Q60 / 170 · Manipal (MET) 2017 · Mathematics
If $f(x+y)=f(x) \cdot f(y), f(3)=3, f^{\prime}(0)=11$, then $f^{\prime}(3)$ is equal to
- A$11 \cdot e^{33}$
- Correct answerB$33$
- C11
- D$\log 33$
Solution
$\begin{aligned}
& f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(3) f(h)-f(3)}{h} \quad[\because f(x+y)=f(x) f(y)]
\end{aligned}$
Now, $f(x+0)=f(x) \cdot f(0)$
$\Rightarrow f(x)[f(0)-1]=0$
Either $f(x)=0$ or $f(0)=1$
$\begin{aligned}
\therefore f^{\prime}(3) & =f(3) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\
& =f(3) f^{\prime}(0)=f(3) \cdot 11 \\
& =3 \times 11=33
\end{aligned}$
Hence, the correct answer is option: 'B'.