Given that \(f(a+b)=f(a) \cdot f(b)\) Putting \(a=b=0\), we have \(f(0+0)=f(0) \times f(0)\) \[\] \(\Rightarrow f(0)=1\) or 0 \[\] Now \(f^{\prime}(0)=\operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(0+h)-f(0)}{h}\right]\) \[\] \(\operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(0) \cdot f(h)-f(0)}{h}\right]=f(0) \operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(h)-1}{h}\right]\) \[\] Since \(f^{\prime}(0)=3\), hence \(f(0)\) cannot be 0 , thus \(f(0)=1\). \[\] \(\Rightarrow f(0) \operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(h)-1}{h}\right]=3\) or \(\operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(h)-1}{h}\right]=3\) \[\] Now \(f^{\prime}(5)=\operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(5+h)-f(5)}{h}\right]\) \[\] \(=f(5) \operatorname{Lim}_{h \rightarrow 0}\left[\frac{f(h)-1}{h}\right]=2 \times 3=6\). \[\] Choice (C)\[\]