Question

Let $f\left(x\right)$ be a differentiable function on $x\in R$ such that $f\left(x + y\right)=f\left(x\right)\cdot f\left(y\right)$ for all $x,y$ . If $f\left(0\right)\neq 0, \, f\left(5\right)=12$ and $f^{'} \left(0\right) = 16$ , then $f^{'} \left(5\right)$ is equal to

• A. $190$
• B. $186$
• C. $196$
• D. $192$

Answer 1

Answer: (D)

$f^{'} \left(5\right) = \underset{h \rightarrow 0}{\text{lim}} \frac{f \left(5 + h\right) - f \left(5\right)}{h}$
$=\underset{h \rightarrow 0}{l i m}\frac{f \left(5 + h\right) - f \left(5 + 0\right)}{h}$
$=\underset{h \rightarrow 0}{l i m}\frac{f \left(5\right) \cdot f \left(h\right) - f \left(5\right) \cdot f \left(0\right)}{h}$
$\left[\because f \left(x + y\right) = f \left(x\right) . f \left(y\right) f o r a l l x , y\right]$
$=\left(\underset{h \rightarrow 0}{l i m} \frac{f \left(h\right) - f \left(0\right)}{h}\right)\cdot f\left(5\right)$
$f^{'} \left(0\right) \times f \left(5\right) = 16 \times 12 = 192$