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\mathrm{\text { Let } f: R \rightarrow R} be a function such that

\mathrm{f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}, f(0)=3, f^{\prime}(0)=3 \text {, then }}

Option: 1

f(x)  is continuous but not differentiable in R.


Option: 2

f(x) is differentiable but not continuous in R.


Option: 3

f(x)  is both continuous and differentiable in R 


Option: 4

\mathrm{\frac{f(x)}{x} \text { is differentiable in } R \text {. }}


Answers (1)

best_answer

\mathrm{\text { Given, } f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}}        ...[i]

Replacing x by 3x & y by zero in (i), we get

\mathrm{\begin{aligned} & f\left(\frac{3 x+0}{3}\right)=\frac{f(3 x)+f(0)}{3} \\ & \Rightarrow f(3 x)-3 f(x)=-f(0) \end{aligned}}            ...[ii]

\mathrm{\text { and } f^{\prime}(x)=\lim _{h \rightarrow 0^{+}} \frac{f(x+h)-f(x)}{h}}

\mathrm{\begin{aligned} & =\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)}{3}-f(x)}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{f(3 x)+f(3 h)-3 f(x)}{3 h} \end{aligned}}

\mathrm{=\lim _{h \rightarrow 0^{+}} \frac{f(3 x)+f(3 h)-3 f(x)}{3 h}}

\mathrm{=\lim _{h \rightarrow 0^{+}} \frac{f(3 h)-f(0)}{3 h}}

\mathrm{\begin{aligned} & =f^{\prime}(0)=3 \text { (given) } \\ & \Rightarrow f(x)=3 x+k \\ & \Rightarrow f(x)=3(x+1) \end{aligned}}

which is a linear function which is continuous &
differentiable ∀ x ∈ R.

 

Posted by

manish painkra

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