• # question_answer Let $\mathbf{f}\left( \mathbf{x}+\mathbf{y} \right)=\mathbf{f}\left( \mathbf{x} \right).\mathbf{f}\left( \mathbf{y} \right),\forall \mathbf{x},\mathbf{y}$; where $f(0)\ne 0.\,if\,f(5)=2$and $\mathbf{f}'\left( \mathbf{0} \right)=\mathbf{3}$, then f'(5) is equal to A) 2                     B) 4              C) 6                                 D) 8

[c] $f'(5)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5+h)-f(5)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5+h)-f(5+0)}{h}$ $=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5).f(h)-.f(0)}{h}[\because \,(x+y)=f(x).f(y)]\,$ $=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(5)[f(h)-f(0)]}{h}=f(5)\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}$ $=f\left( 5 \right).f'\left( 0 \right)=2\times 3=6$ Hence, option [c] is correct.