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If $f(x+y)=f(x) f(y)$ for all $x$ and $y$ and $f(5)=2, f^{\prime}(0)=3$, then $f^{\prime}(5)$ is
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The correct answer is:
6
We have, $f(x+y)=f(x) f(y)$
Putting $x=y=0$
$$
\begin{array}{ll}
\Rightarrow & f(0+0)=f(0) f(0) \\
\Rightarrow & f(0)=[f(0)]^{2} \\
\Rightarrow & f(0)=1 \\
\text { Now, } f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\
= & \lim _{h \rightarrow 0} \frac{f(5)}{f(h)-f(5)}=\lim _{h \rightarrow 0} f(5)\left[\frac{f(h)-1}{h}\right] \\
= & f(5) \lim _{h \rightarrow 0}\left[\frac{f(h)-f(0)}{h}\right]\quad \text{f(0)=1} \\
= & 2 \times f^{\prime}(0)=2 \times 3=6
\end{array}
$$
Putting $x=y=0$
$$
\begin{array}{ll}
\Rightarrow & f(0+0)=f(0) f(0) \\
\Rightarrow & f(0)=[f(0)]^{2} \\
\Rightarrow & f(0)=1 \\
\text { Now, } f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\
= & \lim _{h \rightarrow 0} \frac{f(5)}{f(h)-f(5)}=\lim _{h \rightarrow 0} f(5)\left[\frac{f(h)-1}{h}\right] \\
= & f(5) \lim _{h \rightarrow 0}\left[\frac{f(h)-f(0)}{h}\right]\quad \text{f(0)=1} \\
= & 2 \times f^{\prime}(0)=2 \times 3=6
\end{array}
$$
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