer

Take \mathrm{f(x)=\operatorname{sgn} x\: then\: g(x)=\lfloor x \mid}, hence \mathrm{g} need not be differentiable. Take \mathrm{f(x)=x}, then \mathrm{g(x)=x^{2}} which is not one-one.

Let f be continuous at 0 , then for \mathrm{h \neq 0}.

\mathrm{\frac{g(0+h)-g(0)}{h}=\frac{h f(h)}{h}=f(h). Therefore, g^{\prime}(0)=f(0)}.

Posted by

Ritika Kankaria

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