Q74 / 150 · MHT CET 2011 · Mathematics
Let $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$, suppose that $f(3)=3$ and $f^{\prime}(0)=11$, then $f^{\prime}(3)$ is given by
- A22
- B44
- C28
- Correct answerD33
Solution
$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$
$$
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} \\
\Rightarrow \quad f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f(h)-1}{h}
\end{aligned}
$$
$[\because f(3)=3]$
Use L'Hospital rule,
$$
\begin{aligned}
\Rightarrow \quad f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1} \\
&=3 f^{\prime}(0)
\end{aligned}
$$
$\left[\because f^{\prime}(0)=11\right]$
$$
=3 \times 11=33
$$
Hence, the correct answer is option: 'D'.