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 $f\left(0\right)\ne 0,f\left(5\right)=12$ and $f^{{}^{\prime }}\left(0\right)=16$ , then $f^{{}^{\prime }}\left(5\right)$ is equal to

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

Answer 1

Answer: (D)

$f^{{}^{\prime }}\left(5\right)=\underset{h\to 0}{\lim }\frac{f\left(5+h\right)-f\left(5\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{f\left(5+h\right)-f\left(5+0\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{f\left(5\right)\cdot f\left(h\right)-f\left(5\right)\cdot f\left(0\right)}{h}$
$\left[\because f\left(x+y\right)=f\left(x\right).f\left(y\right)forallx,y\right]$
$=\left(\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}\right)\cdot f\left(5\right)$
$f^{{}^{\prime }}\left(0\right)\times f\left(5\right)=16\times 12=192$