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If f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}for all real x and y and f'(0)=2, then determine f(x)

Option: 1

2x+2


Option: 2

2x+3


Option: 3

2x-2


Option: 4

3x+2


Answers (1)

best_answer

\\\text {Given equation is } f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(i) \\\text {Putting } x=0,y=0 { \,\,in (i), we\,\, have, } \\ 3f(0)= 2+2f(0) \Rightarrow f(0)=2 \\\\\text {Putting } y=0 \text { and } f(0)=2 \text { in (i), we have, } \\f\left(\frac{x}{3}\right)=\frac{1}{3}[f(x)+4] \quad \Rightarrow f(x)=3 f\left(\frac{x}{2}\right)-4\;\;\;\;\;\;\;\;\;\;\;\;\;...(ii) \\\\\text{Now,}\;\;\\\begin{aligned} 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)+2}{3}-f(x)}{h} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[\text { using }(\mathrm{i})]\\ &=\lim _{h \rightarrow 0} \frac{f(3 x)+f(3 h)+2-3 f(x)}{3 h} \\ &=\lim _{h \rightarrow 0} \frac{3 f(x)-4+f(3 h)-3f(x)}{3 h} \;\;\;\;\;\;\;\;\;\;\;[\text { using }(\mathrm{ii})]\end{aligned}\\\\\\\mathrm{\Rightarrow \lim_{h\rightarrow 0}\frac{f(3h)-2}{3h}=\lim_{h\rightarrow 0}\frac{f(3h)-f(0)}{3h}=f'(0)}\\\mathrm{\therefore \;\;f'(x)=f'(0)}\\f'(x)= 2\\ Integrating\\ f(x)= 2x+c\\ Using\,\,f(0)= 2\\ f(x)= 2x+2 

 

Alternate method 

\\f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}\\\\

Taking y as constant and differentiating the given equation w.r.t x

f'\left (\frac{x+y}{3} \right )\cdot \frac{1}{3}=\frac{f'(x)}{3}

\\ \text{put x=0 and y=3x then }\\\\ f'(x)=2\\\\ \text{integrate it}

\\ f(x)=2x+c\\\\ \text{put x=y=0}\\\\ 3f(0)=2+2f(0)\\\\3f(0)-2f(0)=2\\\\ f(0)=2\\\\ f(x)=2x+2

Posted by

sudhir.kumar

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