Limit Continuity And Differentiability Ques 49
- Let $f[(x+y) / 2]=\{f(x)+f(y)\} / 2$ for all real $x$ and $y$, if $f^{\prime}(0)$ exists and equals -1 and $f(0)=1$, find $f(2)$.
$(1995,5 M)$
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Answer:
Correct Answer: 49.$(-1)$
Solution: Given, $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}, \forall x, y \in R$
On putting $y=0$, we get
$f\left(\frac{x}{2}\right) =\frac{f(x)+f(0)}{2}=\frac{1}{2}[1+f(x)] \quad \quad[\because f(0)=1] $
$\Rightarrow \quad 2 f\left(\frac{x}{2}\right) =f(x)+1 $
$\Rightarrow \quad f(x) =2 f\left(\frac{x}{2}\right)-1, \forall x, y \in R \quad ……..(i)$
Since, $f^{\prime}(0)=-1$, we get
$ \begin{aligned} & \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=-1 \\ \Rightarrow \quad & \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=-1 \quad ……..(ii) \end{aligned} $
Again, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f\left(\frac{2 x+2 h}{2}\right)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{f(2 x)+f(2 h)}{2}-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2}\left[2 f\left(\frac{2 x}{2}\right)-1+2 f\left(\frac{2 h}{2}\right)-1\right]-f(x)}{h} \quad $ [from Eq. (i)]
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2}[2 f(x)-1+2 f(h)-1]-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(x)+f(h)-1-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(h)-1}{h}=-1 \quad$ [from Eq. (ii)]
$\therefore \quad f^{\prime}(x)=-1, \forall x \in R$
$\Rightarrow \quad \int f^{\prime}(x) d x=\int-1 d x$
$\Rightarrow \quad f(x)=-x+k$, where, $k$ is a constant.
But $\quad f(0)=1$,
therefore $f(0)=-0+k$
$\Rightarrow \quad 1=k$
$\Rightarrow \quad f(x)=1-x, \forall x \in R \quad \Rightarrow \quad f(2)=-1$
BETA
