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  • Let <img alt="\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}" src="https://entrancecor

Let \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}  be a function such that f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}, f(0)=0 \text { and } f^{\prime}(0)=3 \text {, then }

Option: 1

\frac{f(x)}{x^2}  is differentiable in R

 


Option: 2

f(x) is continuous but not differentiable in R

 


Option: 3

f(x) is continuous in R

 


Option: 4

 f(x) is bounded in R

 


Answers (1)

\text { Given, } f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}

replacing x by 3x and y be zero, then f(x)=\frac{f(3 x)+f(0)}{3}

\Rightarrow f(3 x)=3 f(x)-f(0) \text { and } f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f\left(\frac{3 x+3 h}{3}\right)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\frac{f(3 x)+f(3 h)}{h}-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(3 x)+f(3 h)-3 f(x)}{3 h}=\lim _{h \rightarrow 0} \frac{f(3 h)+f(0)}{3 h} \quad \text { [from eq. (i)] }=\mathrm{f}^{\prime}(0)=3

\therefore \mathrm{f}(\mathrm{x})=3 \mathrm{x}+\mathrm{c} \Rightarrow \mathrm{f}(0)=0+\mathrm{c}=0

\therefore c=0

Then, f(x)=3x

Hence, f(x) is continuous and differentiable everywhere.

 

Posted by

Sumit Saini

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