Question

If f((x+y)/2)=(f(x)+f(y))/2 for all real values of x and y and f'(0) = - 1 then find f(5).

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f\bigg[\dfrac{x + y}{2}\bigg] = \dfrac{f(x) + f(y)}{2}$

Now,

By definition of differentiation, we have

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

So, can also be rewritten as

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(x + y) - f(x)}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f\bigg( \dfrac{2x + 2y}{2}\bigg) - f(x)}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{\bigg( \dfrac{f(2x) + f(2y)}{2}\bigg) - f(x)}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{ \dfrac{f(2x) + f(2y)}{2} - f(x)}{y}$

$\boxed{\bf :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(2x) + f(2y) - 2f(x)}{2y} - - - (1)}$

Given that,

$\rm :\longmapsto\:f\bigg[\dfrac{x + y}{2}\bigg] = \dfrac{f(x) + f(y)}{2}$

Replace x by 2x and y by 0, we get

$\rm :\longmapsto\:f\bigg[\dfrac{2x + 0}{2}\bigg] = \dfrac{f(2x) + f(0)}{2}$

$\rm :\longmapsto\:f\bigg[\dfrac{2x}{2}\bigg] = \dfrac{f(2x) + f(0)}{2}$

$\rm :\longmapsto\:f(x) = \dfrac{f(2x) + f(0)}{2}$

$\boxed{\bf :\longmapsto\:2f(x) = f(2x) + f(0) - - - (2)}$

On substituting equation (2) in equation (1), we get

${\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(2x) + f(2y) - (f(2x) + f(0))}{2y}}$

${\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(2x) + f(2y) - f(2x) - f(0)}{2y}}$

${\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{ f(2y) - f(0)}{2y}}$

can be rewritten as

${\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{ f(2y) - f(0)}{2y - 0}}$

${\rm :\longmapsto\:f'(x) = f'(0)}$

As it is given that,

$\rm :\longmapsto\:f'(0) = - \: 1$

So, using this value we get

${\rm :\longmapsto\:f'(x) = - 1}$

So, on integrating both sides w. r. t. x, we get

$\rm : \displaystyle\longmapsto\: \int \: f'(x)dx \: = - \: \int \: 1 \: dx$

$\rm :\longmapsto\:f(x) = - x + c$

where c is arbitrary constant.

So,

$\bf\implies \:\:f(5) = - 5 + c$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$