Limit Continuity And Differentiability Ques 50
- A function $f: R \rightarrow R$ satisfies the equation $f(x+y)=f(x) f(y), \forall x, y$ in $R$ and $f(x) \neq 0$ for any $x$ in $R$. Let the function be differentiable at $x=0$ and $f^{\prime}(0)=2$. Show that $f^{\prime}(x)=2 f(x), \forall x$ in $R$. Hence, determine $f(x)$.
$(1990,4 \mathrm{M})$
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Answer:
Correct Answer: 50.($e^{2 x}$)
Solution: We have, $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$.
$\therefore \quad f(0)=f(0) \cdot f(0) \Rightarrow f(0)\{f(0)-1\}=0$
$\Rightarrow \quad f(0)=1 \quad \quad[\because f(0) \neq 0]$
Since, $f^{\prime}(0)=2 \Rightarrow \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=2$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2 \quad\quad[\because f(0)=1]$ $\quad$ ……..(i)
Also,
$ \begin{aligned} f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h}, \end{aligned} $
[using, $f(x+y)=f(x) \cdot f(y)]$
$ =f(x)\left\{\lim _{h \rightarrow 0} \frac{f(h)-1}{h}\right\} $
$\therefore \quad f^{\prime}(x)=2 f(x) \quad\quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2$
On integrating both sides between $0$ to $x$, we get
$ \int_0^x \frac{f^{\prime}(x)}{f(x)} d x=2 x $
$\Rightarrow \quad \log _e|f(x)|-\log _e|f(0)|=2 x$
$\Rightarrow \quad \log _e|f(x)|=2 x \quad\quad$ $[\because f(0)=1]$
$\Rightarrow \quad \log _e|f(0)|=0$
$\Rightarrow \quad f(x)=e^{2 x}$
BETA
